A parallel-plate capacitor is made of two conducting plates of area A separated

Question

A parallel-plate capacitor is made of two conducting plates of area A separated by a distance d. The capacitor carriesa charge Q and is initially connected to a battery that maintains a constant potential difference between the plates. The battery is then disconnected from the plates and the separation between the plates is doubled 1) Which of the following remains constant? Voltage across the capacitor Capacitance of the capacitor Charge on the capacitor Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. (Survey Question) 2) Briefly explain your answer to the previous question. Submit You currently have 0 submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question.

Explanation / Answer

1) Charge on the capacitor.

2)The conservation of charge applies here and the charge remains conserved even on changing the separation. The capacitance changes as C = epsilon0 A/d and V = Q/C.

3)increase

4)We know that

U = 1/2 C V^2 = 1/2 Q^2/C

C = epsilon0 A/d

with increase in d C decreases.

As we learned, Q is constant. But C decreases. So E will increase since C decreased.

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