Chapter 21.1, Problem 17SSC

To determine

Electric field at the position of test charge.

Answer to Problem 17SSC

  $E=6.25\times {10}^4\text{N/C}$ towards east.

Explanation of Solution

Given:

Positive test charge, $q=2.4\times {10}^{-8}\text{C}$

Force experienced by charge, $F=1.5\times {10}^{-3}\text{N}$ towards east.

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 at test charge is

  $\begin{array}{l}E=\frac{1.5\times {10}^{-3}}{2.4\times {10}^{-8}}\text{N/C}\\ =6.25\times {10}^4\text{N/C}\end{array}$

Since, the charge is positive so the electric field is also directed towards east.

Conclusion:

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

  $E=6.25\times {10}^4\text{N/C}$ and it is directed towards east.