Chapter 21.1, Problem 1PP

To determine

Magnitude of electric field.

Answer to Problem 1PP

  $E=40\text{N/C}$

Explanation of Solution

Given:

Positive test charge, $q=5\times {10}^{-6}\text{C}$

Force of electric field, $F=2\times {10}^{-4}\text{N}$

Formula Used:

Strength of electric field is given by,

  $E=\frac{F}{q}$

Where, $E$ is electric field strength, $F$ is force exerted by electric field and $q$ is the test charge.

Calculation:

So, on putting the values the magnitude of electric field is

  $\begin{array}{c}E=\frac{2\times {10}^{-4}}{5\times {10}^{-6}}\text{N/C}\\ =40\text{N/C}\end{array}$

Conclusion:

Therefore, the magnitude of electric field strength at the location of positive test charge is

  $E=40\text{N/C}$ .