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. $\frac{q}{4\pi\varepsilon_{0} a^{2}}$
• B. $\frac{\sqrt{2}q}{4\pi\varepsilon_{0} a^{2}}$
• C. $\frac{\sqrt{3}q}{4\pi\varepsilon_{0} a^{2}}$
• D. $\frac{2q}{4\pi\varepsilon_{0} a^{2}}$

Answer 1

Answer: (C)

The situation is as shown in the figure.

Electric field intensity at $C$ due to charge $q$ at $A$ is
$E_{1}=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{a^{2}}$ along $AC$
Electric field intensity at C due to charge $q$ at $B$ is
$E_{2}=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{a^{2}}$ along $BC$
As $E_{1}=E_{2}=\frac{1}{4\pi\varepsilon_{0}} \frac{q}{a^{2}}$
$E_{1}$ and $E_{2}$ are inclined at angle $60^{\circ}$, therefore the resultant electric field intensity at $C$ is
$E=\sqrt{E^{2}_{1}+E^{2}_{2}+2E_{1}E_{2}\,cos\,60^{\circ}}$
$=\sqrt{E^{2}_{1}+E^{2}_{1}+2E_{1}E_{2}\,cos\,60^{\circ}}$
$=\sqrt{E^{2}_{1}+E^{2}_{1}+2E^{2}_{1}\,cos\,60^{\circ}}$
$=E_{1}\sqrt{3}$
$=\frac{1}{4\pi\varepsilon_{0}} \frac{q\sqrt{3}}{a^{2}}$