Question

Three point charges $q_1=1\mu \text{C},q_2=-2\mu \text{C}$ and $q_3=3\mu \text{C}$ are placed at $(1\text{m},0,0)$ and $(0,2\text{m},0)$ and $(0,0,3\text{m})$ respectively. The electric potential at origin is

• A. $9\times {10}^3\text{J/C}$
• B. $3\times {10}^3\text{J/C}$
• C. $15\times {10}^3\text{J/C}$
• D. $12\times {10}^3\text{J/C}$

Answer 1

The correct option is A $9\times {10}^3\text{J/C}$

We know potential due to a system of point charges at any point $P$ is given by,
$V_P$ = $\frac{{kq}_1}{r_1}+\frac{{kq}_2}{r_2}+\frac{{kq}_3}{r_3}....\left(i\right)$

Here, $q_1=1\mu \text{C}$ is at placed at $(1\text{m},0,0)$

Therefore the distance of charge $q_1$ from origin $(0,0,0)$ is,
$r_1=\sqrt{(1-0{)}^2+(0-0{)}^2+(0-0{)}^2}$
$\because [r=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}]$
or, $r_1=1\text{m}$
similarly, $q_2=-2\mu \text{C}$ and $r_2=2\text{m}$
$q_3=3\mu \text{C}$ and $r_3=3\text{m}$
From Eq. $\left(i\right)$ the net potential at point $P$ is,
$V_P=9\times {10}^9[\frac{1\times {10}^{-6}}{1}+\frac{(-2\times {10}^{-6})}{2}+\frac{(3\times {10}^{-6})}{3}]$
$\therefore V_P=9\times {10}^3\text{J/C}$

$\begin{array}{c}\text{Why this question}?\\ \text{Concept: Distance between any two points}A(x_1,y_1,z_1)\text{and}B(x_2,y_2,z_2)\\ \text{is given by}r=\sqrt{(x_2-x_1{)}^2+(y_2-y_1^2)+(z_2-z_1{)}^2}\end{array}$