91Ó°ÊÓ

The capacitor is a charge storage device. The capacitance of a capacitor depends upon the value of the charge on each plate and the potential difference between the plates.

The charge stored in a capacitor is given by,

\(Q = CV\) … (i)

Here, Q is the charge stored, C is the capacitance and V is the potential difference between the plates.

The increase in charge on the capacitor is,\({Q_2} - {Q_1} = 15\;{\rm{\mu C}}\).

The initial voltage is,\({V_1} = 97\;{\rm{V}}\).

The final voltage is, \({V_2} = 121\;{\rm{V}}\).

From equation (i), the increase in charge is given by,

\(\begin{aligned}{Q_2} - {Q_1} &= C{V_2} - C{V_1}\\C &= \frac{{{Q_2} - {Q_1}}}{{{V_2} - {V_1}}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}C &= \frac{{15\;{\rm{\mu C}} \times \frac{{{{10}^{ - 6}}\;{\rm{C}}}}{{1\;{\rm{\mu C}}}}}}{{121\;{\rm{V}} - 97\;{\rm{V}}}}\\C &= 6.25 \times {10^{ - 7}}\;{\rm{F}}\end{aligned}\)

Thus, the capacitance of the capacitor is \(6.25 \times {10^{ - 7}}\;{\rm{F}}\).