(I) An 8500-pF capacitor holds plus and minus charges of \({\bf{16}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\;{\bf{C}}\). What is the voltage across the capacitor?

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 capacitance is,\(C = 8500\;{\rm{pF}}\).

The charge on the capacitor is, \(Q = 16.5 \times {10^{ - 8}}\;{\rm{C}}\)

From equation (i), the voltage across the capacitor is given by,

\(V = \frac{Q}{C}\)

Substitute the values in the above expression.

\(\begin{aligned}V &= \frac{{16.5 \times {{10}^{ - 8}}\;{\rm{C}}}}{{8500\;{\rm{pF}} \times \frac{{{{10}^{ - 12}}\;{\rm{F}}}}{{1\;{\rm{pF}}}}}}\\V &= 19.4\;{\rm{V}}\end{aligned}\)

Thus, the voltage across the capacitor is \(19.4\;{\rm{V}}\).