Chapter 21.2, Problem 25PP

To determine

Magnitude of electric field.

Answer to Problem 25PP

  $E=7.86\times {10}^4\text{N/C}$

Explanation of Solution

Given:

Electric potential difference between the two parallel plates, $V=275\text{V}$

Distance between the plates, $d=0.35\text{cm}=3.5\times {10}^{-3}\text{m}$

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,

  $\begin{array}{l}275=E\times 3.5\times {10}^{-3}\\ \Rightarrow E=\frac{275}{3.5\times {10}^{-3}}\\ \Rightarrow E=7.86\times {10}^4\text{N/C}\end{array}$

Conclusion:

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

  $E=7.86\times {10}^4\text{N/C}$ .