Question

Equal charges $q$ each are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of electric intensity at the point $C$ is

• A. $\frac{q}{4\pi {\epsilon }_0{a}^2}$
• B. $\frac{\sqrt{2}q}{4\pi {\epsilon }_0{a}^2}$
• C. $\frac{\sqrt{3}q}{4\pi {\epsilon }_0{a}^2}$
• D. $\frac{2q}{4\pi {\epsilon }_0{a}^2}$

Answer 1

Answer: (C)

The situation is as shown in the figure.

Electric field intensity at $C$ due to charge $q$ at $A$ is
$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}$ along $AC$
Electric field intensity at C due to charge $q$ at $B$ is
$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}$ along $BC$
As $E_1=E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}$
$E_1$ and $E_2$ are inclined at angle $60^{\circ }$, therefore the resultant electric field intensity at $C$ is
$E=\sqrt{E_1^2+E_2^2+2E_1{E}_{2}cos60^{\circ }}$
$=\sqrt{E_1^2+E_1^2+2E_1{E}_{2}cos60^{\circ }}$
$=\sqrt{E_1^2+E_1^2+2E_1^{2}cos60^{\circ }}$
$=E_1\sqrt{3}$
$=\frac{1}{4\pi {\epsilon }_0}\frac{q\sqrt{3}}{a^2}$