Chapter 21.2, Problem 21PP

To determine

Electric potential difference.

Answer to Problem 21PP

  $\Delta V=300\text{V}$

Explanation of Solution

Given:

Electric field intensity between two charged parallel plates, $E=6000\text{N/C}$

Distance between the plates, $d=0.05\text{m}$

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,

  $\begin{array}{l}\Delta V=6000\times 0.05\\ =300\text{V}\end{array}$

Conclusion:

Therefore, the electric potential difference between the parallel plates is,

  $\Delta V=300\text{V}$ .