Chapter 21.1, Problem 9PP

To determine

Magnitude of electric field at given position.

Answer to Problem 9PP

  $E=6.5625\times {10}^3\text{N/C}$

Explanation of Solution

Given:

Point charge, $q=4.2\times {10}^{-6}\text{C}$

Distance from point charge, $d=2.4\text{m}$

Formula Used:

Strength of electric field at distance d away 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 the magnitude of electric field at d is

  $\begin{array}{c}E=\left(9\times {19}^9\right)\times \frac{4.2\times {10}^{-6}}{{2.4}^2}\text{N/C}\\ =6.5625\times {10}^3\text{N/C}\end{array}$

Conclusion:

Therefore, the magnitude of electric field strength at point d from the point charge is

  $E=6.5625\times {10}^3\text{N/C}$ .