Question

Two-point charges $q_1\left(\sqrt{10}\mu C\right)$ and $q_2(-25\mu C)$ are placed on the $x-$axis at $x=1m$ and $x=4m$, respectively. The electric field (in $V/m$) at a point $y=3m$ on $y-$axis is $[\text{take}\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^2]$

• A. $(-81\hat{i}+81\hat{j})\times {10}^2$
• B. $(-63\hat{i}+27\hat{j})\times {10}^2$
• C. $(63\hat{i}-27\hat{j})\times {10}^2$
• D. $(81\hat{i}-81\hat{j})\times {10}^2$

Answer 1

The correct option is C $(63\hat{i}-27\hat{j})\times {10}^2$

Let $E_1$ and $E_2$ be the electric field due to charge $q_1$ and $q_1$ respectively.
Diagram
Given $q_1=\sqrt{10}\mu C,q_2=-25\mu C$

We know that electric field $E$ at a distance $r$ from charge $q$ is given by:

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

Magnitude of $E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\sqrt{10}\times {10}^{-6}}{(1^2+3^2)}$

$=(9\times {10}^9)\times \sqrt{10}\times {10}^{-7}$

$=9\sqrt{10}\times {10}^{-2}$

${\overrightarrow{E}}_1=9\sqrt{10}\times {10}^{-2}\left(\cos {\theta }_1\right(-\hat{i})-\sin {\theta }_2\hat{j})$

(from figure) $\tan {\theta }_1=3$

$E_1=9\times \sqrt{10}\times {10}^2\left[\frac{1}{\sqrt{10}}\right(-\hat{i})+\frac{3}{\sqrt{10}}\hat{j}]$

$E_1=9\times {10}^2[-\hat{i}+3\hat{j}]$

$=[-9\hat{i}+27\hat{j}]\times {10}^2$

Now, $E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{r^2}$

$E_2=\frac{9\times {10}^9\times \left(25\right)\times {10}^{-6}}{(4^2+3^2)}V/m$

$E_2=9\times {10}^{3}V/m$

${\overrightarrow{E}}_2=9\times {10}^3(\cos {\theta }_2\hat{i}-\sin {\theta }_2\hat{j})$

[from figure] $\tan {\theta }_2=\frac{3}{4}$

${\overrightarrow{E}}_2=9\times {10}^3(\frac{4}{3}\hat{i}-\frac{3}{5}\hat{j})$

$=(72\hat{i}-54\hat{j})\times {10}^{2}V/m$

Resultant electric field will be given by the vector addition of $E_1$ and $E_2$

$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

$=(72\hat{i}-54\hat{j})\times {10}^2+[-9\hat{i}+27\hat{j}]\times {10}^2$

$=(63\hat{i}-27\hat{j})\times {10}^{2}V/m$