Question

A parallel plate capacitor has area a and plate separation d

Answer 1

Answer:

Part a)

For a parallel plate capacitor of plate area A and plate separation d,

The capacitance C1=ϵ0Ad

The capacitor is charged to a potential difference V1

So total charge stored in the capacitor Q1=C1V1

When the plate separation is doubled the new capacitance C2=ϵ0A2d=C12

Since the charge is not disturbed total charge remains constant. So the potential difference across the capacitor plates changes to Vf

So Q1=C1V1=C2Vf⟹C1V1=C12Vf

So the new potential difference Vf=2V1

Part b)

Let V be the potential difference applied across the capacitor plates.

Then the work done in giving a charge dq to the capacitor dW=Vdq

So for supplying Q amount of energy to the capacitor we need to integrate the expression from 0 to Q

So the work done W=∫Q0Vdq=∫Q0qCdq=Q22C=12CV2

Initial stored energy E1=12C1V21

Part c)

Final energy stored in the capacitor E2=12C2V22=12×C12×(2V1)2=C1V21

Part d)

Let the positive plate be the fixed plate.

The electric field due to the fixed plate near the movable plate is E=−σ2ϵ0=−Q12ϵ0A

So force acting on the negative plate F=Q212ϵ0A

Now the plate separation increased from d to 2d

So the work done W=Q212ϵ0A×(2d−d)=Q212×(1C2−1C1)=Q212×(2C1−1C1)=Q212C1=12C1V21

PLZ FOLLOW ME

Explanation:

Answer 2

Answer:

Part a)

For a parallel plate capacitor of plate area A and plate separation d,

The capacitance C1=ϵ0Ad

The capacitor is charged to a potential difference V1

So total charge stored in the capacitor Q1=C1V1

When the plate separation is doubled the new capacitance C2=ϵ0A2d=C12

Since the charge is not disturbed total charge remains constant. So the potential difference across the capacitor plates changes to Vf

So Q1=C1V1=C2Vf⟹C1V1=C12Vf

So the new potential difference Vf=2V1

Part b)

Let V be the potential difference applied across the capacitor plates.

Then the work done in giving a charge dq to the capacitor dW=Vdq

So for supplying Q amount of energy to the capacitor we need to integrate the expression from 0 to Q

So the work done W=∫Q0Vdq=∫Q0qCdq=Q22C=12CV2

Initial stored energy E1=12C1V21

Part c)

Final energy stored in the capacitor E2=12C2V22=12×C12×(2V1)2=C1V21

Part d)

Let the positive plate be the fixed plate.

The electric field due to the fixed plate near the movable plate is E=−σ2ϵ0=−Q12ϵ0A

So force acting on the negative plate F=Q212ϵ0A

Now the plate separation increased from d to 2d

So the work done W=Q212ϵ0A×(2d−d)=Q212×(1C2−1C1)=Q212×(2C1−1C1)=Q212C1=12C1V21

PLZ FOLLOW ME

Explanation: