Chapter 21, Problem 100A

To determine

The work it needed to move an additional charge between the capacitor plates.

Answer to Problem 100A

  $W=1.2\times {\text{10}}^{-6}\text{ J}$

Explanation of Solution

Given:

The charge is $q=0.010\mu \text{C = 1}\text{.0}\times {\text{10}}^{-8}\text{ C}$

Potential difference between the capacitor plates is $\Delta \text{V = 120 V}$

Formula used:

Potential difference $\left(\Delta V\right)$ , work or energy $\left(W\right)$ and charge in a capacitor $\left(q\right)$ are related as,

  $\Delta V\text{ = }\frac{W}{q}$

  $\Rightarrow W=q\Delta V$ $\left(1\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $W=\left(\text{1}\text{.0}\times {\text{10}}^{-8}\text{ C}\right)\left(\text{120 V}\right)$

  $=1.2\times {\text{10}}^{-6}\text{ J}$

Conclusion:

The work it needed to move an additional $0.010\mu \text{C}$ charge between the capacitor plates at $\text{120 V}$ is $1.2\times {\text{10}}^{-6}\text{ J}$ .