Question

With $V=0$ at infinity, what is the electric potential at corner A?

• A. $6\times {10}^{4}V$
• B. $-7.8\times {10}^{5}V$
• C. $8.4\times {10}^{5}V$
• D. None of these

Answer 1

The correct option is A $6\times {10}^{4}V$

Rectangle has sides $5cm$ and $15cm$.

$q_1=-5\mu C$ and $q_2=2\mu C$ as shown in the figure.

Given $V=0$ at infinity.

So, we have to find the electric potential at corner $A$.

Since point $A$ is $15cm$ away from $-5\mu C$ charge and $5cm$ away from $2\mu C$ charge.

We get, $V=\frac{1}{4\pi {\epsilon }_0}[\frac{q_1}{r_1}+\frac{q_2}{r_2}]$

$=9\times {10}^9[\frac{-5\times {10}^{-6}}{15\times {10}^{-2}}+\frac{2\times {10}^{-6}}{5\times {10}^{-2}}]$

$=6\times {10}^{4}V$