Question

An electric charge ${10}^{-3}\mu C$ is placed at the origin $(0,0)$ of $(x,y)$ co-ordinate system. Two points $A$ and $B$ are situated at $(\sqrt{2},\sqrt{2})$ and $(2,0)$ respectively. The potential difference between the points $A$ and $B$ will be:

• A. $4.5V$
• B. $9V$
• C. $zero$
• D. $2V$

Answer 1

Answer: (C)

Given : $q={10}^{-3}\mu C={10}^{-9}C$

Distance of A from origin $r_A=\sqrt{(\sqrt{2}-0{)}^2+(\sqrt{2}-0{)}^2}=2$

Distance of B from origin $r_B=2$

Thus potential due to charge at point A $V_A=\frac{Kq}{r_A}$ where $K=\frac{1}{4\pi {ϵ}_o}=9\times {10}^9$

$\therefore$ $V_A=\frac{9\times {10}^9\times {10}^{-9}}{2}=4.5$ V

Potential due to the charge at B

$V_B=\frac{Kq}{r_B}=\frac{9\times {10}^9\times {10}^{-9}}{2}=4.5$ V

Potential difference between A and B

$V_{AB}=V_A-V_B=0$