Question

A charge of 10 micro coulomb is placed at the origin of x-y coordinate system. The potential difference b/w two points (0,a) (a,0) in volt will be?

Answer 1

Answer:

The potential difference between two points will be zero.

Explanation:

Given that,

Charge [tex[q = 10\times10^{-6}[/tex]

First point = (0,a)

Second point = (a,0)

The distance between O and A is

$\vec{r_{1}}=0\kap{i}+a\kap{j}$

The magnitude of $r_{1}$ is

$|\vec{r_{1}}=\sqrt{(0)^2+(a)^2}$

$\vec{r_{1}}=a$

The distance between O and B is

$\vec{r_{2}}=a\kap{i}+0\kap{j}$

The magnitude of $r_{2}$ is

$|\vec{r_{2}}=\sqrt{(a)^2+(0)^2}$

$\vec{r_{2}}=a$

The potential is

$V=\dfrac{q}{4\pi\epsilon_{0}r}$

The potential at point A is

$V_{A}=\dfrac{10\times10^{-6}}{4\pi\epsilon_{0}a}$

The potential at point B is

$V_{B}=\dfrac{10\times10^{-6}}{4\pi\epsilon_{0}a}$

The potential difference will be

$V=V_{A}-V_{B}$

$V=\dfrac{10\times10^{-6}}{4\pi\epsilon_{0}a}-\dfrac{10\times10^{-6}}{4\pi\epsilon_{0}a}$

$V= 0$

Hence, The potential difference between two points will be zero.