Question

A capacitor of capacitance $\mathrm{C}$ is charged to a potential difference $\mathrm{V}$ from a cell and then disconnected from it. A charge $+\mathrm{Q}$ is now given to its positive plate. The potential difference across the capacitor is now _________.

Answer 1

Step 1: Given data

The capacitance of the capacitor $=\mathrm{C}$

The potential difference of a cell $=\mathrm{V}$

The charge given on the positive plate $=+\mathrm{Q}$

We have to find the potential difference across the capacitor.

Step 2: Formula to be used.

Let us consider that the initial charge on the capacitor is ${\mathrm{Q}}_{\circ }$.

So, the charge on a parallel plate capacitor is capacitance times the potential difference is,

${\mathrm{Q}}_{\circ }=\mathrm{CV}$

Now, a charge $\mathrm{Q}$ is given to the positive plate and the new potential is $\mathrm{V}\text{'}$. We get,

$\mathrm{V}\text{'}=\mathrm{Ed}$

$=\left({\mathrm{E}}_1+{\mathrm{E}}_2\right)\mathrm{d}$

Where, ${\mathrm{E}}_1$ and ${\mathrm{E}}_2$ are the field in the gap due to charges $\mathrm{Q}+{\mathrm{Q}}_{\circ }$ and $-{\mathrm{Q}}_{\circ }$ respectively.

Step 3: Find the potential difference across the capacitor.

So, the electric fields are,

${\mathrm{E}}_1=\frac{\mathrm{Q}+{\mathrm{Q}}_{\circ }}{2{\mathrm{\epsilon }}_{\circ }\mathrm{A}}$,

And

${\mathrm{E}}_2=\frac{{\mathrm{Q}}_{\circ }}{2{\mathrm{\epsilon }}_{\circ }\mathrm{A}}$

We get,

$\mathrm{V}\text{'}=\left(\frac{\mathrm{Q}+{\mathrm{Q}}_{\circ }}{2{\mathrm{\epsilon }}_{\circ }\mathrm{A}}+\frac{{\mathrm{Q}}_{\circ }}{2{\mathrm{\epsilon }}_{\circ }\mathrm{A}}\right)\mathrm{d}$

$=\frac{\mathrm{Qd}}{2{\mathrm{\epsilon }}_{\circ }\mathrm{A}}+\frac{{\mathrm{Q}}_{\circ }\mathrm{d}}{{\mathrm{\epsilon }}_{\circ }\mathrm{A}}$

$=\frac{\mathrm{Q}}{2\mathrm{C}}+\mathrm{V}$

So, the potential difference across the capacitor is $\frac{\mathrm{Q}}{2\mathrm{C}}+\mathrm{V}$.

Hence, A capacitor of capacitance $\mathrm{C}$ is charged to a potential difference $\mathrm{V}$ from a cell and then disconnected from it. A charge $+\mathrm{Q}$ is now given to its positive plate. The potential difference across the capacitor is now $\frac{\mathrm{Q}}{2\mathrm{C}}+\mathrm{V}$.