Chapter 21.2, Problem 22PP

To determine

Magnitude of electric field.

Answer to Problem 22PP

  $E=2\times {10}^4\text{N/C}$

Explanation of Solution

Given:

Electric potential difference between the parallel plates, $V=400\text{V}$

Distance between the plates, $d=0.02\text{m}$

Formula Used:

The electric potential difference is given as,

  $\Delta V=Ed$

Where, $\Delta V$ electric potential difference, $E$ is Electric field intensity and $d$ is distance between the parallel plates.

Calculation:

On substituting the given values,

  $\begin{array}{l}400=E\times 0.02\\ \Rightarrow E=\frac{400}{0.02}\\ \Rightarrow E=2\times {10}^4\text{N/C}\end{array}$

Conclusion:

Therefore, the magnitude of electric field between the parallel plates is,

  $E=2\times {10}^4\text{N/C}$ .