Question

A point charge $50\mu C$ is located in the $x\text{-}y$ plane at the position vector ${\overrightarrow{r}}_0=\left(2\hat{i}+3\hat{j}\right)m.$ The electric field at the point of position vector $\overrightarrow{r}=\left(8\hat{i}-5\hat{j}\right)m,$ in vector from is equal to:

• A. $900\left(-3\hat{i}+4\hat{j}\right)V/m$
• B. $90\left(3\hat{i}-4\hat{j}\right)V/m$
• C. $9\left(-3\hat{i}+4\hat{j}\right)V/m$
• D. $900\left(3\hat{i}-4\hat{j}\right)V/m$

Answer 1

The correct option is A $900\left(-3\hat{i}+4\hat{j}\right)V/m$

Distance between the point of charge and point at observation $B$,

$d=\sqrt{{(8-2)}^2+{(-5-3)}^2}=\sqrt{100}=10m$

$\therefore$ Electric field, $E=\frac{KQ}{r^2}=\frac{9\times {10}^9\times 50\times {10}^{-6}}{100}=4500u/m$

In vector form $\overrightarrow{E}=\frac{KQ}{r^2}\hat{r}$

for, $\hat{r}=\frac{\overrightarrow{r}}{\left|\overrightarrow{r}\right|}=\frac{6\hat{i}-8\hat{j}}{10}=\frac{2\{3\hat{j}-4\hat{j}\}}{10}$

$\therefore$ Electric field in vector form is given by,

$\overrightarrow{E}=\left(4500\right)\left(\frac{1}{2}\right)\{3\hat{i}-4\hat{j}\}$

$=900\{3\hat{i}-4\hat{j}\}V/m$

$\therefore$ Option (D) is the correct answer.