JEE Challenger
More from Electrostatic Potential and Capacitance

Charge Distribution on Capacitors After Sequential Switching

In the circuit displayed in the figure, two parallel plate capacitors each having capacitance CC are connected. Initially, switch S1S_1 is pressed to fully charge capacitor C1C_1 and then released. Subsequently, switch S2S_2 is pressed to charge capacitor C2C_2. After some time, S2S_2 is released and then S3S_3 is pressed. After sufficient time has elapsed,

Question Diagram 1

Options

A

the charge on the upper plate of C1C_1 is 2CV02CV_0.

B

the charge on the upper plate of C1C_1 is CV0CV_0.

Correct
C

the charge on the upper plate of C2C_2 is 00.

D

the charge on the upper plate of C2C_2 is −CV0-CV_0.

Correct

Step-by-Step Solution

To determine the charge on the upper plates of capacitors C1C_1 and C2C_2, we follow the sequential switching operations step-by-step:

Step 1: Switch S1S_1 is closed and then opened

  • The capacitor C1C_1 is connected across the DC source of emf 2V02V_0, whose positive terminal is connected to the top wire and negative terminal to the bottom wire.
  • Capacitor C1C_1 gets fully charged to a potential difference of 2V02V_0.
  • The charge on the upper plate of C1C_1 becomes: Q1=C(2V0)=2CV0Q_1 = C(2V_0) = 2CV_0
  • When S1S_1 is opened, capacitor C1C_1 retains this charge of 2CV02CV_0 on its upper plate.

Step 2: Switch S2S_2 is closed and then opened

  • Closing S2S_2 connects the uncharged capacitor C2C_2 in parallel with C1C_1.
  • The total charge on the isolated top plates of C1C_1 and C2C_2 is conserved: Qtotal=Q1+0=2CV0Q_{\text{total}} = Q_1 + 0 = 2CV_0
  • Since both capacitors have identical capacitance CC, the common potential VcV_c is: Vc=QtotalC+C=2CV02C=V0V_c = \frac{Q_{\text{total}}}{C + C} = \frac{2CV_0}{2C} = V_0
  • The charge on each capacitor's upper plate is distributed equally: q1=CVc=CV0q_1 = C V_c = CV_0 q2=CVc=CV0q_2 = C V_c = CV_0
  • When S2S_2 is opened, capacitor C1C_1 is completely isolated from the rest of the circuit. Therefore, its charge remains unchanged at: qupper,C1=CV0q_{\text{upper}, C_1} = CV_0

Step 3: Switch S3S_3 is closed

  • In the circuit diagram, the battery of emf V0V_0 in the rightmost branch is connected with its negative terminal at the top (short bar) and positive terminal at the bottom (long bar).
  • When S3S_3 is closed, the capacitor C2C_2 is connected directly across this battery.
  • After sufficient time has elapsed, the potential of the upper plate of C2C_2 relative to its lower plate becomes −V0-V_0.
  • Consequently, the steady-state charge on the upper plate of C2C_2 is: qupper,C2=C(−V0)=−CV0q_{\text{upper}, C_2} = C(-V_0) = -CV_0

Conclusion:

  • The charge on the upper plate of C1C_1 is CV0CV_0. (Option B)
  • The charge on the upper plate of C2C_2 is −CV0-CV_0. (Option D)

Therefore, the correct options are B and D.