Question

Equal charges $Q$ are placed at the four corners $A,B,C$ and $D$ of a square of length $a$. The magnitude of the force on the charge at $B$ will be

• A. $\frac{3Q^2}{4\pi {\epsilon }_0{a}^2}$
• B. $\frac{4Q^2}{4\pi {\epsilon }_0{a}^2}$
• C. $\frac{k{Q}^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$
• D. $(2+\frac{1}{\sqrt{2}})\frac{3Q^2}{4\pi {\epsilon }_0{a}^2}$

Answer 1

The correct option is C $\frac{k{Q}^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$

$\text{Step 1: Draw the figure and mention the direction of forces at}$ $B$.$\text{[Ref. Fig.]}$

$\text{Step 2: Calculate the force experienced by B due to other charges}$

Applying Coulomb's law,

Force on $B$ due to charge at $A,F_A=\frac{1}{4\pi {\in }_0}\frac{Q^2}{a^2}$

Force on $B$ due to charge at $C,F_C=\frac{1}{4\pi {\in }_0}\frac{Q^2}{a^2}$

Force on $B$ due to charge at $D,F_D=\frac{1}{4\pi {\in }_0}\frac{Q^2}{2a^2}$

$\text{Step 3: Net resultant force on B}$

Resultant of $F_A$ and $F_C,F^{\prime }=\sqrt{F_A^2+F_C^2}$

$=\sqrt{\frac{1}{4\pi {\in }_0}\frac{Q^2}{a^2}+\frac{1}{4\pi {\in }_0}\frac{Q^2}{a^2}}$

$=\frac{\sqrt{2}}{4\pi {\in }_0}\frac{Q^2}{a^2}$

Net force at $B=F^{\prime }+F_D$ {Both are in the same direction}

$=\frac{\sqrt{2}{Q}^2}{4\pi {\in }_0{a}^2}+\frac{1}{4\pi {\in }_0}\frac{Q^2}{2a^2}$

$=\left(\frac{2\sqrt{2}+1}{2}\right)\frac{Q^2}{4\pi {\in }_0{a}^2}$ $=\left(\frac{2\sqrt{2}+1}{2}\right)\frac{K{Q}^2}{a^2}$

Hence, Option(C) is correct.