Question

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of electric field 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. $\frac{q}{2\pi\varepsilon_{0}a^{2}}$

Answer 1

Answer: (C)

$\frac{\sqrt{3}q}{4\pi\varepsilon_{0}a^{2}}$

Answer 2

$|E_{A}| = |E_{B}| = k.\frac{q}{a^{2}}$

So, $E_{net} = \sqrt{E_{A}^{2} + E_{B}^{2} + 2E_{A}E_{B}\cos 60^{o}} = \frac{\sqrt{3}k.q}{a^{2}}$

⇒ $E_{net} = \frac{\sqrt{3}q}{4\pi\varepsilon_{0}a^{2}}$