Chapter 21.1, Problem 2PP

To determine

Magnitude and direction of electric field.

Answer to Problem 2PP

  $E=3\times {10}^6\text{N/C}$ and directed to left.

Explanation of Solution

Given:

Negative test charge, $q=2\times {10}^{-8}\text{C}$

Force experienced by charge, $F=0.06\text{N}$ to the right in electric field

Formula Used:

Strength of electric field is given by,

  $E=\frac{F}{q}$

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

Calculation:

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

  $\begin{array}{c}E=\frac{0.06}{2\times {10}^{-8}}\text{N/C}\\ =3\times {10}^6\text{N/C}\end{array}$

Since, the charge is negative so the electric field is directed to left.

Conclusion:

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

  $E=3\times {10}^6\text{N/C}$ and it is directed to left.