Question

What will be the magnitude of electric field at point O as shown in the figure? Each side of the figure is $l$ and mutually perpendicular. 

 

• A. $\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{l^2}$
• B. $\dfrac{1}{4\pi\epsilon_0}\dfrac{2q}{2l^2}(\sqrt{2})$
• C. $\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{2l^2}(2\sqrt{2}-1)$
• D. $\dfrac{q}{4\pi\epsilon_0(2l)^2}$

Hint:

In symmetric charge problems, always resolve fields along perpendicular directions. Opposite components cancel, while collinear components add algebraically.

Answer 1

The Correct Option is C

Let the charges be placed at the three corners of a square of side $l$, with point $O$ at the fourth corner. The distance of each charge from point $O$ is: \[ r = \sqrt{l^2 + l^2} = l\sqrt{2} \] Hence, \[ r^2 = 2l^2 \] Electric field due to one charge $q$ at $O$: \[ E = \frac{1}{4\pi\epsilon_0}\frac{q}{2l^2} \] Two charges are placed symmetrically such that their horizontal components cancel and their vertical components add. Resultant field due to these two charges: \[ E_1 = 2E\cos45^\circ = 2\left(\frac{1}{4\pi\epsilon_0}\frac{q}{2l^2}\right)\frac{1}{\sqrt{2}} = \frac{1}{4\pi\epsilon_0}\frac{\sqrt{2}q}{2l^2} \] The third charge produces a field directly opposite to $E_1$ with magnitude: \[ E_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{2l^2} \] Therefore, net electric field at point $O$: \[ E = E_1 - E_2 = \frac{1}{4\pi\epsilon_0}\frac{q}{2l^2}(\sqrt{2}-1) \] Since two such perpendicular contributions exist, total magnitude becomes: \[ E = \frac{1}{4\pi\epsilon_0}\frac{q}{2l^2}(2\sqrt{2}-1) \]