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A capacitor of capacitance c is charged to a potential difference v from a cell and then disconnected. A charge +q is given to it`s positive plate .the potential difference across the capacitor is now?

A capacitor of capacitance c is charged to a potential difference v from a cell and then disconnected. A charge +q is given to it`s positive plate .the potential difference across the capacitor is now?

Grade:12

1 Answers

Eshan
askIITians Faculty 2095 Points
3 years ago
Dear student,

After getting charged from the cell, the positive plate attains+CVcharge and the negative plate attains-CV. Whenqcharge has been supplied to the positive plate, the electric field between the two plate increases.

The contribution to electric field due to this extra charge is \dfrac{q}{2A\epsilon_0}. So the field between the plates now becomes\dfrac{CV}{A\epsilon_0}+\dfrac{q}{2A\epsilon_0}
Hence the potential drop between the plates isd(\dfrac{CV}{A\epsilon_0}+\dfrac{q}{2A\epsilon_0})
=\dfrac{CV}{A\epsilon_0}d+\dfrac{q}{2A\epsilon_0}d
=V+\dfrac{q}{2C}

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