Question

Two charges of $4\mu C$ each are placed at the corners of A and B of an equilateral triangle ABC of side length $0.2m$ in air. The electric potential at C is :$(\frac{1}{4\pi {ϵ}_0}=9\times {10}^9)$

• A. $9\times {10}^{4}V$
• B. $18\times {10}^{4}V$
• C. $36\times {10}^{4}V$
• D. $72\times {10}^{4}V$

Answer 1

Answer: (C)

$36\times {10}^{4}V$

We know V $=\frac{kq}{r}$
$\therefore$ Voltage at C$=\frac{k\left(4\mu c\right)}{0.2}+\frac{k\left(4\mu \right)}{0.2}$

$=\frac{2\times (9\times {10}^9)\times 4\times {10}^{-6}}{0.2}=36\times {10}^{4}V$