Chapter 21.1, Problem 15PP

To determine

Magnitude of electric field at opposite corner of square.

Answer to Problem 15PP

  $E=6.364\times {10}^3\text{N/C}$

Explanation of Solution

Given:

Charge of small sphere at one corner of square, $q=5\times {10}^{-6}\text{C}$

Side of square is 5 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, $K=9\times {10}^9{\text{Nm}}^2/{\text{C}}^{\text{2}}$

And the distance of opposite corner of the square is it’s diagonal.

So,

  $\begin{array}{l}d=\sqrt{5^2+5^2}\\ =7.071\text{m}\end{array}$

Therefore, the magnitude of electric field at opposite corner of the square is,

  $\begin{array}{l}E=\left(9\times {19}^9\right)\times \frac{5\times {10}^{-6}}{{7.071}^2}\text{N/C}\\ =6.364\times {10}^3\text{N/C}\end{array}$

Conclusion:

Therefore, the magnitude of electric field strength at the opposite corner of the square is

  $E=6.364\times {10}^3\text{N/C}$ .