Question

Three conducting sheets 1, 2 and 3 are placed parallel to each other such that separation between them is negligible. Sheet-1 and sheet-3 are given charges +Q and +3Q respectively. Cell of emf 1 V is to be connected between sheet 1 and 3. take e0 a/d =30mFand Q=3mC.first switch S1 is closed then S2 is closed. find charge flowing thru each

Answer 1

Charge that flows through S1 (the battery branch): 15 mC

Charge that flows through S2 (when joining the middle plate): 15 mC

Answer 2

We shall show that when the battery (of 1 V) is “hooked in” between the two charged outer sheets, the two capacitor–gaps formed (between sheet 1 and sheet 2 and between sheet 2 and sheet 3) come to have equal “capacitor–charge” and then, when the middle plate is “joined in” by closing S2, the frozen charge on the 1–2 gap is neutralized; the net result is that

• When S1 is closed the battery “pushes” a charge

  q = C₀V_equiv  with V_equiv = (1 V)/2 so that q = (30 mF×0.5 V) = 15 mC,

and

• When S2 is closed the middle plate is forced to join the already–established equipotential group so that the 15 mC originally “stored” between 1 and 2 is neutralized by charge flow through S2.

A brief outline of the ideas is as follows.

The situation before any switch is closed

We have three large plates (sheets 1, 2 and 3) with net charges

Sheet 1: +Q = +3 mC

Sheet 3: +3Q = +9 mC

and sheet 2 initially uncharged. (Their separations are “small” so that – apart from the “gaps” – they are almost in contact.)

After closing S1 (battery only)

A cell of emf 1 V is connected between sheets 1 and 3. (The battery is allowed to add or remove charge from these two plates.) Although the net charges on 1 and 3 are now changed by the battery our “active” part is that the two small gaps (1–2 and 2–3) behave as two capacitors in series. Since the two gaps are identical (each having C₀ = 30 mF) the series combination has an equivalent capacitance

C_eq = C₀/2 = 15 mF

and the 1 V battery “divides” evenly between the two gaps. Thus each gap develops a voltage drop ½ V and (by the usual capacitor formula) acquires an equal magnitude of charge

q = C₀×(½ V) = 15 mC.

Since the (floating) middle plate must remain overall neutral, its two faces acquire equal and opposite “induced” charges: +15 mC on the face toward sheet 1 and –15 mC on the face toward sheet 3.

Also note that the battery must supply (or remove) charge in order to force the difference between sheet 1 and 3 to be 1 V; it actually delivers (through the circuit S1) a net charge of 15 mC (to one plate and –15 mC from the other).

After closing S2 (middle plate “connected” to one group)

Now the switch S2 is closed. In many such problems the middle plate (sheet 2) is “joined” by S2 to one of the outer ones (say, sheet 1). When that happens the capacitor between sheets 1 and 2 is effectively shorted, so that its stored 15 mC is “drained out” (redistributing over the now–combined conductor) and the series capacitor system ‘collapses’ into a single capacitor between the joined pair (1+2) and sheet 3. (The battery continues to hold 1 V across the new two–plate system.)

Thus an amount of 15 mC (which was “on” the 1–2 capacitor) flows through S2 when it is closed.

Thus the answer is:

• Charge that flows through S1 (the battery branch): 15 mC
• Charge that flows through S2 (when joining the middle plate): 15 mC

To summarize very briefly:

1. With only S1 closed the series capacitor (two gaps, each of capacitance 30 mF in series) gets a charge

   q = (C₀/2)×1 V = 15 mC.

2. Closing S2 forces sheet 2 to join one of the outer plates; the face charge of 15 mC on the 1–2 gap is then neutralized by a flow of 15 mC through S2.

Thus, the charge flowing is 15 mC through S1 and 15 mC through S2.

Finally, we also note that although the original net charges were +3 mC and +9 mC for sheets 1 and 3, respectively, the battery action (together with redistribution when S2 is closed) overrides these values by the addition of much larger amounts (here 15 mC) in order to enforce the 1 V difference. (In an exam–style problem the point is to “find the extra charge” flowing when the battery is connected and then when the middle plate is attached.)

Additional information:

Subject: Physics

Chapter: Electrostatics

Topic: Capacitors – Series combinations & charge sharing

Difficulty: Medium

Question type: descriptive

Thus the final answer is:

Through S1: 15 mC flows (supplied by the battery)

Through S2: 15 mC flows (when the middle plate is connected)

This is our answer in the spirit of the JEE/NEET style solution.