Question

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

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

Answer 1

The correct option is D $\frac{\sqrt{3}q}{4\pi {\epsilon }_0{\alpha }^2}$

Let $E_{A}and{E}_B$ be the electric field due the the charges placed at point A and B respectively then direction of electric field is as shown:

$|E_A|=|E_B|=k.\frac{q}{{\alpha }^2}$




So,$E_{net}=\sqrt{E_A^2+E_B^2+2E_A{E}_{B}cos{60}^0}$

$=\frac{\sqrt{3}kq}{{\alpha }^2}=\frac{\sqrt{3}q}{4\pi {\epsilon }_0{\alpha }^2}$

Hence, option (c) is correct.