A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now.
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Answer:
Explanation:
Initially charge on the capacitor is q=CV.
Thus, positive plate of the capacitor will get charge +q and negative plate get charge −q.
As battery is disconnected so charge is constant. Thus, when charge Q is given to positive plate, the will equally distribute into two plates. So the charge on positive plate becomes q
′
=q+Q/2 and charge on negative plate −q
′
.
Now the potential difference between the plates is: V
′
=
C
q
′
=
C
q+Q/2
=
C
CV+Q/2
=V+
2C
Q
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