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Correct option is B)

$Q_{i}=C_{i}V$

$U_{i}=21βC_{i}V_{2}$

$U_{f}=21βC_{f}Q_{f}β$

After the capacitor is disconnected from the battery, charge on the capacitor cannot change.

$β΄Q_{f}=Q_{i}$

$β΄U_{f}=21βC_{f}Q_{f}β=21βKC_{i}C_{i}V_{2}β$

$ΞU=U_{f}βU_{i}=21βKC_{i}C_{i}V_{2}ββ21βC_{i}V_{2}=21βC_{i}V_{2}(K1ββ1)=21βCV_{2}(K1ββ1)$ since $C_{i}=C$ as given

Option(A) is correct.

Option (B) :

Charge on the capacitor remains same since battery is disconnected.

Option (C):

C increases K times.

Since $Q=CV$ and Q remains same, V decreases K times.

Option D:

The energy stored in the capacitor $U_{f}$ decreases K times.

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