Question

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

Short answer

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

Step-by-step solution

Understanding the energy stored in a capacitor

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.

Given data

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

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

Determination of stored energy in capacitor

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 \(5.9 \times {10^{ - 4}}{\rm{J}}\).