Question

Two electric charges $12\mu C$ and $-6\mu C$ are placed 20 cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from $-6\mu C$ charge is

• A. 0.10 m
• B. 0.15 m
• C. 0.20 m
• D. 0.25 m

Answer 1

The correct option is C 0.20 m

Point P will lie near the charge which is smaller in magnitude i.e. $-6\mu C$. Hence potential at P

$V=\frac{1}{4\pi {ϵ}_0}\frac{(-6\times {10}^{-6})}{x}+\frac{1}{4\pi {ϵ}_0}\frac{(12\times {10}^{-6})}{(0.2+x)}=0\Rightarrow x=0.2m$