Question

A point charged $q=50\mu \text{C}$ is located in the $xy$ plane at the point of position vector $\overrightarrow{r_o}=2\hat{i}+3\hat{j}$. What is electric field at the point of position vector $\overrightarrow{r}=8\hat{i}-5\hat{j}$? (In S.I unit)

• A. $900{\text{Vm}}^{-1}$
• B. $1800{\text{Vm}}^{-1}$
• C. $4500{\text{Vm}}^{-1}$
• D. $2700{\text{Vm}}^{-1}$

Answer 1

The correct option is C $4500{\text{Vm}}^{-1}$

Position of point in space relative to position of charge , $\overrightarrow{r_1}=\overrightarrow{r}-\overrightarrow{r_0}=(8\hat{i}-5\hat{j})-(2\hat{i}+3\hat{j})=6\hat{i}-8\hat{j}$

$\therefore |\overrightarrow{r_1}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{m}$

Electric field at this point, $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_1^2}=\frac{9\times {10}^9\times 50\times {10}^{-6}}{{10}^2}=4500{\text{Vm}}^{-1}$

Hence, option (c) is the correct answer.