Question

A parallel plate capacitor of capacitance
100 uF is charged to 500 V. The plate
separation is then reduced to half its original
value. Then
(A) the potential on the capacitor becomes
1000 V
(B) the potential on the capacitor becomes
250 V
(©) the change in stored energy is
3.75x10*9
(D) the change in stored energy is
6.25x10minus6]​

Answer 1

Answer:

Q=CV

C=εA

d

Qinitial = Qfinal = Q = 5×10-²C

Here,the capacitance becomes twice. Since the capacitor is not connected to the battery the charge will remain same. The potential difference must decrease to maintain the same charge.

Qfinal= 2C×V'

5×10-² = 200×10-⁶ ×V'

=> V' = 250V

U = Q²/2(1/V²–1/V'²) = 6.25J

So, the answer is B

Answer 2

Answer:

your answer

Explanation:

hope it's help you XD