Question

A point charge $q=-8nC$ is located at the origin. Find the electric field vector at the point $x=1.2m,y=-1.6m$ :

• A. $(5.4\hat{j}-7.2\hat{i})N/C$
• B. $(7.2\hat{j}-5.4\hat{i})N/C$
• C. $(10.8\hat{j}-14.4\hat{i})N/C$
• D. $(-10.8\hat{i}+14.4\hat{j})\text{N/C}$
  

Answer 1

The correct option is D $(-10.8\hat{i}+14.4\hat{j})\text{N/C}$

$\text{Step 1: Finding Position vector and its magnitude}$

$x=1.2m,y=-1.6m$

Position vector of the point with respect to the charge is given by:

$\overrightarrow{r}=\overrightarrow{r_{point}}-\overrightarrow{r_{charge}}$

$=(1.2\hat{i}-1.6\hat{j})-(0\hat{i}+0\hat{j})$

$=1.2\hat{i}-1.6\hat{j}$

Thus, $r=|\overrightarrow{r}|=\sqrt{{1.2}^2+{1.6}^2}=2\text{m}$

$\text{Step 2: Electric field at a distance}r$

The electric field at a distance $r$ due to a point charge $q$ is given by:

$\overrightarrow{E}=\frac{q}{4\pi {ϵ}_0|\overrightarrow{r}{|}^3}\overrightarrow{r}$

$\Rightarrow \overrightarrow{E}=\frac{(-8\times {10}^{-9}C)\times 9\times {10}^9}{{\left(2m\right)}^3}(1.2\hat{i}-1.6\hat{j})=(-10.8\hat{i}+14.4\hat{j})\text{N/C}$

Hence, Option $\left(D\right)$ is correct.