Question

A charge of 10 $\mu C$ is placed at the origin of (x-y) coordinate system. The potential difference between two points $(0,a)$ and $(a,0)$ in volts is

• A. $9\times {10}^4/a$
• B. $9\times {10}^4/a\sqrt{2}$
• C. $9\times {10}^{4}2a$
• D. $0$

Answer 1

The correct option is D $0$

Given that,

Charge, $q=10\times {10}^{-6}C$

Two points $=(0,a)(a,0)$

A charge is placed at the origin of $x-y$ coordinate system. The potentail difference on $A$ is,

$V_A=\frac{1}{4\pi {ϵ}_0}\times \frac{q}{r}$

$V_A=\frac{10\times {10}^{-6}\times 9\times {10}^9}{a}$

The potential difference on $B$ is

$V_B=\frac{10\times {10}^{-6}\times 9\times {10}^9}{a}$

The potential difference between two points is,

$V_{AB}=V_A-V_B$

$V_{AB}=\frac{10\times {10}^{-6}\times 9\times {10}^9}{a}-\frac{10\times {10}^{-6}\times 9\times {10}^9}{a}$

$⟹V_{AB}=0$

Hence, the potential difference two points is zero.