Question

A point charge $q=-8.0nC$ is located at the origin. The electric field (in NC${}^{-1}$) vector at the point $x=1.2m,y=-1.6m$, as shown in figure, is :

• A. $-14.4\hat{i}+10.8\hat{j}$
• B. $-14.4\hat{i}-10.8\hat{j}$
• C. $-10.8\hat{i}+14.4\hat{j}$
• D. $-10.8\hat{i}-14.4\hat{j}$

Answer 1

Answer: (C)

$-10.8\hat{i}+14.4\hat{j}$

Electric field due to point charge at $\overrightarrow{r}$ $\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\hat{r}$
$\hat{r}=\frac{\overrightarrow{r}}{r}=\frac{(1.2-0)\hat{i}+(-1.6-0)\hat{j}}{\sqrt{(1.2{)}^2+(1.6{)}^2}}$$=\frac{1}{2}(1.2\hat{i}-1.6\hat{j})$

$\overrightarrow{E}=9\times {10}^9\times \frac{(-8\times {10}^{-9})}{2^2}\times \left[\frac{1}{2}\right(1.2\hat{i}-1.6\hat{j}\left)\right]$

$=-10.8\hat{i}+14.4\hat{j}$