Question

A positive point charge 50 $\mu$C is located in the plane xy at a point with radius vector ${\overrightarrow{r}}_0=2\hat{i}+3\hat{j}$. The electric field vector $\overrightarrow{E}$ at a point with radius vector $\overrightarrow{r}=8\hat{i}-5\hat{j}$, where $r_0$ and r are expressed in meter, is :

• A. $(1.4\hat{i}-2.6\hat{j})kN{C}^{-1}$
• B. $(1.4\hat{i}+2.6\hat{j})kN{C}^{-1}$
• C. $(2.7\hat{i}-3.6\hat{j})kN{C}^{-1}$
• D. $(2.7\hat{i}+3.6\hat{j})kN{C}^{-1}$

Answer 1

The correct option is C $(2.7\hat{i}-3.6\hat{j})kN{C}^{-1}$

As electric field due to charge q at distance r from $r_0$
$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{|\overrightarrow{r}-{\overrightarrow{r}}_0{|}^3}(\overrightarrow{r}-{\overrightarrow{r}}_0)$
$\overrightarrow{MN}=\overrightarrow{r}-{\overrightarrow{r}}_0=(8\hat{i}-5\hat{j})-(2\hat{i}+3\hat{j})=(6\hat{i}-8\hat{j})$
$|\overrightarrow{r}-{\overrightarrow{r}}_0|=\sqrt{6^2+8^2}=10m$
$\overrightarrow{E}=9\times {10}^9\times \frac{50\times {10}^{-6}}{(10{)}^3}(6\hat{i}-8\hat{j})$
$=(2.7\hat{i}-3.6\hat{j})kN{C}^{-1}$