Chapter 21.2, Problem 46SSC

To determine

The magnitude of net charge on each conductor plate of a $0.47\mu \text{F}$ .

Answer to Problem 46SSC

  $5.64\times {10}^{-6}\text{}\text{C}$

Explanation of Solution

Given:

The capacitance of the capacitor, $C=0.47\mu \text{F}$

The potential applied across the capacitor, $\Delta V=12\text{V}$

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 charge on the capacitor will be,

  $\begin{array}{c}q=C\Delta V\\ =\left(0.47\times {10}^{-6}\right)\left(12\right)\\ =5.64\times {10}^{-6}\text{}\text{C}\end{array}$

Conclusion:

Hence, the charge on the capacitor will be $5.64\times {10}^{-6}\text{}\text{C}$ .