Question

Four electric charges $+q,+q,-q$ and $-q$ are placed at the corners of a square of side $2L$ (see figure). The electric potential at point $A$ (midway between the two charges $+q$ and $-q$ ) is

• A. $\frac{1}{4\pi {\in }_0}\frac{2q}{L}(1+\sqrt{5})$
• B. $\frac{1}{4\pi {\in }_0}\frac{2q}{L}(1+\frac{1}{\sqrt{5}})$
• C. $\frac{1}{4\pi {\in }_0}\frac{2q}{L}(1-\frac{1}{\sqrt{5}})$
• D. Zero

Answer 1

The correct option is C $\frac{1}{4\pi {\in }_0}\frac{2q}{L}(1-\frac{1}{\sqrt{5}})$


Let $V_A=$ potential at $A$ due to all charges
$V_{MA}=$ potential at $A$ due to charge at $M$
$V_{NA}=$ potential at $A$ due to charge at $N$
$V_{OA}=$ potential at $A$ due to charge at $O$
$V_{PA}=$ potential at $A$ due to charge at $P$
Distance between $N$ and $A$ $=$ Distance between $O$ and $A$ $=$ $\sqrt{(2L{)}^2+L^2}=L\sqrt{5}$
$V_A=V_{MA}+V_{PA}+V_{NA}+V_{OA}$
$\Rightarrow V_A=\frac{kq}{L}+\frac{kq}{L}-\frac{kq}{L\sqrt{5}}-\frac{kq}{L\sqrt{5}}\Rightarrow V_A=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{L}(1-\frac{1}{\sqrt{5}})$