Question

Three identical charges are placed at the vertices of an equilateral triangle. The force experienced by each charge, (if k = 1/4$\pi {\epsilon }_0$) is

• A. $2k\frac{q^2}{r^2}$
• B. $\frac{k{q}^2}{2r^2}$
• C. $\sqrt{3}k\frac{q^2}{r^2}$
• D. $\frac{k{q}^2}{\sqrt{2}{r}^2}$

Answer 1

Answer: (C)

Force on charge q at A:
$\left(i\right)$ Due to B, f = kq × $\frac{q}{r2}$
or $f=\frac{k{q}^2}{r^2}$ along BA
$\left(ii\right)$ Due to C, $f=\frac{k{q}^2}{r^2}$ along CA
$\therefore$ Resultant force = F
$F^2=f^2+f^2+2f\times f\times cos60^{\circ }$
$F^2=2f^2+\frac{2f^2\times 1}{2}=3f^2$ or $F=\sqrt{3}f$
or Resultant force = $\sqrt{3}\frac{k{q}^2}{r^2}$