Q. Three identical charges are placed at the vertices of an equilateral triangle. The force experienced by each charge, (if k = 1/4$\pi\varepsilon_{0}$) is
Solution:
Force on charge q at A:
$\left(i\right) \quad$ Due to B, f = kq × $\frac{q}{r2}$
or $f=\frac{kq^{2}}{r^{2}}$ along BA
$\left(ii\right)\quad$ Due to C, $f=\frac{kq^{2}}{r^{2}}$ along CA
$\therefore \quad$ Resultant force = F
$F^{2}=f^{ 2}+f^{ 2}+2f \times f \times cos 60^{\circ}$
$F^{2}=2f^{ 2}+\frac{2f^{ 2}\times1}{2}=3f^{ 2}$ or $\quad F=\sqrt{3}f$
or$\quad$ Resultant force = $\sqrt{3} \frac{kq^{2}}{r^{2}}$
