A parallel plate capacitor of capacitance C_0 has plates of area A with separati

Question

A parallel plate capacitor of capacitance C_0 has plates of area A with separation d between them. When it is connected to a battery of voltage V_0, it has charge of magnitude O_0 on its plates. The plates are pulled apart to a separation 2d while the capacitor remains connected to the battery. After the plates are 2d apart, the capacitance of the capacitor and the magnitude of the charge on the plates are a. 1/2 C_0, 1/2 Q_0 b. 1/2 C_0, Q_0 d. 2C_0, Q_0 e. 2C_0, 2Q_0 Determine the equivalent capacitance of the combination shown when C = 12 pF.

Explanation / Answer

capacitance=epsilon*area/distance between plates

charge=capacitance*voltage

Q1. Q0=C0*V0

when plates are pulled apart, capacitance will be halved to C0/2

as voltage is still connected, charge on plates=C0*V0/2=Q0/2

so option a is correct.

Q2.to the right side, C and C are in parallel.

net capacitance=C+C=2*C

so C,2*C and 2*C are in series.

if equivalent capacitance is C_net,

then 1/C_net=(1/C)+(1/2C)+(1/2C)=2/C

==>C_net=C/2=12 pF/2=6 pF

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