Question

A point charge $50\mu C$ is located in the $XY$ plane at the point of position vector ${\overrightarrow{r}}_0=2\hat{i}+3\hat{j}$. What is the electric field at the point vector ${\overrightarrow{r}}_1=8\hat{i}-5\hat{j}$

• A. $1200V/m$
• B. $0.04V/m$
• C. $900V/m$
• D. $4500V/m$

Answer 1

The correct option is D $4500V/m$

$\text{Step 1: Distance of charge from the point at which we have to find electric field [Refer Fig. 1]}$

Position vector of Charge Q, $\overrightarrow{r_0}=2\hat{i}+3\hat{j},$

Position vector of Point where Electric Field has to be calculated, $\overrightarrow{r_1}=8\hat{i}-5\hat{j}$

So, Distance between Point and Charge

$\overrightarrow{r}=|△\overrightarrow{r}|=|\overrightarrow{r_1}-\overrightarrow{r_0}|=|8\hat{i}-5\hat{j}-2\hat{i}-3\hat{j}|$

$=|6\hat{i}-8\hat{j}|$
$r=\sqrt{6^2+8^2}=10m$

$\text{Step 2: Magnitude of Electric field [Refer Fig. 2]}$

$Q=50\mu C$

By Coulomb's Law, Electric field of a point charge at a distance $r$

$E=\frac{kQ}{r^2}$

$=\frac{9\times {10}^9\times (50\times {10}^{-6}C)}{(10m{)}^2}=4500V/m$

Hence Correct Option id $D$