Question

Two point charges $q_1(\sqrt{10}\mu C)$ 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, $\left[\text{take}\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2{C}^{-2}\right]$

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

Answer 1

Answer: (C)

Let ${\vec{E}}_1$ & ${\vec{E}}_2$ are the vaues of electric field due to $q_1$ & $q_2$ respectively magnitude of $E_2=\frac{1}{4\pi {\in }_0}\frac{q_2}{r^2}$
$E_2=\frac{9\times 10^9\times \left(25\right)\times 10^{-6}}{\left(4^2+3^2\right)}V/m$
$E_2=9\times 10^{3}V/m$
$\therefore {\vec{E}}_2=9\times 10^3\left(\cos {\theta }_2\hat{i}-\sin {\theta }_2\hat{j}\right)$
$\because \tan {\theta }_2=\frac{3}{4}$
$\therefore {\vec{E}}_2=9\times 10^3\left(\frac{4}{5}\hat{i}-\frac{3}{5}\hat{j}\right)=\left(72\hat{i}-54\hat{j}\right)\times 10^2$
$E_1=\frac{1}{4\pi {\in }_0}\frac{\sqrt{10}\times 10^{-6}}{\left(1^2+3^2\right)}$
$=\left(9\times 10^9\right)\times \sqrt{10}\times 10^{-7}$
$=9\sqrt{10}\times 10^2$
$\therefore {\vec{E}}_1=9\sqrt{10}\times 10^2\left[\cos {\theta }_1\left(-\hat{i}\right)+\sin {\theta }_1\hat{j}\right]$
$\therefore \tan {\theta }_1=3$
$E_1=9\times \sqrt{10}\times 10^2\left[\frac{1}{\sqrt{10}}\left(-\hat{i}\right)+\frac{3}{\sqrt{10}}\hat{j}\right]$
$E_1=9\times 10^2\left[-\hat{i}+3\hat{j}\right]=\left[-9\hat{i}+27\hat{j}\right]10^2$
$\therefore \vec{E}={\vec{E}}_1+{\vec{E}}_2=\left(63\hat{i}-27\hat{j}\right)\times 10^{2}V/m$