Question

Position vector of a point charge $3.7\mu C$ with respect to origin is $(1\hat{i}+2\hat{j}+3\hat{k})\text{m}$. The potential due to point charge at origin is

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

Answer 1

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

We know electric potential due to a point charge $q$ at a point $P$ is given by,
$V=\frac{kq}{r}$ where $r$ is the distance of point $P$ from the charge.

Given the position vector ( $\overrightarrow{r}$) of charge,
$\overrightarrow{r}=1\hat{i}+2\hat{j}+3\hat{k}$
So, $r=\sqrt{1^2+2^2+3^2}$
$\therefore r=\sqrt{14}\text{m}=3.7\text{m}$
Further the potential due to charge is ,
$V_P=\frac{9\times {10}^9\times 3.7\times {10}^{-6}}{3.7}$
$\because k=9\times {10}^9,q=3.7\times {10}^{-6}\mu C$
$\Rightarrow V_P=9\times {10}^3\text{J/C}$

$\begin{array}{c}\text{Why this question}?\\ \text{Concept: For any vector}\overrightarrow{A}=x\hat{i}+y\hat{j}+z\hat{k}\\ \text{its magnitude is given by}|A|=\sqrt{x^2+y^2+z^2}\end{array}$