A parallel plate capacitor with plates of area A and plate separation d is charg

Question

A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to d/2, what happens to its capacitance?

The capacitance is one half of its original value.

The capacitance is eight times its original value.

The capacitance is twice its original value.

The capacitance is four times its original value.

The capacitance is unchanged.

2. What happens to the potential difference between the plates ?

3. The magnitude of the charge on the plates of an isolated parallel plate capacitor is doubled. Which one of the following statements is true concerning the capacitance of this parallel-plate system?

The capacitance is one half of its original value.

The capacitance is eight times its original value.

The capacitance is twice its original value.

The capacitance is four times its original value.

The capacitance is unchanged.

2. What happens to the potential difference between the plates ?

3. The magnitude of the charge on the plates of an isolated parallel plate capacitor is doubled. Which one of the following statements is true concerning the capacitance of this parallel-plate system?

Explanation / Answer

C = e0 A / d

now d -> d/2

hence C -> 2C

Ans:Capacitance is twice the original value.

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2 . As capacitor is isolated hence charge on plates will remain constant.

Q = C V

C - > 2C

so V -> V/2

Potential difference is one half of its original value.

---------------

3. Q = C V

charge is doubled then potential diff will also get doubled.

(and there is no change in C until area of plate or separation is changed )

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