Question

A point charge $50\mu \text{C}$ is located in the $\text{XY-}$plane at the point of position vector ${\overrightarrow{r}}_0=2\hat{i}+3\hat{j}$. Find the electric field ($\text{V/m}$) due to this point charge at the point of position vector $\overrightarrow{r}=8\hat{i}-5\hat{j}$

Answer 1

Given,
Point charge, $q=50\mu \text{C}=50\times {10}^{-6}\text{C}$

Position vector of point charge,
$\overrightarrow{r_0}=2\hat{i}+3\hat{j}$

Position vector of unit test charge located at point $P$, $\overrightarrow{r}=(8\hat{i}-5\hat{j})$

The magnitude of electric field at any point $P$ due to a point charge is given by,$$|E_P|=\frac{Kq}{|r^{{}^{\prime }}{|}^3}{\overrightarrow{r}}^{\prime }...\left(1\right)$$ Where, ${\overrightarrow{r}}^{\prime }$ position vector of $P$ with respect to the charge $q$ is

${\overrightarrow{r}}^{{}^{\prime }}=\overrightarrow{r}-{\overrightarrow{r}}_0$

$\Rightarrow {\overrightarrow{r}}^{{}^{\prime }}=(8\hat{i}-5\hat{j})-(2\hat{i}+3\hat{j})$

$\Rightarrow {\overrightarrow{r}}^{{}^{\prime }}=(6\hat{i}-8\hat{j})$

So, $|\overrightarrow{r^{\prime }}|=\sqrt{(6{)}^2+(-8{)}^2}=10$

Substituting the values in equation $\left(1\right)$, we get

$|E_P|=\frac{9\times {10}^9\times 50\times {10}^{-6}}{{10}^2}$

$\therefore E_P=4500\text{V/m}$

Accepted answer : $4500$