Question

Four charges +q, -q, +q and- q are placed in order on the four consecutive corners of a square of side a.The work done in interchanging the positions of any two neighbouring charges of the opposite sign is :

• A. $\frac{q^2}{4\pi {\epsilon }_{0}a}(-4+\sqrt{2})$
• B. $\frac{q^2}{4\pi {\epsilon }_{0}a}(4+2\sqrt{2})$
• C. $\frac{q^2}{4\pi {\epsilon }_{0}a}(4-2\sqrt{2})$
• D. $\frac{q^2}{4\pi {\epsilon }_{0}a}(4+\sqrt{2})$

Answer 1

Answer: (C)

$\frac{q^2}{4\pi {\epsilon }_{0}a}(4-2\sqrt{2})$

For square , $AB=BC=CD=AD=a$ and $AC=BD=\sqrt{a^2+a^2}=\sqrt{2}a$

Initial potential energy of the system is

$U_i=k[\frac{-q^2}{AB}+\frac{-q^2}{BC}+\frac{-q^2}{CD}+\frac{-q^2}{AD}+\frac{q^2}{AC}+\frac{q^2}{BD}]$

$=k[\frac{-q^2}{a}+\frac{-q^2}{a}+\frac{-q^2}{a}+\frac{-q^2}{a}+\frac{q^2}{\sqrt{2}a}+\frac{q^2}{\sqrt{2}a}]$

$=\frac{k{q}^2}{a}[\sqrt{2}-4]$

After interchanging the charges in between B and C ,the final potential energy is

$U_f=k[\frac{q^2}{AB}+\frac{-q^2}{BC}+\frac{q^2}{CD}+\frac{-q^2}{AD}+\frac{-q^2}{AC}+\frac{-q^2}{BD}]$

$=k[\frac{q^2}{a}+\frac{-q^2}{a}+\frac{q^2}{a}+\frac{-q^2}{a}+\frac{-q^2}{\sqrt{2}a}+\frac{-q^2}{\sqrt{2}a}]$

$=\frac{k{q}^2}{a}(-\sqrt{2})$

Work done $=$ change in potential energy$=U_f-U_i=\frac{k{q}^2}{a}\left[\right(-\sqrt{2})-\sqrt{2}+4]=\frac{q^2}{4\pi {ϵ}_{0}a}[4-2\sqrt{2}]$