Questão 6 (Física)

**Calculating capacitance values** I'm considering how ideal parallel capacitances behave at a certain time, t2. The voltage reading at this point is typically fully charged before another switch closes. I'll use this reading to work out the values, which suggests they could represent numbers like 3/4, 2/3, or 27/64. It seems I need to finalize a standard solution that simplifies this with an implicit approximation and no complex masses involved.**Analyzing electrical attraction and gravity** I'm being quantitative about the initial electrical attraction between capacitors. When considering the distance \(d\), it seems \(C \sim 1/d\) and force \(F \sim 1/d^2\). This leads to some dependencies involving mass and length, which I might need to clarify but are not directly provided in the request. I shouldn't overcomplicate it, though I could introduce some constants. Still, I need to focus on providing a clear expression without introducing unnecessary elements that might confuse things.**Examining forces on plates and wires** I'm looking into the forces at play with plates held by springs and wires. The plates are mentioned as remaining parallel, which makes me wonder if rigid conducting wires are attached to a horizontal plate. The description suggests these are identical conducting wires suspended from a non-conductive plate in space, but they might not be perfectly taut—more like flexible wires that can stretch a bit. This stretching could affect the tension and movement of the system, so I need to clarify that their stiffness can influence the overall dynamics.

**Analyzing pendulum parameters** I need to parse the problem about pendulum wires regarding mass, length, and equilibrium force, especially since there's missing data. I should consider if the identical conductors are intentionally without these parameters—this inconsistency needs investigating. The setup may imply that I can derive parameters through energy equilibrium instead of considering gravity directly. I'm looking into the relationship of capacitance, electric forces, and other details to compute k, while also checking how gravity plays into the forces acting on the wires.**Understanding circuit description** I need to read the provided text carefully since it describes a specific circuit setup. It talks about identical conducting wires suspended on a non-conducting plate, linking two neutral, parallel metal plates with a given capacitance and area. The initial setup describes equilibrium with a certain distance between the plates, a spring connecting them, and switches that manage the connection to a power source. I should pay attention to the sequence of events regarding the capacitor's charging process.**Analyzing equilibrium state** At a moment t3 after t2, the system reaches equilibrium where the voltage measured by the voltmeter is one-fourth of that at t2. I have to consider that the conducting plates remain parallel throughout this process. For this t3 situation, I'm tasked with finding specific parameters, but there are small angle considerations that are relevant. It's striking that I don't have mass or length provided, which makes it tricky to define necessary variables, particularly regarding the gravitational effects.**Analyzing tension and torque** I need to consider an ideal massless thread supporting gravity, which doesn't ignore the wire's mass being zero. I must examine the geometry and torque related to vertical assumptions while applying horizontal tension at the top. This could lead to tilting unless there's a constraint to "remain parallel." If both plates are tilted opposite ways, they might break their parallel condition. It’s also important that the electric forces are distributed at the center, as a spring is attached there. This horizontal tension could cause torque.**Evaluating delivery of content** I’m considering how to deliver the main message effectively. It looks like doing it in a typical manner where wires just support panels seems artificial and creates physical inconsistencies. So, I’m leaning towards ensuring a more rigorous approach. I could mention that there’s no numerical determination as stated, which could guide toward providing a clearer final summary without relying on invalid assumptions. Timing and clarity are essential in getting this right!

**Calculating capacitor loss** I’m analyzing the loss in the capacitor, where the loss per side is Q0 - Q_C, which equals 3Q0/4. That leads me to conclude that the charge magnitude can be represented as 3Cε/4. It feels like a straightforward calculation, yet it’s fascinating how these equations break down the behavior of capacitors. I’ll make sure to double-check to ensure accuracy and clarity in this context for those who might need it.**Calculating voltage and force** I need to derive the final voltage related to the capacitance ratio. Let’s set the gap at t3 as d' and use the capacitor plates’ capacitance as Cp=ε0 A/d'. By applying charge conservation at the upper node, I realize that Q0=Cε equals (C+Cp)V3. Given that V3=ε/4, I deduce Cp=3C and d'=d/3. Now, for the initial charge Cε and the final charge ε/4, the difference is indeed 3Cε/4, which correctly matches charge on each movable conductor. Then, spring compression relates as d-d'=2d/3.

I'm working through the circuit setup: a source, resistor, and switch charging a capacitor, with a voltmeter and second switch connecting to parallel plates (capacitance matching the circuit capacitor) suspended by a spring at separation d. I'm noting that when S1 closes first, the capacitor charges up to some charge Q. Then when S2 closes and S1 opens, charge redistributes between the two capacitors, pulling the plates together and compressing the spring until equilibrium is reached at voltage ε/4. Using charge conservation, I find the new plate separation shrinks to d/3, meaning the compression is x = 2d/3, and the charge transferred to the plates is 3Cε/4. For the spring constant, I'm computing the electrostatic attractive force between the plates using the transferred charge and plate area, setting it equal to the spring force at equilibrium to solve for k. I'm double-checking this by computing the force a second way using the field between the plates and the charge, confirming F = 9Cε²/(32d) matches the earlier result, giving k = 27Cε²/(64d²). I'm now reconsidering the physical setup—whether the plates hang from wires like pendulums and whether the small-angle approximation sin θ ≈ θ applies here due to the wires tilting as the plates are pulled together. Since no length L or mass is given for the wires, I conclude that effect can't be included quantitatively, so I'll proceed with the spring force balance as the intended solution, treating the plates as staying parallel and the voltmeter as ideal.

Using conservation of charge, the capacitor starts with Q0 = Cε and settles at ε/4, so solving Cε = Cε/4 + C'ε/4 gives C' = 3C, meaning 3Cε/4 charge flowed. Since C' = ε0A/d' equals 3C while C = ε0A/d, the new separation d' comes out equal to d. I briefly consider whether the suspension wires' tilt and tension matter, but without given mass or length I'll neglect that and treat it as a pure spring problem. Computing the electrostatic force F = Q²/(2ε0A) with Q = 3Cε/4 gives F = 9Cε²/(32d), and setting that equal to k times the compression 2d/3 gives k = 27Cε²/(64d²).

I'm working through the circuit: capacitor C charged to ε giving Q0 = Cε, then switching to connect it to the parallel plates which share charge until equilibrium, where the voltmeter reads ε/4. I'm setting up the capacitance change as the plate separation shrinks from d to d−x due to the attractive force compressing a spring. Using charge conservation, I solve ε = V(1 + d/(d−x)) with V = ε/4, giving d/(d−x) = 3, so x = 2d/3, meaning the plates displace by two-thirds of the original gap. Then I compute the charge transferred to the plates (3Cε/4), confirm it matches the capacitor losing charge down to Cε/4, and use the electrostatic attraction force F = q²/(2ε0A) to solve for the spring constant k, expressing it first in terms of C and d, then rewriting it using ε0 and the plate area a² to get k = 27ε0a²ε²/(64d³).

The problem is in Portuguese from the 2027 ITA physics exam. The circuit contains a capacitor, resistor, battery, two switches, and a voltmeter. Two identical conducting wires suspend a non-conducting rigid plate with two neutral parallel metallic plates of area A and capacitance C. The plates are separated by distance d in equilibrium and connected by an ideal non-conducting spring at its natural length. Switch S1 closes at t1 to charge the capacitor from ε. At t2 it opens and S2 closes. At t3 the system reaches equilibrium with voltmeter reading one quarter of the t2 value. The conducting plates remain parallel throughout. The task is to find the free charge that flowed from the capacitor to the plates at t3. The problem also asks for the spring deformation and its elastic constant. The plates are suspended by wires and connected by a spring at natural length. Closing S2 charges the capacitor and causes charge to flow onto the plates. Opposite charges attract, which should compress the spring. Gravity may also be involved since the plates hang. The figure shows the plates hanging vertically from the non-conducting plate. The spring connects them horizontally at their bottom edges. The spring is likely compressed rather than stretched, since the plates attract when oppositely charged. Gravity may be negligible in the initial equilibrium. The plate capacitor equals the circuit capacitor: ε₀a²/d = C. The charged capacitor connects in parallel with the plates at t2, but as charge flows and the plates attract, their separation changes and the total capacitance shifts. At equilibrium the voltage is one quarter of the t2 value. At t2 the voltage across the capacitor is still ε. After S2 closes, charge flows and the plates move. At equilibrium at t3 the voltage drops to ε/4. The fixed circuit capacitor C and the variable plate capacitor C_plates are in parallel. The total charge is conserved at Q_total = Cε. The spring's natural length equals the initial plate separation d, so the deformation is δ = d − x. The equilibrium distance between the plates is x = d/3. The plate capacitor becomes C' = C·(d/x) = 3C, and charge conservation gives Q_c + Q_p = Cε, confirming the result. The free charge that flowed from the capacitor to the plates is (3/4)Cε. This equals the charge on the plates at equilibrium, confirming the result is consistent. The hint about approximating sin θ ≈ θ for small angles suggests the plates swing like pendulums. This introduces a new degree of freedom beyond simple compression or stretching of the spring. The figure shows the plates hanging vertically from wires attached to the non-conducting plate. The spring connects them at the bottom edge, and the distance d is marked there. The wire lengths are not given. The plates swing inward when charged, but the suspension wire lengths are not given. The problem likely expects the answer in terms of the provided variables. The small-angle approximation suggests the geometry must be considered. The voltage relation V(1 + d/x) = ε confirms x = d/3. The small-angle approximation likely accounts for the plates' pendulum motion, since the spring deformation may not simply be d − x when the wires swing inward. The inward displacement of each plate is (d − x)/2, so the suspension wire makes an angle θ with the vertical. The wire length l is not given, which complicates applying the small-angle approximation. No mass is provided, so gravity likely does not affect the equilibrium. The elastic constant can be found by equating the spring force to the electrostatic force between the plates. The permittivity ε₀ can be eliminated using C = ε₀A/d. The suspension wires may not provide horizontal support, which complicates the equilibrium analysis. The small-angle approximation is puzzling because the problem provides no wire lengths or masses. The earlier charge and distance calculations appear correct, but the parallel-plate assumption may be oversimplified. The figure may show the non-conducting plate tilted, which would explain the small-angle approximation. The wires likely connect the circuit to the plates through this suspended setup. The figure shows the non-conducting plate as a parallelogram, suggesting it is tilted. The small-angle approximation is likely just a general exam instruction and may not be needed for this specific problem. The pendulum analysis still requires the wire length l and plate mass m, which are not provided. Without these values the spring constant cannot be determined. The problem likely intends for gravity to be ignored or for the suspension geometry to be treated differently. The figure shows the plates as vertical bars with wires coming from above. The fixed non-conducting plate sits above them. Gravity may be negligible. The plates could slide horizontally. The spring would then compress directly. Equilibrium would balance the spring force against the electrostatic force. The suspension wires might remain vertical. This is confusing because the fixed plate cannot move. The spring compresses by 2d/3. The new plate separation is d/3, so the deformation equals the initial distance minus the final distance. The equilibrium position is determined by the voltage condition, not by the spring constant. Once the position is found from charge conservation and the measured voltage, the deformation follows directly. The spring constant can then be calculated from the force balance between the spring and the electrostatic attraction. The spring constant can be found by equating the spring force to the electrostatic attraction at the equilibrium position. The presence of suspending wires raises a question about whether other forces must be considered. The suspension wires are likely flexible and do not constrain the plates horizontally. The spring is the only restoring force, so the equilibrium can be solved by balancing the spring force against the electrostatic attraction directly. The attractive force between the plates is F = Q²/(2ε₀A), independent of separation under the infinite-plate approximation. The distance d/3 remains small compared to the plate area a, so this formula applies. The spring constant is k = 9Cε²/(32d). The derivation uses the plate charge Q_p = (3/4)Cε and the relation ε₀A = Cd. The spring constant is k = (27/64)Cε²/d². The derivation checks out. The author now doubts whether gravity or horizontal wire tension should be included in the force balance. The author is reconsidering the physical setup. The problem states both plates are suspended, which is unusual for a spring-and-plate arrangement. They wonder if perhaps only one plate is movable, or if the spring is vertical. The figure and the lack of wire lengths make the intended geometry unclear. The spring constant must be expressed only in terms of the given parameters C, ε, d, a, A, and ε₀. The resistance R is irrelevant for this static problem. The spring constant calculation assumes the suspending wires remain vertical, but the fixed plate complicates this. The displacement of d/3 may be too large for a small-angle approximation. The author wonders if the plates are constrained to move vertically while only the spring deforms. The author suspects the suspension wires are merely idealized indicators of free motion, not physical constraints. The flat non-conducting plate in the figure suggests a horizontal setup with vertical wires, which would make the small-angle approximation unnecessary. They also wonder if the voltmeter reading introduces a different geometric constraint. The author considers whether the suspension wires are very long, which would make gravity negligible and keep the plates nearly horizontal. They then realize this assumption is flawed. If the suspension wires are very long, gravity provides almost no horizontal restoring force. The plates remain nearly vertical, so the spring and electric forces balance directly. The author concludes the spring constant is k = (27/64)Cε²/d². They note the small-angle approximation is likely just a general exam instruction and not needed here, since no wire length is given. The problem statement that the plates remain parallel suggests they do not tilt, so the pendulum effect can be ignored. The author questions whether the plates can truly remain parallel while moving closer, since tilting would be required for pure translation. They note that even if the suspension wires allow this motion, an angle θ would still exist. The author assumes two wires, one per plate, and accepts the parallelism constraint. Without masses or wire lengths, gravity cannot be included in the force balance; only the spring and electric forces matter. They still question why the plates are described as suspended. The suspension likely indicates an isolated setup with no external supports. The author is now reconsidering the figure to see if the deformation is purely vertical. The figure shows a horizontal non-conducting plate with two vertical wires hanging down to the tops of the metallic plates. The spring connects the two plates horizontally, and d is the horizontal distance between them. The author is now checking whether equal-length wires would keep the plates vertical during their approach. The author realized that equal-length wires would naturally tilt the plates inward from gravity. The plate faces can still remain parallel while approaching each other. The horizontal separation becomes x = d − 2l sin θ, so the spring deformation is 2l sin θ. The vertical drop and other geometric relationships are also being considered. The author is now analyzing the horizontal force balance on one plate. The electric attraction pulls the plates together, while the compressed spring pushes them apart. They are checking whether the tilted suspension wires contribute a horizontal component that must also be included in the equilibrium equation. The author now sees that the tilted suspension wires contribute a horizontal outward force T sin θ. At equilibrium, the inward electric force must balance both this wire tension and the outward spring force. The author realized that including the suspension wires in the force balance introduces unknown quantities: mass, wire length, and angle. Since none are given, these cannot be eliminated. The problem would need to specify mass or wire length to proceed. The problem text contains no mass, wire length, or gravitational acceleration. The spring constant must therefore be expressed only in terms of the given parameters C, ε, d, and A. Gravity is excluded from the force balance. The author confirms the charge on each plate is Q = 3Cε/4, since the plates are metallic and carry free charge. They also wonder whether the capacitors are connected in series. The circuit is confirmed to be parallel. The voltmeter reads ε at t2 and ε/4 later, which matches the charge flow. The voltmeter measures the capacitor voltage continuously as the plates move. The circuit contains no resistors, so charge redistribution would oscillate without damping. At t3 the system reaches static equilibrium, so the final charge and voltage are constant. This approach is standard for problems combining capacitors with springs. The author now imagines the suspension wires as limp and flexible, so they only resist vertical gravity. They would not contribute any horizontal force during the plates' approach. This would leave only the spring and electric forces to balance, confirming kΔ = F_e. The author is now questioning whether the final distance between the plates is exactly d/3. They note that the problem asks for the deformation of the spring, which would be 2d/3 if the distance shrinks from d to d/3. The author derived the final plate separation x = d/3 by solving the voltage equation (ε/4)(C + C·d/x) = Cε. They confirmed the free charge that flowed onto the plates is 3Cε/4, and verified the spring deformation is 2d/3 and the elastic constant is 27Cε²/(64d²). The author now sees that the fixed suspension points and fixed non-conducting plate make it impossible for the wires to remain vertical while the plates move horizontally. This means the horizontal tension force from the wires cannot be ignored, and the equilibrium must account for both the spring and electric forces together. The author now sees that including the suspension wires in the equilibrium equation introduces unknowns—mass, wire length, and angle—that cannot be eliminated from the given data. This makes the problem unsolvable for the spring constant without additional information. They suspect the problem may have been created without fully accounting for this geometric constraint. The author suspects this may be a composed problem rather than an official ITA 2027 exam question. They note that the figure includes a radius R, but it appears unused in the problem statement. All requested answers must therefore be expressed solely in terms of the given parameters: ε, C, d, a, A, and any numerical constants. The author realizes the requested answers contain no mass, wire length, or gravity, so the suspension wires must not contribute a horizontal force. They are now re-reading the full problem text to check for any overlooked symbols. The full problem text is now being read carefully. It describes the circuit with two identical conductive wires suspending the plates from a rigid non-conductive plate, and notes that initially both switches are open. The problem text is now fully read. The circuit starts with both switches open, then S1 closes to charge the capacitor to ε, S1 opens and S2 closes, and the system reaches static equilibrium at t3 with the voltmeter reading ε/4. The author has identified the three requested quantities: the charge that flows onto the plates, the spring deformation, and the spring constant. The phrase "fixa no espaço" suggests the system is in outer space, where gravity is absent. This would explain why mass and wire length do not appear in the answer. The phrase "fixa no espaço" means the plate is rigid and immobile, not that it is in outer space. Gravity would still be present, so mass would remain an unknown parameter. The author concludes that the problem must be solved by equating the electrostatic force directly to the spring force. This ignores the suspension wires and gravity, which would otherwise introduce unknown mass and wire length parameters. The figure shows the suspension wires as vertical, which would make any horizontal tension component vanish. A student might still question whether the wires truly stay vertical during the plates' horizontal displacement. The author confirmed that at t3 the circuit is in steady state with no current, so the resistance R does not appear in the final equations. The radius R shown in the figure is likely a red herring. The author now doubts the problem can be solved as written. Without mass or wire length, the spring constant cannot be determined. They suspect the figure may contain an unmentioned wire length, or that the small distance d is meant to allow a simplifying assumption. The author re-reads the Portuguese description and confirms the wires are suspended from a rigid plate, meaning the plates hang from them. They now wonder if the plates might actually be horizontal rather than vertical, and whether the upper plate is supported by those two wires. The author is now trying to reconcile the two wires with the two plates. They consider whether both plates could be horizontal and parallel, with the spring connecting them vertically between a rigid non-conductive plate. This raises the question of how the wires suspend the upper plate, since there are two wires but two plates. The figure shows vertical plates with a horizontal spring between them. This confirms the plates are hanging vertically, so the suspension wires would contribute horizontal tension and the angle θ must be considered. Perhaps the candidate should report the answer without the tension component, or perhaps the small-angle approximation is required. The figure might contain an angle, which would allow solving for k. I am reconsidering whether the problem can actually be solved. The figure description confirms no wire length, angle, mass, or gravity are labeled. The problem remains unsolvable for the spring constant. The problem must be solvable without the unknowns introduced by the suspension wires. The small-angle approximation is not required here because no angle is given. The author will proceed with the direct calculation. The author now suspects the wires are loose conducting connections, which would explain why no horizontal tension force appears. A typical candidate would solve the problem by first finding the charge that flows onto the plates and the new plate separation from the voltage conservation, then equating the spring force directly to the electrostatic force to determine the elastic constant. The author now accepts that the suspension wires are purely electrical connections, and the plates are guided to move parallel without friction. Gravity is balanced separately, so no horizontal tension force from the wires needs to be considered. The spring constant will therefore be found by equating the electrostatic force directly to the spring force. The author will now present the step-by-step solution in Portuguese, as requested. The author is now drafting the step-by-step Portuguese solution. They begin by noting that at t2 the capacitor holds charge Q0 = Cε and the plates are neutral at distance d. When S2 closes, the circuit reaches equilibrium with voltage ε/4 and unknown separation x. The new capacitance becomes C' = ε0A/x = C(d/x), and the charge on the capacitor is q_c = C·V = Cε/4. The charge that flows onto the plates is (3/4)Cε. The new plate separation is x = d/3, so the spring compression is δ = 2d/3. The spring constant will be found by equating the spring force to the electrostatic attraction between the plates. The spring constant is k = 27Cε²/(64d²). The author notes that the problem likely assumes no other horizontal forces act on the plates. The author now doubts the direct force balance is valid. Without specifying wire length or mass, gravity and horizontal tension from the suspension wires cannot be ignored. The problem does not explicitly ask to neglect these forces. The author now suspects the problem may be a fictional ITA 2027 test item designed to check whether the candidate correctly identifies missing parameters. They are exploring alternative configurations, such as horizontal plates or a vertical spring deformation, in order to find an interpretation that yields a complete answer without introducing unknown variables. The author is now trying to reconcile the two suspension wires with the two plates in a configuration that uses all given parameters. They are examining the figure to determine how the wires connect to the plates and whether the spring sits between them. The author decides to proceed with the direct force balance, assuming the suspension wires do not introduce horizontal tension. The solution will omit any angle and focus solely on the mechanical and electrical equilibrium. The author confirmed that the force balance is equivalent to the energy minimum condition. The derivative of the total energy with respect to displacement yields the same equation, so both approaches are consistent. The author decided to express the spring constant using only the given symbols C, ε and d. They noted that A = a² was likely provided to indicate square plates and justify neglecting edge effects, but it does not appear explicitly in the final answer. The fraction 27/64 is a clean result that suggests the problem was deliberately designed with these specific numbers. The voltage being one quarter and the final separation being d/3 produce simple integer ratios that simplify to exactly this value.

We need answer in Portuguese, solve physics problem. Need parse. Let's restate problem. We have circuit: voltage source epsilon, resistor R? switch S1, capacitor C, voltmeter V, switch S2 connects to two identical conducting wires suspended from nonconducting rigid plate fixed in space, to two parallel metallic neutral plates area A=a^2 and capacitance C equal to capacitor in circuit. Plates parallel distance d (d << a) initially equilibrium and connected by ideal nonconducting spring at natural length. Initially switches S1 S2 open. At t1 S1 closed, capacitor charges powered by source epsilon. After fully charged, at t2 S1 opened and S2 closed. At t3 > t2 system at equilibrium, and voltage measured by voltmeter is one quarter of that at t2. Consider plates remain parallel. For t3 find a) free charge that flowed from capacitor to conducting plates; b) deformation of spring; c) spring constant. Need understand setup: Circuit capacitor and voltmeter? Diagram: source ε in series with resistor R, switch S1. Then parallel with capacitor C and voltmeter V? Right side switch S2 connects to external plates? Let's inspect image: Left branch: resistor R and battery ε in series, switch S1 top. Then two parallel vertical lines labeled C (capacitor) maybe between top node and bottom node. Middle: voltmeter V between same nodes? Actually diagram: left branch R and ε with S1; parallel branch with C; another branch with V; top node after S1, then switch S2 connects top node to external conducting plates? Bottom common. Yes. So at t1, S1 closed, capacitor charges to voltage ε. After fully charged, no current, voltage across capacitor = ε. Voltmeter reads ε at t2? At t2 S1 opens, S2 closes; capacitor connected to external plates? The external plates are connected by two wires to somewhere? Text: circuit is connected by two identical conducting wires, suspended from a nonconducting rigid plate fixed in space, to two parallel metallic neutral plates... The two wires connect from circuit (after S2?) to plates. At t2, S2 closed, so capacitor is connected to the two parallel plates (one wire to each plate). The plates form a capacitor in parallel with C? But they are neutral initially, so charge redistribution between C and external plates. Voltmeter measures voltage across capacitor C (same as across plates since connected). At t3 equilibrium, voltage is ε/4 (one quarter of that at t2). Need charge flowed from capacitor to plates, deformation of spring, spring constant. Need model: Initially capacitor C charged Q0 = C ε. External plates initially uncharged? But both plates neutral, connected by spring, distance d, maybe they attract due to charges? At t2, S2 closed, circuit now has capacitor C and external parallel plate capacitor (with capacitance C as well? "capacitância C igual à do capacitor do circuito" meaning external plates capacitance also C) connected in parallel. But external plates are movable? They are connected by a spring at natural length; after charging, electrostatic attraction pulls plates together, compressing? Or repulsion? Need determine. The plates are neutral initially, but when connected to charged capacitor, they receive charges: one plate +Q, other -Q. Opposite charges attract; plates move toward each other, compressing spring? The spring is between plates, nonconducting, at natural length initially. If they attract, distance decreases, spring compressed? If they repel? But connected to capacitor, plates acquire opposite charges, so attract, so move closer; spring compressed (if spring resists compression? A spring at natural length connecting plates: if plates move closer, spring compresses? Actually spring between plates, if plates move closer, spring length decreases -> compressed. It will exert force pushing plates apart, opposing attraction. Equilibrium at smaller distance). Need find deformation and k. Need use energy? Charge redistribution between capacitor and external capacitor? Initially C has Q0, external capacitance C uncharged (voltage 0). When connected in parallel, total charge Q0 conserved? But wait external plates are neutral but conductive; when connected to capacitor, charge flows from both plates of capacitor to external plates, resulting voltage V = Q0/(C + C_ext) = Q0/(2C) = ε/2, if C_ext = C. But measured voltage at t3 is ε/4, not ε/2. So something else: energy dissipated? Resistive? There is resistor R in charging branch, but S2 branch maybe no resistance? But after S1 opened, only capacitor and external plates; charge redistribution would conserve charge if isolated, but there is also a voltmeter? Voltmeter ideal infinite resistance, so yes. So voltage should be ε/2 if C_ext = C and no energy loss. But measured at t3 is ε/4, perhaps due to work done by electric field pulling plates together: mechanical work done by electrostatic force increases stored energy? Wait charge conserved; capacitance of external plates changes as distance d changes? The external capacitance depends on separation x: C_ext = ε0 A / x (or if using C = ε0 A/d initially, so C_ext(x) = C d / x). At equilibrium, voltage V = Q0 / (C + C_ext(x)). If x < d, C_ext larger, so V lower. Measured V = ε/4 = Q0/(C + C_ext) with Q0=C ε => 1/4 = C/(C+C_ext) => C+C_ext = 4C => C_ext = 3C. Thus external plates capacitance at equilibrium is 3C. But their initial capacitance C_ext = C. Since C_ext = C d / x, get x = d/3? Wait C_ext = ε0 A / x = C d/x. If C_ext = 3C, then d/x = 3 => x = d/3. So plates compressed from d to d/3, deformation Δx = d - d/3 = 2d/3. Spring compressed? Maybe spring natural length d, final length d/3, so deformation (compression) = 2d/3. But need check if mechanical equilibrium: electrostatic force between plates equals spring force. Need derive spring constant k. Use equilibrium: electric force between plates = k Δx? But springs? There are maybe two springs? Diagram shows a spring between plates? It says connected by a spring ideal nonconducting at natural length. The spring may be one spring between centers? Yes. So force F_electric = k Δx. Need compute electric force between plates at charge Q? The plates have charge? At equilibrium, capacitor C and external plates both have voltage V = ε/4. Charge on external plates? If connected in parallel, total charge Q0 = C ε. The external plates charge Q_ext = C_ext V = 3C * (ε/4)= 3C ε/4. The circuit capacitor charge Q_C = C V = C ε/4. Total = C ε. So external plates have charge ±Q_ext = ±3Cε/4. But note: capacitor C also has charge; the "plates condutoras" receive free charge from capacitor? Item a) asks "a carga livre que fluiu do capacitor para as placas condutoras" (free charge that flowed from capacitor to conducting plates). That is the charge that transferred from capacitor plates to external plates during redistribution. Need compute magnitude? Initially capacitor had +Q0 on one plate, -Q0 on other. At final, capacitor has +Q_C = C ε/4 on one plate, -Q_C on other. So charge flowed from capacitor to external plates = Q0 - Q_C = C ε - C ε/4 = 3C ε/4. But careful: "do capacitor para as placas condutoras" maybe total free charge on the plates (each?) perhaps 3Cε/4? Yes. But need account if external plates initially neutral; after connecting, one plate gets +Q_ext, other -Q_ext. The charge that flowed from capacitor to external plates equals Q_ext = 3Cε/4. However if question asks "a carga livre que fluiu do capacitor para as placas condutoras" maybe they want charge on the plates, not total? Let's inspect wording: "a carga livre que fluiu do capacitor para as placas condutoras" = free charge that flowed from capacitor to conducting plates. In parallel, charge redistributes; the capacitor's plates lose free charge? The external plates are two conducting plates, each receives charge of opposite sign from the corresponding capacitor plate via wires? But there are two wires, so two external plates. The circuit capacitor has two plates; one plate connected to one external plate, other to other. The charge that flows from one capacitor plate to corresponding external plate is Q_ext on that external plate? Actually final external plates have charges +Q_ext on one plate, -Q_ext on other. Total charge on both external plates? Sum external plates =0. "carga livre que fluiu do capacitor para as placas condutoras" singular? Maybe the amount of charge that moved through each wire, i.e., Q_ext = 3Cε/4 (absolute value on each external plate). They might ask magnitude. So answer a) 3Cε/4. But wait if they mean total free charge that left capacitor (sum of absolute from both plates) = 2 * (3Cε/4)? Let's parse. In Portuguese: "a carga livre que fluiu do capacitor para as placas condutoras" literally "the free charge that flowed from the capacitor to the conducting plates." It could be interpreted as the amount of charge that transferred from one plate to another? In capacitors, "carga livre" often means excess charge on one conductor. The external plates received free charge from capacitor; one external plate received positive charge, the other negative. The total free charge (absolute) on the conducting plates? Usually they ask "carga livre que fluiu" = magnitude of charge that moved, i.e., through connecting wires, equal to final charge on one plate? But perhaps they might expect total charge transferred from capacitor: since capacitor lost charge on both plates, the total charge lost (absolute sum) = 2 * (Q0 - Qf) = 2*(3Cε/4)=3Cε/2. But physically "charge flow" through a wire is the amount of charge that passes a cross-section. Through each wire, charge Q_ext flows from capacitor plate to external plate. So each wire carries Q_ext = 3Cε/4. If they ask "a carga livre que fluiu do capacitor para as placas condutoras", maybe singular total charge? Let's see. Let's solve fully. But need be careful: The voltmeter at t2 reads ε? At t2 S1 open, S2 closed. At t2 immediately after closing? "Em um instante t3 > t2, o sistema se encontra no ponto de equilíbrio, e a tensão medida pelo voltímetro é um quarto daquela em t2." At t2, before redistribution? S1 opened, S2 closed; voltage across capacitor at t2 is ε (since fully charged, no current, S1 open? At t2 S1 just open, S2 just closed, external plates uncharged, voltage across capacitor initially still ε? Yes). Then relaxation to t3 with plates moving; final voltage ε/4. So charge redistribution and mechanical equilibrium. Potential issue: The external plates are initially neutral and separated by d; when connected to charged capacitor, the capacitance of the system is not simply C_ext = ε0 A / x because the capacitor C is also in parallel, but there may be a "voltmeter" in parallel; ideal voltmeter draws no current. So yes parallel combination. But there is a resistor R? It is in the source branch before S1. After S1 open, resistor disconnected? At t1 charging through R; at t2 S1 open, resistor no longer in circuit. So no dissipative element besides perhaps wires? But redistribution between ideal capacitors should conserve charge and energy? Wait energy loss due to mechanical work? Let's calculate energy: initially capacitor energy U_i = Q0^2/(2C) = C ε^2/2. Final: capacitor C has Qf=Cε/4, energy C (ε/4)^2 /2 = C ε^2/32. External capacitor (if capacitance 3C) has Qf = 3Cε/4, energy = (3Cε/4)^2/(2*3C) = 9C^2 ε^2/16 / (6C)= (9/96) Cε^2 = 3Cε^2/32. Total electrical final = 4Cε^2/32 = Cε^2/8. Initial electrical = 16Cε^2/32. Difference = 15Cε^2/32. This energy could be mechanical work done by electric field (attraction) plus maybe kinetic dissipated? But equilibrium, maybe mechanical energy stored in spring. Work done by electric field = change in electrical energy + work done? Let's derive. The system includes capacitors and spring. Initial: electrical only. Final: electrical + spring potential. If charge redistribution and plate motion occur with no dissipation, energy conservation: Initial electrical = final electrical + spring potential + kinetic energy. At equilibrium after oscillations damped? But ideal spring no damping; the system may oscillate. However "instante t3 > t2, o sistema se encontra no ponto de equilíbrio" suggests equilibrium, likely after damping? But spring ideal, no damping; maybe oscillations? Maybe there is some mechanism? But the problem says after t2, at t3 equilibrium; perhaps energy conservation including final kinetic zero? There must be dissipation? But maybe the plates are suspended in air, damping? Not specified. Maybe we should not use energy conservation because charge redistribution? But we can use equilibrium force balance and capacitance relation, not energy. However, spring constant k may be determined by force balance alone, not energy. Let's verify: Electrostatic force between plates depends on charge Q_ext and area, separation. Final voltage gives charge Q_ext = 3Cε/4. Force between two parallel plates with charges +Q and -Q is F = Q^2/(2 ε0 A) (attractive). Need express in terms of C, ε, x? Use C_ext = ε0 A / x, and C = ε0 A / d => ε0 A = C d. Then F = Q_ext^2 / (2 ε0 A) = (3Cε/4)^2 / (2 C d) = (9 C^2 ε^2 /16) / (2 C d)= 9 C ε^2 / (32 d). At equilibrium, spring force = k Δx = k (d - x) = k (d - d/3)= 2k d/3. Equate: 2k d/3 = 9 C ε^2 / (32 d) => k = (9 C ε^2 /(32 d)) * (3/(2d))? Wait: 2k d/3 = 9 C ε^2/(32 d) => k = (9 C ε^2/(32 d)) * (3/(2 d)) = 27 C ε^2 / (64 d^2). Is this answer? Let's check dimensions: C ε^2 / d^2 = F? C = ε0 A/d = F? C ε^2 = energy *? ε^2 C = energy? Energy = C V^2 = C ε^2; divided by d^2 => energy/length^2 = force/length? Actually energy = force*length, so energy/length^2 = force/length = N/m? Spring constant N/m, yes energy/length^2 = (N*m)/m^2=N/m. Good. But wait: Electric force between plates formula F = Q^2/(2 ε0 A) is force on one plate due to other? For parallel plate capacitor, force of attraction = Q^2/(2 ε0 A) = (1/2) Q E? Let's derive: Field due to one plate is σ/(2ε0)=Q/(2A ε0); force on other plate charge Q is Q * Q/(2A ε0)=Q^2/(2 ε0 A). Yes. So k = 27 C ε^2/(64 d^2). But need ensure d << a so edge effects negligible, and formula for C_ext = ε0 A / x? But if plates remain parallel, yes. But wait item a asks "a carga livre que fluiu do capacitor para as placas condutoras"; item b "a deformação da mola"; item c "a constante elástica da mola." We need maybe express deformation as d/3 or 2d/3? Let's check sign: deformation (compression) = d - x = d - d/3 = 2d/3. But if "deformação" maybe magnitude? In Portuguese "deformação" often means change in length (could be positive). Since plates approach each other, the spring is compressed, so deformation = 2d/3. But if the spring originally natural length d, final separation d/3. So deformation = 2d/3. Need be cautious: The plates are connected by a spring "ideal non-conducting at natural length" maybe the spring itself may be between the plates and its length at natural (equilibrium) is d. When plates attract, they move closer, so the spring compresses (if it's a compression spring? A spring at natural length with no force; if plates move closer, length decreases, so it resists compression). Yes. Need consider whether the external plates initially have capacitance C and are neutral. At t2, S2 closed; before charge redistribution, capacitor C has voltage ε. But perhaps the external plates are connected through the wires, and the two wires "suspended" from a rigid nonconducting plate; the plates are free to move? The wires are identical conductors, maybe they are flexible? The plates are suspended by wires; because wires may be vertical, plates can move? They are suspended from a fixed plate, but the wires might be like pendulums? Hmm "suspensos numa placa não condutora rígida e fixa no espaço" = suspended from a rigid nonconducting plate fixed in space. The plates are connected by a spring; they can move? Probably the wires allow vertical movement? They are conducting wires, maybe rigid? But they might be vertical wires and plates can move horizontally? Let's parse original: "o qual está ligado por dois fios idênticos e condutores, suspensos numa placa não condutora rígida e fixa no espaço, a duas placas planas paralelas metálicas, neutras, de área A..." The circuit is connected by two identical conducting wires, suspended from a nonconducting rigid plate fixed in space, to two parallel metallic neutral plates. So the wires suspend the plates? The wires are attached to fixed plate above and to the top of each plate? Then plates can swing? But they remain parallel? The spring between them ensures parallel? Hmm "placas condutoras permanecem paralelas em todas as situações" perhaps they remain parallel due to constraints. The movement may be vertical? Let's visualize: two identical conducting wires suspended from a nonconducting plate, like two pendulums, each wire attached to a conductive plate. The plates are horizontal? But problem says parallel plates initially separated by d, area A. They may move closer by swinging? But if wires are identical and parallel, the plates might remain parallel and move horizontally? They are suspended, perhaps the wires act as hinges? Hmm. But for force calculation, distance between plates changes; area A remains; capacitance changes as 1/x. Need know electric force direction: opposite charges attract; they pull plates together, compressing spring. The wires might apply tension? The plates maybe keep orientation? Let's see: if plates are suspended by wires from fixed plate, the wires could be flexible or rigid? If wires are conducting and "suspend" plates, they might be vertical and allow the plates to move horizontally due to electrostatic attraction? But then spring between plates? The deformation may involve rotation? Maybe not; the problem likely ignores gravity? It says "suspensos" but no mention of gravity; maybe the suspension is just to hold plates in place. The spring connects them; electrostatic force compresses spring; equilibrium with spring force. So okay. Need also consider after S1 opened and S2 closed, the circuit's capacitor C remains in parallel with external plates; voltmeter measures voltage. But at t3, voltage is ε/4; charge conservation total Q = C ε. We used that. However, is there any charge initially on external plates? They are neutral, so total zero, but connected in parallel to capacitor. After closing, charge redistributes: total charge on capacitor C and plates = initial charge on capacitor = Cε. Yes. Need maybe include work done by battery? No battery disconnected at t2 (S1 open). So no external source. Voltmeter ideal. Need solve item a maybe "carga livre que fluiu do capacitor para as placas condutoras" = charge on each external plate? Let's compute final charges: Q_C = C V = C * ε/4 = Cε/4 on capacitor. Q_plates = C_ext V = 3C * ε/4 = 3Cε/4 on external plates (each? Actually external capacitor has charge Q = C_ext V = 3Cε/4; one plate +, other -). So charge that flowed from capacitor to external plates = initial capacitor charge - final capacitor charge = Cε - Cε/4 = 3Cε/4. This is exactly Q_plates. So answer a = 3Cε/4. But if they ask "free charge that flowed from capacitor to conducting plates" maybe they may want the sum of magnitudes? Let's check wording: "a carga livre que fluiu do capacitor para as placas condutoras" Could be "a carga livre (total) que fluiu do capacitor" = the amount of free charge that left the capacitor. On a capacitor, "carga" often means charge on one plate (magnitude). The capacitor initially had charge Q0 (on positive plate). Finally it has Qf. The amount of charge that flowed out of that plate into the external circuit is Q0 - Qf = 3Cε/4. The negative plate also loses 3Cε/4 electrons? Actually total charge transferred through external circuit? Since each wire carries same current, charge that flows from positive plate to one external plate is 3Cε/4; from negative plate to other external plate also 3Cε/4 (opposite sign). So the total free charge that flowed from the capacitor (sum of absolute charges moved from both plates) would be 2 * 3Cε/4 = 3Cε/2. But in circuit analysis, "carga que fluiu do capacitor" might be charge on one plate? Let's see standard: A capacitor with charge Q connected to another capacitor; "carga que passa" through connecting wire = |ΔQ| = Q0 - Qf =? Actually if both capacitors share charge, the amount of charge that flows through the wire from one capacitor to the other equals change in charge of one capacitor: ΔQ = Q0 - Qf = (C_ext V) = 3Cε/4. Because through the wire, charge Q_ext transferred. The other wire carries opposite? But current in both wires? The circuit is a loop; charge flows from one plate of C to one plate of external capacitor, and from other plate of external capacitor to other plate of C. The amount of charge passing any cross-section in the loop is ΔQ = 3Cε/4. So answer a likely 3Cε/4. We'll state magnitude. Need double-check if there is any possibility that external plates each have area A and capacitance C equal to capacitor C. But when connected in parallel, there is also the original capacitor C. The total initial charge is Cε. Final voltage V = ε/4. Let C_ext = ? Total capacitance = C + C_ext. Charge conservation: Cε = (C + C_ext)(ε/4) => C + C_ext = 4C => C_ext = 3C. Then external plates capacitance increased by factor 3, so distance reduced by factor 3 (x = d/3). Deformation = d - d/3 = 2d/3. Good. Now need derive spring constant more rigorously maybe using force. But maybe there is also a factor due to the two wires? The wires may have tension? But force balance on plates includes electrostatic attraction, spring force, and maybe tension from suspending wires? If wires are vertical and flexible, tension vertical, not in direction of motion? Hmm if plates move horizontally? Let's think geometry: The two plates are suspended from a fixed nonconducting plate by two conducting wires. The wires attach at points? The plates are horizontal? The spring connects between them? The electric force is attraction along horizontal? Let's parse: "suspensos numa placa não condutora rígida e fixa no espaço" = suspended from a rigid nonconducting plate. The wires maybe vertical, each holding one plate? The two plates are "placas planas paralelas metálicas", likely vertical? For a parallel-plate capacitor, plates could be vertical and separated horizontally. They are suspended by wires from above, perhaps like two pendulums; the spring connects them horizontally; electrostatic attraction pulls them horizontally. The wires are conducting and maybe flexible, maintaining electrical connection but allowing movement? The problem states plates remain parallel in all situations, so they translate horizontally maintaining parallel. The wires may swing? But the force analysis: The electrostatic force is horizontal; spring force horizontal; tension from wires maybe at an angle? If wires are vertical and rigid? Not enough. Maybe they are suspended such that wires are always vertical? Perhaps the wires are flexible and the plates hang horizontally? Hmm. Maybe a more standard setup: two identical conducting wires are attached to the plates and to fixed points, acting as parallel pendulums, but they are suspended so that the plates can move while remaining parallel. The spring provides restoring force. The electrostatic force is F_e = Q^2/(2ε0 A), and spring force F_s = k Δx. Tension in wires? If wires are vertical and the plates are heavy, there may be gravitational force; but problem doesn't mention mass or gravity. Could be ignored? Usually in such problems, they ignore gravity or the suspension just constrains the plates to move horizontally. The potential energy method might be more robust: The force between plates is derivative of electrical energy with respect to separation if charge constant? For isolated plates with constant charge, electrical energy U_e = Q^2/(2C_ext) = Q^2 x/(2ε0 A), so force F = -dU_e/dx = - Q^2/(2ε0 A) (attractive, tends to decrease x). This gives same magnitude. Mechanical equilibrium: spring force equals F_e. So okay. But let's consider energy: At constant charge, if plates move under electric force, electrical energy decreases; that energy goes into spring. The final equilibrium can be found by minimizing total energy: U_total(x) = Q^2 x/(2ε0 A) + (1/2) k (d - x)^2, with Q determined by charge sharing? But Q on external plates depends on x because capacitor C in parallel changes voltage. Since the system includes capacitor C, total charge Q0 = C V + Q_ext = C V + (ε0 A / x) V = constant, and V = Q0 / (C + ε0 A / x). Total electrical energy U_e = (1/2)(C + C_ext) V^2 = (1/2) Q0^2 / (C + C_ext) = (1/2) Q0^2 / (C + ε0 A / x). Then force on plates at constant charge? But total charge constant and connected conductors, if one plate moves, charge redistributes? Actually if plates move slowly, charge can redistribute through wires, so the system is not at constant Q_ext; total charge constant, but voltage changes. The force on the movable plates at constant total charge and fixed C? The electrical energy depends on x, and mechanical force can be found by -∂U_e/∂x at constant total charge. Let's compute: U_e(x) = Q0^2/[2(C + ε0 A/x)]. Derivative: dU/dx = -(Q0^2/[2]) * ( -ε0 A/x^2 ) / (C + ε0 A/x)^2 = Q0^2 ε0 A/(2 x^2 (C + ε0 A/x)^2). Force on plates (tending to increase x? Since derivative positive, force = -dU/dx negative = attractive). At equilibrium, spring force balances. We can use final state V = ε/4 to determine x then force. This avoids assuming Q_ext fixed? Actually at equilibrium we already know x from capacitance ratio, and Q_ext = C_ext V; force formula with Q_ext and x? Let's test: U_e derivative at final x = d/3, with Q0=Cε, gives F_e = Q0^2 ε0 A/(2 x^2 (C + ε0 A/x)^2). Using ε0 A = C d; x=d/3; C_ext=ε0 A/x = 3C; denominator C+3C=4C; numerator C^2 ε^2 * C d / (2 (d^2/9) * 16C^2) = (C^3 ε^2 d) /(2 * d^2/9 * 16 C^2)= (C ε^2 d)/(2 * 16 d^2/9?) Let's compute: C^3/C^2 = C. Numerator C ε^2 d. Denominator 2 * x^2 * (4C)^2 = 2*(d^2/9)*16C^2 = (32/9) C^2 d^2. So F = C ε^2 d / ((32/9) C^2 d^2)= (9 ε^2)/(32 C d)? Wait C ε^2 d / ( (32/9) C^2 d^2 ) = (9 ε^2)/(32 C d). This is different from force using Q_ext^2/(2ε0A) = (3Cε/4)^2/(2 C d)=9C^2ε^2/(16 * 2? wait (3Cε/4)^2 = 9C^2ε^2/16; divide by 2 C d = 9 C ε^2/(32 d). Same? My derivative result (9 ε^2)/(32 C d) has C in denominator, so not same; wrong because derivative formula? Let's recalc derivative carefully. U_e = (1/2) Q0^2 / (C + C_ext), C_ext = ε0 A / x = k_e/x? Let K = ε0 A = C d. C_ext = K/x. U = Q0^2/[2(C + K/x)] = Q0^2 x/[2(C x + K)] = Q0^2/[2(C + K/x)]. Derivative dU/dx = Q0^2/2 * d/dx [ (C + K/x)^{-1} ] = Q0^2/2 * [ -1/(C+K/x)^2 * (-K/x^2) ] = Q0^2 K /(2 x^2 (C+K/x)^2). Yes. At x=d/3: C+K/x = C + C d /(d/3)= C+3C=4C. x^2 = d^2/9. K = C d. dU/dx = Q0^2 * C d / (2 * d^2/9 * (4C)^2) = Q0^2 C d *9 / (2 d^2 *16 C^2)= 9 Q0^2 /(32 C d). Q0=Cε, so =9 C^2 ε^2 /(32 C d)=9 C ε^2/(32 d). That matches force magnitude Q_ext^2/(2K)? Q_ext = 3Cε/4, K=C d. Force = (9C^2ε^2/16)/(2 C d)=9 C ε^2/(32 d). Yes same. Good. My earlier expression with ε0A? okay. Then spring force = k Δx = k (2d/3). Equate: 9 C ε^2/(32 d) = 2k d/3 => k = (9 C ε^2/(32 d)) * (3/(2d)) = 27 C ε^2/(64 d^2). Good. But wait in derivative expression, if Q0 is total charge on capacitor initially, after connection, total charge Q0 conserved. But does the battery disconnected? Yes S1 open. However at t2, before closing S2, the capacitor charge is Q0=Cε. Then S2 closed; charge redistributes. The total charge on both capacitors C and external is conserved if no other connections. But note: The external plates are connected through wires, but their movement might change capacitance; charge conservation still holds if no charge leaks. So yes. Now, could there be charge on the voltmeter? Ideal voltmeter infinite resistance, no. So okay. Need perhaps mention assumption d << a so edge effects neglected, uniform field. Also assume spring obeys Hooke, natural length d. Let's formulate final answer in Portuguese, step-by-step. We need produce final answer with items a, b, c. Let's derive in a clean way. Given: - Circuit capacitor: capacitance C, initial charge Q0 = C ε. - External parallel plates: area A = a^2, initial separation d, initial capacitance C_ext0 = ε0 A / d = C (given). Since same as C, indeed ε0 A = C d. - At t2, S1 open, S2 closed. Charge redistributes between C and external plates. Voltmeter reads final V_f = ε/4. - Charge conservation: Q0 = C V_f + C_ext V_f = (C + C_ext) V_f. C ε = (C + C_ext)(ε/4) => C + C_ext = 4C => C_ext = 3C. - Since C_ext = ε0 A / x_f = C d / x_f, then 3C = C d / x_f => x_f = d/3. Therefore deformation of spring: Δx = d - x_f = d - d/3 = 2d/3 (compression). - Charge that flowed from capacitor to plates: Q_ext = C_ext V_f = 3C * ε/4 = 3Cε/4. - Force equilibrium at final separation x_f: Electrostatic attraction between plates: F_e = Q_ext^2 / (2 ε0 A) = (3Cε/4)^2 / (2 C d) = 9 C ε^2 / (32 d). Spring force: F_s = k Δx = k (2d/3). Equate: k (2d/3) = 9Cε^2/(32d) => k = 27 C ε^2 / (64 d^2). Need maybe note "em módulo" because charge flowed maybe positive. Potential subtlety: At t2, the voltmeter reading is ε? At t2, before S2 closes? Actually "Depois de o capacitor estar completamente carregado, no instante t2, a chave S1 é aberta, e a chave S2 é fechada. Em um instante t3 > t2, ... a tensão medida pelo voltímetro é um quarto daquela em t2." At t2 immediately after opening S1 and closing S2, voltage might still be ε if no charge has moved yet; but if S2 close, plates connected, voltage may instantaneously? But charge redistribution is not instantaneous; at t2 exactly the voltage across capacitor is still ε because no charge has moved yet? However, if switches ideal, at t2 S1 opens and S2 closes simultaneously? It says at t2 S1 opened, and S2 closed; then at t3 > t2 equilibrium. So at t2, before redistribution, V=ε; final V=ε/4. Good. Need maybe include that voltmeter measures voltage across C (and external plates in parallel). "tensão medida pelo voltímetro é um quarto daquela em t2" => V(t3)=ε/4. Let's consider if there is any effect of the resistor R after S1 open? No. Now, let's ensure item a: "a carga livre que fluiu do capacitor para as placas condutoras" Could they expect the charge on the external plates after redistribution, maybe 3Cε/4. But if they ask "carga livre" maybe "carga livre" = "free charge" not "total charge"? In Portuguese physics, "carga livre" is the excess charge on a conductor. The free charge that flowed from the capacitor to the plates might be the charge that each plate acquired, i.e., 3Cε/4. We'll state magnitude. Let's maybe phrase: "A carga livre que passou do capacitor para as placas é, em módulo, Q_p = C_ext V_f = 3Cε/4." Good. Let's also maybe derive charge on each external plate? "cada placa condutora recebeu 3Cε/4 em módulo (uma positiva, outra negativa)." Good. Now, final answer maybe: a) \(3C\varepsilon/4\) b) \(2d/3\) (compressão) c) \(27 C \varepsilon^2/(64d^2)\) Need ensure units in final: c) N/m? Yes. Let's maybe include "em módulo" for a. Let's think if there is any possibility that the external plates are connected in series? The circuit has capacitor C and external plates connected via two wires. Are they in parallel with C? Yes, because the two wires connect across the capacitor terminals. So yes parallel. But wait: The diagram shows switch S2 in top wire; after S2 closes, the external plates are connected across the capacitor C? The bottom wire common. Yes. Potential issue: The voltmeter is in parallel with capacitor C; at t3, voltage ε/4. But if external plates capacitance changes due to plate motion, and C is in parallel, total capacitance C + C_ext. Charge conservation gives C_ext=3C. Good. Need perhaps account that the two parallel plates are "neutras" initially, but when connected to charged capacitor, the two plates become charged. However, the connecting wires themselves may have capacitance? Negligible. Now, let's see if there is an alternative interpretation: The external plates are neutral, but when connected to capacitor, charge flows until voltage equal. But because plates attract, they move, changing capacitance; perhaps the system at t3 is in equilibrium, but not necessarily all charge redistribution completed? It says at t3 equilibrium, so yes. Now, item c maybe using energy instead of force? Let's verify with energy method perhaps yields same k? Let's check energy conservation? Initial electrical energy = Cε^2/2. Final electrical energy = C V_f^2/2 + C_ext V_f^2/2 = (1/2)(C+3C)(ε/4)^2 = (1/2)*4C*ε^2/16 = Cε^2/8. Difference = Cε^2/2 - Cε^2/8 = 3Cε^2/8. Spring potential energy = (1/2) k (2d/3)^2 = (1/2) k (4d^2/9)=2k d^2/9. Equate to difference: 2k d^2/9 = 3Cε^2/8 => k = (3Cε^2/8)*(9/(2d^2)) = 27 C ε^2 /(16 d^2). This is different by factor? Let's compute: 3Cε^2/8 * 9/2 = 27 Cε^2/16 divided by d^2? Wait 2k d^2/9 = 3Cε^2/8 => k = (3Cε^2/8)*(9/(2 d^2)) = 27 C ε^2/(16 d^2). That's not same as force method k = 27 Cε^2/(64 d^2). Factor 4 difference. Hmm! Which is correct? Let's investigate. Important! Energy conservation may be invalid because charge redistribution involves energy loss? Let's understand. At t2, connect charged capacitor C to uncharged external plates. If no mechanical work, charge redistributes between two capacitors C and C_ext. The energy lost is ΔU = (1/2)Q0^2(1/C - 1/(C+C_ext)) = (1/2)Q0^2 * C_ext/(C(C+C_ext)). For C_ext=C, ΔU = (1/2)Q0^2 * C/(C*2C)=Q0^2/(4C)=Cε^2/4. This energy is dissipated as heat/radiation/sparks. In our system, the plates also move; some electrical energy may be converted to mechanical work, but there is still energy loss due to charge redistribution (like connecting capacitors). The final state minimizes total energy? The actual final equilibrium should minimize total free energy (electrical + mechanical), but the charge distribution and plate position adjust; the energy dissipated depends on dynamics. However, the force balance method is correct for mechanical equilibrium; energy conservation cannot be used naively because charge redistribution is dissipative. But is there a way to derive k from energy? The system may have an energy minimization principle if the charge redistribution is reversible? It isn't reversible if connecting two capacitors at different voltages; current flows through wires with resistance (even if small) causing dissipation. But in equilibrium, force balance must hold; energy minimization with respect to plate position at fixed total charge? Let's examine. If the system is isolated and evolves quasi-statically (maybe via external control), the total energy might be conserved if we include the work done by the battery? No battery disconnected, but charge redistribution at constant total charge and variable capacitance: The electrical energy U_e = Q0^2/[2(C+C_ext)]; mechanical energy spring = (1/2)k(d-x)^2. The total potential energy function U_total(x) = U_e(x)+spring. The equilibrium position is where derivative zero, which yields force balance. This is independent of energy loss? Actually if the system is conservative with generalized coordinate x and charge distribution always in equilibrium (quasi-static), then energy conservation would imply that the work done by electric force equals change in electrical energy? Wait in a conservative electrostatic system, if charge distribution is allowed to redistribute through ideal conductors without resistance, the system is conservative? But connecting two capacitors with different voltages through ideal wires is not physically possible without infinite current; usually there is dissipation. However, if you move the plates slowly while allowing charge to redistribute, the process might be reversible if you also adjust a battery? But no. Let's derive force from energy with respect to x at constant total charge, but the equilibrium condition is dU_total/dx = 0. This gives force balance, not energy conservation. The total potential energy U_total(x) = U_e(x) + U_spring(x), and equilibrium is at minimum of U_total if stable. So k can be found by setting derivative zero. That's what we did? Wait force balance is equivalent to derivative of U_total? Let's check: U_e = Q0^2/[2(C+K/x)]. Derivative dU_e/dx = F_e? We computed positive dU_e/dx = Q0^2 K/(2x^2(C+K/x)^2) = force? But electric force tends to decrease x, so force F_e = -dU_e/dx negative. Spring force = -dU_spring/dx = k(d-x) (if U_spring=1/2 k(d-x)^2, force = -dU/dx = -k(d-x)*(-1)? Let's compute: U_s = 1/2 k (d-x)^2. dU_s/dx = -k(d-x). Force = -dU_s/dx = k(d-x). Equilibrium: F_e + F_s =0? F_e = -dU_e/dx = -Q0^2 K/(2x^2(...)). F_s = k(d-x). Sum =0 => k(d-x)= Q0^2 K/(2x^2(...)). At x=d/3, RHS =? We computed dU_e/dx = 9 C ε^2/(32 d)? Wait that's magnitude? dU_e/dx = 9 C ε^2/(32 d) positive, force F_e = - that. Then F_s = +k(2d/3). Sum F_e + F_s = -9Cε^2/(32d)+ k(2d/3)=0 => k = 27Cε^2/(64d^2). Yes same as force balance. This is equilibrium condition. Energy conservation would be: initial total energy = final total energy + dissipated energy. Not helpful. So force method correct. My force calculation with U_e derivative was correct; energy balance including spring kinetic? Wait if total mechanical energy conserved, then initial electrical energy (before closing) = final electrical energy + spring energy + kinetic energy + dissipated. At final equilibrium, kinetic zero, but dissipated unknown. However, if we consider final equilibrium reached after damping, dissipated energy = initial - final electrical - spring. Unknown. So cannot determine k from energy alone without dissipation. However, equilibrium force balance independent of dissipation. So force method is right. But let's double-check force equilibrium using the actual charge on external plates Q_ext = 3Cε/4. The electric force on a plate with charge Q_ext due to the other plate is \(F = \frac{Q_ext^2}{2\epsilon_0 A}\). We used A = a^2 = C d / ε0? Wait C = ε0 A / d => ε0 A = C d. Yes. This is correct. Force magnitude = (3Cε/4)^2 / (2 C d)=9 Cε^2/(32 d). Spring force = k (2d/3). Equate: k= (9 Cε^2/(32d)) * (3/(2d)) = 27 Cε^2/(64d^2). Good. But wait: Does the electric force formula include both plates? The force on one plate is Q_ext times field due to the other plate, which is Q_ext/(2 ε0 A). So F = Q_ext^2/(2 ε0 A). Yes. This is the force pulling one plate; the spring force on that plate must balance. Good. Potential issue: The external plates have charges ±3Cε/4; but the capacitor C is also connected, so the electric field between the external plates maybe affected by the field from capacitor? No, capacitor C is separate; only the external plates produce field between them. The wires connect, but no other charges between the external plates. So okay. Now, let's consider if the moving plates' capacitance initially C and area A; when they move to x=d/3, area same, capacitance = ε0 A / x = 3C. Good. Now, maybe the "deformação da mola" could be \(d/3\) if they define deformation as final length? But "deformação" usually change in length = d - d/3 = 2d/3. Let's verify in Portuguese physics: "deformação da mola" = deformation of spring = \(x\) final? Some might call deformation \(x = l - l_0\), final length minus natural length, negative if compressed. They might expect magnitude "2d/3". We'll provide "2d/3 (compressão)". Let's think of possible hidden issue: The plates are neutral initially, but after connecting, the capacitor C and external plates may not simply share charge because the external plates are suspended and their capacitance is C at distance d; however, when connected, the initial voltage across external plates is zero; charge flows until voltages equal. But if the plates are free to move, they may start moving immediately; the final equilibrium voltage is ε/4. We used charge conservation after redistribution. But at the instant t2, the external plates might have capacitance larger/smaller due to motion? They are at initial separation d until charge flows and force acts; but after transient, final x determined. Charge conservation from initial t2 to final t3: total charge on C + external plates conserved? Yes, because S1 open, no battery; but is there any charge on the connecting wires? negligible. Total charge initially on C = Cε; external initially zero. So final total = Cε. Good. Let's maybe derive final charge on capacitor C and external plates: V_f = ε/4; Q_C = Cε/4; Q_ext = 3Cε/4. Now, maybe the "carga livre que fluiu do capacitor para as placas condutoras" could be \(Q_ext = 3Cε/4\), but since there are two wires, the total free charge that moved maybe \(2 \times Q_ext = 3Cε/2\). Let's look at similar ITA problems? They often ask "a carga livre que fluiu do capacitor para as placas condutoras" maybe they want the charge that each plate received? Let's search memory: In capacitor charge redistribution, "carga livre" on a conductor is its net charge. The external plates are two conductors; initially both neutral. After connection, one has +q, the other -q. The free charge that flowed from the capacitor to the plates? The capacitor's positive plate lost q charge; negative plate lost q electrons? But "free charge" maybe refers to the net charge on each plate. The total free charge on the two plates is zero, so they wouldn't ask total. They likely mean the magnitude of charge transferred through the connecting wires, \(q = Q_ext = 3Cε/4\). Let's phrase "carga livre (em módulo) que passou para cada placa condutora é \(3C\varepsilon/4\)." But if they specifically say "placa condutoras" plural maybe "as placas" receive charges; one could answer "as placas adquirem cargas +3Cε/4 e -3Cε/4". That is likely. Let's maybe include both: "Em módulo, a carga livre que passou do capacitor para as placas é \(3C\varepsilon/4\) (uma placa recebe +, a outra −)." Good. Now, let's ensure no use of the small-angle approximation? The initial text "Para θ << 1 assuma senθ ≈ θ." That is included before question? The text at top: "Quando necessário, use a seguinte informação: Para θ << 1 assuma senθ ≈ θ." This might be from the exam cover; is it relevant? Maybe this problem might involve small angle? But our problem has no angles. It might be a generic instruction for the whole exam, not used. But maybe there is a pendulum aspect? The wires suspended? "suspensos numa placa não condutora rígida e fixa no espaço" maybe if the plates are suspended by wires and move like pendulums, there is an angle? But the problem says plates remain parallel and distance d << a; our force balance ignored tension. Let's revisit possible geometry to see if small angle might be needed. Maybe the setup is: Two identical conducting wires are suspended from a nonconducting rigid plate fixed in space, and each wire supports one of the parallel plates. Initially the wires are vertical and the plates are separated by d. When the plates attract, they move closer, causing the wires to tilt? Wait if the wires are attached to the plates at their centers, the plates might move horizontally while remaining parallel, and the wires remain vertical? But if wires are flexible and attached at top, for plates to move horizontally, the wires would tilt (like pendulums). The problem says "placas condutoras permanecem paralelas em todas as situações" but they could tilt? Actually if wires are identical and each plate attached to one wire, the plates could rotate? But "permanecem paralelas" means no rotation, so maybe the wires allow translation only (e.g., wires are horizontal? no). Let's parse original: "o qual está ligado por dois fios idênticos e condutores, suspensos numa placa não condutora rígida e fixa no espaço, a duas placas planas paralelas metálicas..." Maybe the two conducting wires connect the circuit to the two plates; the wires are suspended from a nonconducting plate. The plates are "suspensas" (suspended) by the wires. They may act like pendulums, so when electrostatic attraction occurs, the plates approach each other and the wires tilt. The problem includes small-angle approximation? Hmm maybe the force component along the wire? Let's see. If the plates are suspended by wires of length \(L\) from fixed points, and the spring connects them; the electrostatic force pulls horizontally; the wires tilt at angle θ. The tension in the wires has a horizontal component that balances electric + spring forces? But if the wires are conductors, they conduct current and also support plates. The mass of plates maybe? No mass given, so maybe neglect gravity? But if wires tilt, tension exists even without gravity? A massless string can only have tension along its length; without gravity or mass, the tension would be zero? Hmm. Maybe the small-angle instruction is for another question; not necessarily this. But maybe there is an angle due to the wires? Let's see if the final deformation could be derived by force balance including wire tension with angle θ and maybe small angle approximation? Let's attempt. Suppose each plate is suspended by a wire of length L from a fixed support, so the wire can swing. The two wires are identical, so both plates swing symmetrically. The spring connects the plates horizontally. The electrostatic force is horizontal attraction. The spring force horizontal. If we ignore gravity, the only forces on each plate are electric force, spring force, and tension along wire. Since there is no vertical force (gravity? maybe nonconducting plate horizontal; but "suspensos" implies gravity? But no mass data). If no gravity, tension can have any direction? A massless string can only transmit force along its length; if the plate has no mass and no other vertical force, the vertical component of tension must be zero, so tension along horizontal? impossible because wire has vertical extent. So there must be gravity or some force. The problem likely ignores gravity and wire tension; the wires are just electrical connections, not mechanical supports? Actually "suspensos" might mean they hang vertically but are not necessarily supporting weight? But wires can't exert horizontal force if vertical and massless. Hmm. Maybe the "two wires" are actually rigid conducting rods, not flexible? Or they are suspended from a fixed plate with the plates hanging below; the spring length changes, but the wires might remain vertical? If the wires are rigid and vertical, the plates can't move horizontally unless wires tilt. But if wires are flexible, they could tilt; tension would have horizontal component balancing electric force and spring force. Since no angles given, perhaps the plates remain parallel and can move horizontally; the wires may be not straight? Maybe "suspensos" means they are hung, but the electrical connection wires are not mechanical constraints; the plates are constrained by the spring? Hmm. Let's search memory: This might be an ITA problem involving capacitor with movable plates and spring, suspended by wires; the force between plates is balanced by spring, not by wire tension. The small-angle approximation may not be needed because there's no angle; the wires just hang vertically and the plates move? Actually if the plates are suspended by vertical wires and move horizontally, the wires would tilt. But if the wires are long and the displacement is small (d/3 might not be small relative to L unknown), then small angle? But no L given. So likely the suspension is not relevant; maybe the wires are "fios idênticos e condutores" that connect electrically, not mechanically; they are limp and maybe the plates are supported by the spring? Let's not overcomplicate. The problem explicitly: "suspensos numa placa não condutora rígida e fixa no espaço" meaning the wires are suspended from a fixed nonconducting plate (like a ceiling), so they hang down and hold the plates. When the plates attract, they may approach, and the wires may swing; but the problem says "as placas condutoras permanecem paralelas em todas as situações" likely means no rotation, so the wires may remain parallel? If wires are identical and suspended vertically, for plates to move horizontally, the wires must tilt by same angle, so plates remain parallel. But then there is tension and possibly gravity? Let's model with small angle: Each plate of mass m? No mass given. Maybe the plates are light; gravity negligible; then the wire tension cannot have vertical component (no other vertical force), so tension must be zero; impossible. So maybe the plates are not massive; the only force along wire? Maybe the wire is just a conductor and doesn't support; it's "suspensos" but maybe the nonconducting plate is above, and the wires are vertical; the plate hangs from the wire, but the wire can only pull along its length; the vertical equilibrium requires tension equal weight. If weight not given, cannot solve. Therefore likely the wires are not mechanical constraints; perhaps the nonconducting plate is just a support for the wires, but the plates are held by the spring? Or the plates are fixed in orientation by some other means (e.g., the wires are part of a rigid pendulum apparatus), and gravity is neglected because the electric force is much larger? But no mass/weight given. Thus ignore. The small-angle approximation note might be for another part of the exam, not used. We'll not use it. Let's consider if "d << a" is used to approximate uniform field and ignore edge effects, and also maybe to approximate force with derivative of capacitance. We'll mention. Now, let's see if item a maybe should be "a carga livre que fluiu do capacitor para as placas condutoras" maybe as function of C, ε. Answer \(3Cε/4\). But if they wanted "carga livre total" maybe \(3Cε/2\). Let's determine by exact wording. In Portuguese, "carga livre" is the net charge on a conductor. "A carga livre que fluiu do capacitor para as placas condutoras" = the free charge that flowed from the capacitor to the conducting plates. Since the capacitor and plates are connected by two wires, charge flows through each wire. The "carga livre que fluiu" could mean the amount of charge that passed through the connection, i.e., the charge on each plate. But if they ask "para as placas condutoras" (plural), maybe they consider the plates collectively as receiving charge from the capacitor. The capacitor initially has charge Q0 on one plate and -Q0 on the other. After redistribution, the conducting plates have charges +q and -q. The charge that flowed from the capacitor's positive plate to the positive conducting plate is q; from negative plate to negative conducting plate is also q (electrons flow opposite). So each wire carries q. The total free charge that "left" the capacitor's plates (in magnitude, both plates) is 2q. But typical answer in such problems: "a carga que passa pelo circuito é \(3Cε/4\)". Let's look at similar examples: If two capacitors C1 and C2 share charge, the charge that flows through the connecting wire is \(Q = \frac{C_1 C_2}{C_1+C_2} V_0\)? Actually if C1 charged to V0, C2 uncharged, final voltage V = C1 V0/(C1+C2); charge transferred from C1 to C2 = C1(V0 - V) = C1*(C2 V0/(C1+C2)) = C1 C2 V0/(C1+C2). For C1=C, C2=C, final V=V0/2, charge transferred = C V0/2. They would not say total 2*C V0/2 = C V0. So likely \(3Cε/4\). Let's verify with total charge: If initial charge on capacitor C = Cε, final charge on C = Cε/4. The charge that flowed out of the capacitor (one plate) = Cε - Cε/4 = 3Cε/4. Yes. So answer a = 3Cε/4. Now, let's think if the spring constant might be asked in terms of \(a\) and \(d\)? We have C, ε. But C is given as equal to external plates, maybe can express C = ε0 A/d = ε0 a^2/d. But the problem asks "c) a constante elástica da mola." It might expect in terms of given quantities: \(C, \varepsilon, d\) or maybe \(a, \varepsilon, \epsilon_0, d\). Since C is given, use C. But if they want using \(a\), replace C = ε0 A/d = ε0 a^2/d. Let's see: in statement: "duas placas planas paralelas metálicas, neutras, de área A (A = a^2) e capacitância C igual à do capacitor do circuito." So C is given symbol; perhaps final answer can include C. They may want using C, ε, d. We'll do. Let's maybe express spring constant as \(k = \frac{27 C \varepsilon^2}{64 d^2}\). If substituting A, \(k = \frac{27 \epsilon_0 a^2 \varepsilon^2}{64 d^3}\)? Wait C = ε0 A/d = ε0 a^2/d, so C/d^2 = ε0 a^2/d^3. Then k = 27 ε0 a^2 ε^2/(64 d^3). But they gave \(A\) and \(C\). Use C. Let's consider whether the force should include the fact that the electric field between external plates is affected by the charge on capacitor C? No, they are separate; but the capacitor C is in parallel, meaning its plates are connected across the same voltage. The external plates and capacitor C are physically separated, but their electric fields don't interact (ideally). So force between external plates depends only on their charge. Now, maybe there is a factor 2 because there are two identical wires, and the plates are suspended, so each plate experiences force; but force balance we wrote on one plate: electric force = spring force. But if the spring is connected between the two plates, its force on each plate has same magnitude. The electric force on each plate has same magnitude. So okay. Let's also consider if the spring itself may be conductive? It says nonconducting spring, so okay. Now, let's maybe derive in final answer step-by-step: 1. Capacitor initial charge. 2. Charge conservation and final voltage. 3. Find external capacitance and separation. 4. Charge flowed. 5. Force balance to find k. Let's format with equations. Potential need to mention "em módulo" for charge and deformation. Let's draft final response in Portuguese: "Sejam \(Q_0 = C\varepsilon\) a carga inicial do capacitor e \(V_f = \varepsilon/4\) a tensão final. Como a chave S1 está aberta, a carga total do sistema capacitor + placas externas se conserva..." Then: \[ Q_0 = C V_f + C_{\rm ext} V_f = (C + C_{\rm ext})\frac{\varepsilon}{4} \] \[ C\varepsilon = (C + C_{\rm ext})\frac{\varepsilon}{4} \Rightarrow C + C_{\rm ext}=4C \Rightarrow C_{\rm ext}=3C. \] \[ C_{\rm ext} = \frac{\varepsilon_0 A}{x_f} = \frac{C d}{x_f} \Rightarrow 3C = \frac{C d}{x_f} \Rightarrow x_f = d/3. \] Logo deformação: \[ \Delta x = d - d/3 = 2d/3 \quad \text{(compressão)}. \] Carga que fluiu: \[ Q_{\rm ext}=C_{\rm ext} V_f = 3C\cdot \frac{\varepsilon}{4} = \frac{3}{4}C\varepsilon. \] Força: \[ F_e = \frac{Q_{\rm ext}^2}{2\varepsilon_0 A} = \frac{(3C\varepsilon/4)^2}{2Cd} = \frac{9C\varepsilon^2}{32d}. \] Como \(\varepsilon_0 A = Cd\). Mola: \[ F_s = k\Delta x = k\frac{2d}{3}. \] Equilíbrio: \[ k\frac{2d}{3} = \frac{9C\varepsilon^2}{32d} \Rightarrow k = \frac{27C\varepsilon^2}{64d^2}. \] Final answer: Resposta final: a) \(3C\varepsilon/4\); b) \(2d/3\) (compressão); c) \(27C\varepsilon^2/(64d^2)\). Need ensure final line exactly as requested: "Resposta final: <o resultado de cada item, de forma curta>" So I'll do that. Let's consider if using \(C\) in item c is okay; if the problem wanted only given symbols, yes. If they wanted in terms of \(a\), maybe substitute \(C=\epsilon_0 a^2/d\). Let's include optional note? The final answer line maybe include both? It says "de forma curta"; use given symbols. I can include in final line \(27Cε^2/(64d^2)\). But if they expect \(27 \epsilon_0 a^2 \varepsilon^2/(64 d^3)\), maybe not. Let's decide. The problem statement defines C as capacitance of external plates equal to circuit capacitor. So C is a given. So okay. Let's also maybe state "a carga livre em módulo" in final line. Good. Now, let's think if there is any issue with using \(C_{\rm ext} = \varepsilon_0 A/x_f\) at final state even though the capacitor C is also connected; total charge conservation includes both. The external plates charge \(Q_ext = C_ext V_f\). But the voltage across external plates is same as capacitor C. Yes. Let's verify units of k: \(C\) in F = C/V; ε in V; C ε^2 = F V^2 = J = N m. Divide by d^2 m^2 => N/m. Good. Let's maybe derive the electric force expression using energy derivative to avoid ambiguity: \(U_e = Q_ext^2/(2C_ext) = Q_ext^2 x/(2ε0 A)\). At constant Q_ext, \(F_e = -dU/dx = -Q_ext^2/(2ε0 A)\) (attractive). But since charge redistributes with capacitor C, at equilibrium the force on plates can also be obtained from total energy derivative, but final state known. We'll use direct formula. Now, let's consider if the external plates are initially neutral but after connection, the free charge that flowed from capacitor to plates might be distributed on both surfaces? The plates are metallic; charge resides on inner surfaces because of attraction; negligible thickness. So okay. Let's think about possible "carga livre" definition: In a conductor, free charge is the net charge; it may reside on surfaces. The external plates each get \(±3Cε/4\). So answer a "a carga livre que fluiu do capacitor para as placas condutoras" = \(3Cε/4\) (em módulo). If they wanted "carga total que saiu do capacitor" maybe maybe \(3Cε/4\) is the charge that left one plate; but the capacitor as a whole lost \(3Cε/4\) from positive plate and gained? Wait the negative plate also lost \(3Cε/4\) electrons? Let's analyze: Initial capacitor: plate A +Cε, plate B -Cε (electrons excess on B). After: plate A +Cε/4? Actually if final charge on C is Cε/4 positive on A, then plate A lost +3Cε/4 (i.e., 3Cε/4 electrons? Wait positive charge less means electrons added? Let's not). Plate B final has -Cε/4 (excess electrons less negative? Actually -Cε to -Cε/4 means it lost negative charge (electrons) of magnitude 3Cε/4, so 3Cε/4 electrons left plate B to external plate? Through the external circuit, electrons flow from plate B to the negative external plate? The "charge that flowed" as conventional current magnitude is 3Cε/4 in both wires. So yeah. Let's formulate final response elegantly. Potential hidden factor: The voltmeter at t3 reads one quarter of t2. But at t2, before S2 closed, the capacitor is fully charged to ε. However, at t2, S2 is closed and S1 open; the external plates are connected. The voltmeter reads voltage across capacitor C. Since the external plates are neutral and connected, initially voltage across external plates is zero; after transient, final voltage \(V_f\). The voltmeter might have internal resistance? No, ideal. Let's maybe include "desprezando efeitos de borda (d << a)". Now, final answer in Portuguese with steps. Let's write: "Como \(d\ll a\), o campo entre as placas é uniforme e \(C_{\rm ext}=\varepsilon_0 A/x\). Inicialmente o capacitor \(C\) está carregado com \(Q_0=C\varepsilon\). Após S2 fechar, a carga total se conserva: \[ C\varepsilon = C V_f + C_{\rm ext}V_f \] com \(V_f=\varepsilon/4\). Logo..." Then solve. Let's ensure no mistake in algebra: \[ C\varepsilon = (C+C_{\rm ext})\frac{\varepsilon}{4} \implies 4C = C+C_{\rm ext} \implies C_{\rm ext}=3C. \] \[ C_{\rm ext} = \frac{\varepsilon_0 A}{d'} = \frac{C d}{d'} = 3C \implies d'=d/3. \] Deformation \(d-d'=2d/3\). Charge flowed: \[ Q_{\rm fl} = Q_0 - C V_f = C\varepsilon - C\varepsilon/4 = 3C\varepsilon/4. \] Alternatively \(Q_{\rm ext}=C_{\rm ext}V_f=3C\varepsilon/4\). Good. Force: \[ F_e = \frac{Q_{\rm fl}^2}{2\varepsilon_0 A} = \frac{(3C\varepsilon/4)^2}{2Cd} = \frac{9C^2\varepsilon^2/16}{2Cd} = \frac{9C\varepsilon^2}{32d}. \] Mola: \[ k\frac{2d}{3} = \frac{9C\varepsilon^2}{32d} \Rightarrow k = \frac{27C\varepsilon^2}{64d^2}. \] Now final answer line. Let's ensure final answer line exactly: "Resposta final: a) \(3C\varepsilon/4\); b) \(2d/3\) (compressão); c) \(27C\varepsilon^2/(64d^2)\)." Maybe include "em módulo" in a. Good. Let's also maybe include "a carga livre que fluiu para cada placa é 3Cε/4 (uma positiva, outra negativa)" in item a explanation. Good. Let's now craft final response.

**Analyzing Exam Question** I am currently focused on understanding the specifics of the ITA exam question you've provided. My immediate goal is to carefully read and interpret all instructions and given information to accurately address the physics problem. **Deconstructing Circuit Dynamics** I'm delving into the circuit's behavior, specifically how the charged capacitor interacts with the spring-loaded plates and subsequent discharge through $S_2$. I'm evaluating the energy transformations and equilibrium conditions that arise after $S_1$ opens and $S_2$ closes. **Interpreting Physics Problem** I'm now examining the detailed parameters of the physics problem. My focus is on precisely understanding the initial conditions and the relationships between capacitance, distance, and voltage at different time points ($t_1$, $t_2$, $t_3$). **Evaluating Energy and Charge Flow** I am analyzing the energy conservation within the system, particularly how the initial charge on the capacitor is redistributed. I'm also calculating the free charge that moves to the conductive plates at $t_3$, considering the spring's deformation and its elastic constant. **Considering Gravitational Force** I am now contemplating the potential influence of gravity on the suspended plates, as this detail seems critical for understanding the equilibrium state and how it might be affected by electrostatic forces, thus impacting my calculations. **Clarifying Physics Context** I'm re-examining the context of the problem, particularly questioning whether the "wire pendulum" exerts a restoring force or if it's solely the spring. The phrase "Para $\theta \ll 1$ assuma $\text{sen}\theta \approx \theta$" appears to be a crucial, but potentially misplaced, hint. **Investigating Pendulum Influence** I am now considering the implications of the plates being suspended by wires. I need to determine if this suspension introduces a pendulum-like behavior and if the provided small angle approximation is relevant to this aspect, rather than just the spring's influence. **Differentiating Force Sources** I am now focused on distinguishing between forces. Specifically, I need to ascertain if the wire suspension itself provides a restoring force, or if the spring is the sole contributor to restoring equilibrium in the system. **Confirming Formula Relevance** I am trying to confirm if the small angle approximation for sine is truly applicable here, given the description of the plates being suspended by wires. I am questioning its origin and direct relevance to the current setup. **Re-evaluating Wire Suspension** I am now considering how the wire suspension of the plates might be influencing the system dynamics. I need to determine if this introduces any pendulum-like effects, and if the small angle approximation mentioned is indeed relevant to the physics at play here. **Questioning Angle Approximation** I'm questioning the relevance of the small angle approximation ($\sin \theta \approx \theta$) given the lack of information on wire length, plate mass, or gravity. I suspect this hint might be from a different part of the exam. **Analyzing Suspension Method** I'm now scrutinizing the wire suspension of the plates. I need to confirm if this configuration implies a pendulum effect or if the assumption of parallel plates overrides such dynamics, potentially making the small angle approximation irrelevant. **Clarifying Problem Assumptions** I am now re-examining the core assumptions about the system's behavior. The lack of explicit parameters for a pendulum suggests that this dynamic might be intentionally excluded from the problem, making the small angle approximation irrelevant to the core physics. **Analyzing Wire Support** I am now questioning the role of the wires suspending the plates. The explicit statement that the plates remain parallel seems to contradict the possibility of them swinging like a pendulum, which makes the small angle approximation feel out of place for this specific problem. **Deducing Missing Parameters** I'm realizing that the problem likely doesn't require knowledge of the plates' mass or the gravitational force for determining the spring constant. The given parameters for capacitance, area, and initial voltage seem sufficient to solve for $k$. **Analyzing Wire Connection** I'm now investigating the precise role of the wires described. The phrasing about suspension from a non-conducting plate, combined with the diagram showing electrical connections, suggests the wires are primarily electrical leads and not intended to create a pendulum motion. **Interpreting Wire Function** I'm now focused on the role of the wires suspending the plates. The Portuguese phrasing suggests the wires connect the circuit to the plates, rather than implying pendulum motion. This clarifies their function as electrical leads. **Clarifying Information Sources** I'm realizing that the small angle approximation hint comes from a general information section, not specifically tied to this problem's diagram. This strongly suggests it's a red herring and not relevant to the current physics scenario. **Discarding Irrelevant Information** I have concluded that the small angle approximation hint is irrelevant. The problem's description explicitly states the plates remain parallel, negating any pendulum motion, and this hint originates from a general section, not the specific problem. **Discarding Extraneous Details** I'm now confident that information regarding wire length, plate mass, and gravity is not required. The spring's role appears to be balancing the electrostatic force, indicating the small angle approximation is a distraction. **Discarding Extraneous Details** I'm now confident that information regarding wire length, plate mass, and gravity is not required. The spring's role appears to be balancing the electrostatic force, indicating the small angle approximation is a distraction. **Confirming Charge Flow** I have confirmed the charge flow calculation independently through two distinct paths. The results are consistent and show a clean, unambiguous value for the charge transfer, which is independent of mechanical parameters like the spring constant. **Calculating New Separation** I have determined the new separation between the parallel plates at $t_3$ to be $d/3$. This result stems from the derived capacitance $C' = 3C$, and confirms the plates have moved closer together. **Analyzing Wire Force** I am now considering if the suspension wires exert any force when the plates move closer. If they deflect, this could introduce a new force component acting on the plates. **Analyzing Wire Force** I am now considering if the suspension wires exert any force when the plates move closer. If they deflect, this could introduce a new force component acting on the plates. I'm re-examining the problem description for any mention of wire length or mass, as this could be crucial. **Analyzing Wire Behavior** I am now analyzing the specific behavior of the suspension wires. The phrasing suggests they primarily serve as electrical leads, supporting the plates without introducing significant pendulum-like restoring forces, especially given the parallel plate assumption. **Considering Wire Behavior** I am now analyzing the behavior of the suspension wires, wondering if they act as ideal flexible strings. The presence of the small angle approximation hint is causing me to reconsider their role, but the context strongly suggests they are primarily electrical leads, not a pendulum system. **Calculating Electrostatic Force** I'm now calculating the electrostatic force using the derived charge on the plates. This force, $F_e = \frac{Q_p^2}{2Cd}$, is essential for determining the spring constant, $k$. **Calculating Spring Constant** I have successfully calculated the electrostatic force and equated it to the spring force to determine the spring constant, $k$. The new separation $d/3$ and derived charge confirm the result. **Confirming Mathematical Structure** I've confirmed the beautiful mathematical structure arising from the problem. The derived spring constant $k = \frac{27}{64} \frac{C \varepsilon^2}{d^2}$ presents a clean coefficient, $(3/4)^3$, which I am now investigating for its underlying algebraic origins. **Confirming Algebraic Structure** I've confirmed the specific choice of parameters leading to the elegant $(3/4)^3$ coefficient for the spring constant. The algebraic simplification arises precisely because the chosen $\alpha = 1/4$ makes the $2(1-2\alpha)$ term equal to 1. **Confirming Problem Structure** I've confirmed the wires act solely as electrical leads, and the spring is the only horizontal restoring force. The problem's design with $\alpha = 1/4$ leads to elegant cancellation. I'm now verifying spring deformation and plate movement, noting the initial spring length is $d$. **Verifying Spring Mechanics** I am now confirming the spring's behavior. The deformation is indeed $2d/3$, and the force exerted on each plate is $k \times (2d/3)$, aligning with Newton's third law for a compressed spring and the electrostatic attraction. **Analyzing Wire Role** I've confirmed the wires function solely as electrical leads, not as pendulum components. The spring provides the horizontal restoring force. The problem's parameters simplify elegantly due to a specific choice leading to cancellations. **Confirming Charge Calculation** I have finalized the calculation for the free charge flow to the conductive plates. My derived value of $\frac{3}{4} C \varepsilon$ appears correct based on the charge conservation principle and the given voltage changes. **Verifying Spring Deformation** I am now verifying the spring deformation. The problem's parameters indicate the spring's initial rest length is $d$, and my subsequent calculations suggest a deformation leading to a new separation of $d/3$ at $t_3$. This implies a deformation of $2d/3$. **Confirming Spring Compression** I have confirmed the spring is compressed by $2d/3$. This compression arises from the electrostatic attraction pulling the plates closer to $d/3$, relative to the initial spring length of $d$. This is now a definite conclusion. **Verifying Spring Force Calculation** I am now verifying the calculation of the spring force, $k \times (2d/3)$, against the derived electrostatic force. This step is crucial to confirm the spring constant's value and its relationship to the initial conditions. **Confirming Final Formula** I've confirmed the calculation for $k$ can be expressed using either $C$ or $\varepsilon_0 A$. Given both are provided, I will express it in terms of $C$ and $\varepsilon$ as $k = \frac{27}{64} \frac{C \varepsilon^2}{d^2}$. I'm now reviewing the voltmeter's role for any subtleties. **Analyzing Charge Stability** I'm re-evaluating the equilibrium condition. With $S_2$ closed, the charge on the plates isn't constant but varies with distance, meaning I need to re-derive the net force equation to check for stability. **Confirming Force Calculation** I've confirmed the electrostatic force formula $F_e = \frac{Q_p^2}{2 \varepsilon_0 A}$ is always applicable regardless of the charging method. This simplifies the equilibrium condition at $t_3$, meaning the spring constant derivation is now robust. **Analyzing Force Derivations** I'm cross-referencing electrostatic force calculations using potential energy derivatives with charge. The initial formula $F_e = Q_p^2 / (2Cd)$ is confirmed valid, and I'm now investigating an alternative derivation via $F = -(\partial U/\partial y)_Q$. **Confirming Force Derivations** I've confirmed that electrostatic force calculations derived via potential energy differentiation with respect to distance and charge yield the same result. This consistent $F_e = \frac{9}{32} \frac{C \varepsilon^2}{d}$ is crucial for verifying the spring constant, and stability is not a required parameter for this problem. **Confirming Physics Terminology** I've confirmed that "carga livre" refers to free charge, meaning the net charge transferred. The spring deformation is indeed a compression of $2d/3$, and the spring constant is $k = \frac{27}{64} \frac{C \varepsilon^2}{d^2}$. The area $A=a^2$ is not directly needed for the final results. **Revising Edge Effect Justification** I'm re-examining the justification for neglecting edge effects. The statements $A = a^2$ and $d \ll a$ confirm that the plates can be treated as ideal parallel plates throughout the process, maintaining their parallelism and uniform field.

**Analyzing the Problem** I'm starting to carefully read the physics problem from the ITA exam. I'm focusing on understanding the circuit diagram and the information provided about the identical wires. My goal is to grasp the core concepts before attempting a solution. **Interpreting the Scenario** I'm now diving deeper into the problem's setup, particularly the capacitor's interaction with the metallic plates and the spring. I'm trying to visualize the impact of closing and opening the switches on the system's equilibrium. I'm focusing on how the voltage changes relate to the charge distribution and forces involved. I need to figure out how the system reaches a new equilibrium state after the switch operations. **Calculating Spring Deformation** I'm now focusing on calculating the spring's deformation and its spring constant. I'm considering the parallel plates after the switches' change, and how charge and voltage change relates to the spring's mechanical equilibrium. I'm looking at how the capacitance of the plates changes with distance and applying the equations for electrostatic force and spring force to determine the new equilibrium state. **Clarifying Equilibrium Position** I'm now zeroing in on the phrase "se encontra no ponto de equilíbrio." It seems clear that, in this scenario involving electrostatic forces and a spring, this refers to the point where the net force on each plate is zero, which means mechanical equilibrium. The text's lack of information about the plates' mass or the wires' length reinforces this interpretation. The gravitational restoring force is therefore irrelevant. **Examining Given Parameters** I'm now methodically checking the provided parameters: $A$, $C$, $d$, $\varepsilon$, $R$. I noticed $\varepsilon_0$ wasn't explicitly stated, but since $C$ is given as $\frac{\varepsilon_0 A}{d}$, I can express everything in terms of $C$, $\varepsilon$, $d$. This should simplify things. I'm focusing on item 'a', calculating the free charge that flowed from the capacitor to the plates. I've determined that, at $t_3$, the remaining charge on the capacitor is $\frac{1}{4} C \varepsilon$. **Determining Capacitor Charge** I've calculated the charge $q$ that flowed from the capacitor to the plates as $\frac{3}{4}C\varepsilon$ after opening $S_1$ and closing $S_2$. The capacitance $C$ and voltage $\varepsilon$ are indeed explicitly provided. Now, I'm shifting to calculating the spring deformation. At $t_3$, the parallel plates have the same voltage as the voltmeter, which is $\frac{\varepsilon}{4}$, so I'm trying to find the capacitance of parallel plates at time $t_3$. **Calculating Spring Constant** I've determined the spring's deformation at time $t_3$ to be (2/3)d. Since the text contains the standard instruction for ITA questions, I've confirmed that the calculation should rely solely on electrostatic and spring forces. Now I will focus on applying these insights to solve for the spring constant. **Revisiting Electrostatic Forces** I'm revisiting the electrostatic force calculation, making absolutely sure I have the $1/2$ factor correct when calculating the force between the capacitor plates. I've verified the formula using both electric field and the virtual work method, confirming that $F_{el} = \frac{q^2}{2\varepsilon_0 A}$ is indeed correct. Now, I'm expressing $\varepsilon_0 A$ in terms of the capacitance $C$ and the distance $d$. **Deriving the Spring Constant** Now I have a complete expression for the spring constant. Using the calculated electrostatic force and spring deformation, I've solved for *k* as $\frac{27 C \varepsilon^2}{64 d^2}$. I've confirmed the equation, and I'm now exploring if the answer can be expressed in terms of $\varepsilon_0$ and $A$. **Justifying Given Information** I'm now understanding why the problem provided $A$ ($A=a^2$) and $d \ll a$. It ensures the validity of the parallel plate capacitor formula, $C = \frac{\varepsilon_0 A}{d}$, by making the fringe effects negligible. The parameters $C$, $\varepsilon$, and $d$ are fundamental, so I'll be sure to express my solution in their terms. I'll include the equivalent form using $\varepsilon_0$ and $a$, as well.

**Evaluating plate alignment forces** I'm examining the forces involved with aligned plates, focusing on vertical suspension wires and parallel conductor squares that attract each other due to equal charges. The center of the electrostatic force is at the center as well. I've noted how the spring attaches at the center and how tension is involved. In equilibrium, torques about the center must be considered, especially when the angle of the wire changes. It seems like there’s a contradiction with the electric field being uniform.**Evaluating assumptions in mechanics** I’m thinking we should emphasize that 'c' isn't uniquely determined, rather than simply saying we can solve for it. The developer mentioned we need to arrive at a solution, but I wonder if we can fully solve it without making assumptions. We might use equilibrium principles with standard parallel geometry, considering the forces and small angles, especially if gravity cancels out with parallel plates and wire geometry. The horizontal forces involved are crucial. Yet, it doesn’t guarantee a clear answer.**Exploring wire tension arrangements** I'm considering whether there's a way to achieve zero horizontal tension with the setup of wires described. The scenario involves identical conductive wires suspended on a rigid, non-conductive plate. These wires are fixed at one end, hanging down and possibly needing to allow for movement. I wonder if I can manipulate the horizontal position of the wire ends while keeping them vertical due to the attachments. It's an interesting puzzle!