Chapter 21.2, Problem 27PP

To determine

Magnitude of electric field.

Answer to Problem 27PP

  $E=2.083\times {10}^4\text{N/C}$

Explanation of Solution

Given:

The given figure is,

  

Electric potential difference between the two parallel plates, $\Delta V=500\text{V}$

Distance between the plates, $d=2.4\text{cm}=0.024\text{m}$

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,

  $\begin{array}{l}500=E\times 0.024\\ \Rightarrow E=\frac{500}{0.024}\\ \Rightarrow E=2.083\times {10}^4\text{N/C}\end{array}$

Conclusion:

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

  $E=2.083\times {10}^4\text{N/C}$ .