Question

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

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

Answer 1

Answer: (C)

$\sqrt{3}\,k \frac{q^{2}}{r^{2}}$

Answer 2

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 = $\sqrt{3} \frac{kq^{2}}{r^{2}}$