Question

A dielectric is inserted between the plates of a changed and isolated parallel plate air capacitor. Then the change on the plates and the potential difference respectively....... a)increases, remains the same b) remains the same, increased c) remains the same, decrease d)decreases, remains the same​

Answer 1

Answer:

hope it will help you

Explanation:

(i) The capacitance increases as the dielectric constant K>1.

(ii) Potential difference V=

C

Q

. As C increases and Q remains the same since the battery is disconnected, the p.d. between the plates decreases.

(iii) Electric field E=

d

V

where V is the p.d. and d the separation between the plates. As V decreases and d remains the same, electric field also decreases.

(iv) Energy stored in a capacitor U=

2

1

C

Q

2

. As Q is constant and C increases, U decreases.