Chapter 21.1, Problem 12PP

To determine

Distance of test charge from point charge.

Answer to Problem 12PP

  $d=7.746\text{m}$

Explanation of Solution

Given:

Point charge, $q=2.4\times {10}^{-6}\text{C}$

Strength of electric field,

Formula Used:

Strength of electric field at distance d from point charge is given by,

  $E=K\frac{q}{d^2}$

Where, $E$ is electric field strength, $K$ is Coulomb’s constant, $q$ is the point charge and $d$ is distance from the point charge.

Calculation:

Since, $K=9\times {10}^9{\text{Nm}}^2/{\text{C}}^{\text{2}}$

So, on putting the values,

  $\begin{array}{c}360=\left(9\times {19}^9\right)\times \frac{2.4\times {10}^{-6}}{d^2}\\ d^2=\frac{\left(9\times {19}^9\right)\times 2.4\times {10}^{-6}}{360}\\ d=\sqrt{\frac{\left(9\times {19}^9\right)\times 2.4\times {10}^{-6}}{360}}\text{m}\\ =\text{7}\text{.746}\text{m}\end{array}$

Conclusion:

Therefore, the distance of test charge from point charge is $d=7.746\text{m}$ .