Question

The electric field at $(30,30)cm$ due to a charge of $-8nC$ at the origin in $N{C}^{-1}$ is

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

Answer 1

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

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

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

Put the value above.

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

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