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 A 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

As shown in figure Net electric field at A

$E = \sqrt { E _ { B } ^ { 2 } + E _ { C } ^ { 2 } + 2 E _ { B } E _ { C } \cos 60 }$ $E_{B} = E_{C} = \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{a^{2}}$

So, $E = \frac{\sqrt{3}Q}{4\pi\varepsilon_{0}a^{2}}$