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?

Question

Eshan · Answer

Dear student,

After getting charged from the cell, the positive plate attains $+CV$ charge and the negative plate attains $-CV$ . When $q$ charge 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 is $d(\dfrac{CV}{A\epsilon_0}+\dfrac{q}{2A\epsilon_0})$
$=\dfrac{CV}{A\epsilon_0}d+\dfrac{q}{2A\epsilon_0}d$
$=V+\dfrac{q}{2C}$

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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?

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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?

1 Answers

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