Chapter 21.2, Problem 40PP

To determine

The capacitance of the capacitor.

Answer to Problem 40PP

  $1.0\times {10}^{-5}\text{F}$

Explanation of Solution

Given:

The potential difference across the capacitor, $\Delta V=\left(14.5-12\right)\text{V}=2.5\text{V}$

The net charge on each plate of the capacitor, $q=2.5\times {10}^{-5}\text{}\text{C}$

Formula used:

The capacitance is given by,

  $C=\frac{q}{\Delta V}$

Where,

  $\Delta V$ is the potential difference and q is the charge stored in the capacitor.

Calculation:

Using the above formula, the capacitance of the capacitor will be,

  $\begin{array}{c}C=\frac{q}{\Delta V}\\ =\frac{2.5\times {10}^{-5}}{2.5}\\ =1.0\times {10}^{-5}\text{F}\end{array}$

Conclusion:

Hence, the capacitance of the capacitor is $1.0\times {10}^{-5}\text{F}$ .