Question: (I) 650 V is applied to a 2800-pF capacitor. How much energy is stored?

Capacitor is an energy storage device. The stored energy is proportional to the square of the potential difference between the plates.

The expression for stored energy in the capacitor is given as:

\({\rm{PE}} = \frac{1}{2}C{V^2}\)

Here, C is the capacitance and V is the potential difference.

The potential difference between the plates is,\(V = 650\;{\rm{V}}\).

The capacitance of the capacitor is, \(C = 2800\;{\rm{pF}} = 2800 \times {10^{ - 12}}\;{\rm{F}}\).

The stored energy in the capacitor is,

\({\rm{PE}} = \frac{1}{2}C{V^2}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{\rm{PE}} &= \frac{1}{2} \times \left( {2800 \times {{10}^{ - 12}}\;{\rm{F}}} \right) \times {\left( {650\;{\rm{V}}} \right)^2}\\ &= 5.9 \times {10^{ - 4}}{\rm{J}}\end{aligned}\)

Hence, the energy stored in the capacitor is \(5.9 \times {10^{ - 4}}{\rm{J}}\).