Question

The electric field intensity at $(30,30)cm$ due to a charge of $-8nC$ at the origin in $N/C$ is

• A. $-400(i+j)$
• B. $400(i+j)$
• C. $-200\sqrt{2}(i+j)$
• D. $200\sqrt{2}(i+j)$

Answer 1

The correct option is C $-200\sqrt{2}(i+j)$

Electric field $\overrightarrow{E}=\frac{1}{4\times {\epsilon }_0}\frac{q_1{q}_2}{r^2}\hat{r}$

$\hat{r}=\frac{(30-0)\hat{i}+(30-0)\hat{j}}{\sqrt{(30-0{)}^2+(30-0{)}^2}}=\frac{30\hat{i}+30\hat{j}}{42.42}$

Put the value in the above equation.

$=9\times {10}^9\times \frac{-8\times 1}{(42.42{)}^2}\times \frac{30\hat{i}+30\hat{j}}{42.42}$

$\overrightarrow{E}=-200\sqrt{2}(\hat{i}+\hat{j})$