Chapter 21, Problem 84A

To determine

Electrical energy stored.

Answer to Problem 84A

  $PE=0.45\text{J}$

Explanation of Solution

Given:

Capacitance of capacitor, $C=10\mu \text{F}$

Electrical potential difference, $\Delta V=3\times {10}^2\text{V}$

Formula Used:

The potential energy of the capacitor is,

  $PE=\frac{1}{2}C\Delta V^2$

Where, $C$ is capacitance and $\Delta V$ is electrical potential difference.

Calculation:

On substituting the values,

  $\begin{array}{c}PE=\frac{1}{2}\times 10\times {10}^{-6}\times {\left(3\times {10}^2\right)}^2\\ =0.45\text{J}\end{array}$

Conclusion:

Therefore, the electrical energy stored in capacitor is $PE=0.45\text{J}$ .