Q. Equal charges $q$ are placed at the four corners $A,B,C,D$ of a square of length $a$. The magnitude of the force on the charge at $B$ will be
A) $\frac{3q^{2}}{4\pi\varepsilon_{0}a^{2}}$
B) $\frac{4q^{2}}{4\pi\varepsilon_{0}a^{2}}$
C) $\left( \frac{1 + 2\sqrt{2}}{2} \right)\frac{q^{2}}{4\pi\varepsilon_{0}a^{2}}$
D) $\left( 2 + \frac{1}{\sqrt{2}} \right)\frac{q^{2}}{4\pi\varepsilon_{0}a^{2}}$
Explanation
(c) $F_{net} = F_{AC} + F_{D} = \sqrt{F_{A}^{2} + F_{C}^{2}} + F_{D}$
Since $F_{A} = F_{C} = \frac{kq^{2}}{a^{2}}$ and $F_{D} = \frac{kq^{2}}{\left( a\sqrt{2} \right)^{2}}$
$F_{net} = \frac{\sqrt{2}kq^{2}}{a^{2}} + \frac{kq^{2}}{2a^{2}} = \frac{kq^{2}}{a^{2}}\left( \sqrt{2} + \frac{1}{2} \right) = \frac{q^{2}}{4\pi\varepsilon_{0}a^{2}}\left( \frac{1 + 2\sqrt{2}}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.