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 = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, $[take\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{,}^2{C}^{-2}]$

• 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)

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

Let $\overrightarrow{E_1}$ & $\overrightarrow{E_2}$ are the values of electric field due to $q_1$ and $q_2$ respectively magnitude of $E_2=\frac{1}{4\pi {\in }_0}\frac{q_2}{r^2}$

$E_2=\frac{9\times {10}^9\times \left(25\right)\times {10}^{-6}}{(4^2+3^2)}V/m$
$E_2=9\times {10}^{3}V/m$
$\therefore \overrightarrow{E_2}=9\times {10}^3(cos{\theta }_2\hat{i}-sin{\theta }_2\hat{j})$
$\therefore tan{\theta }_2=\frac{3}{4}$
$\therefore \overrightarrow{E_2}=9\times {20}^3(\frac{4}{5}\hat{i}-\frac{3}{5}\hat{j})=(72\hat{i}-54\hat{j})\times {10}^2$
Magnitude of $E_1=\frac{1}{4\pi {\in }_0}\frac{\sqrt{10}\times {10}^{-6}}{(1^2+3^2)}$
$=(9\times {10}^9)\times \sqrt{10}\times {10}^{-7}$
$=9\sqrt{10}\times {10}^2$
$\therefore \overrightarrow{E_1}=9\sqrt{10}\times {10}^2\left[cos{\theta }_1\right(-\hat{i})+sin{\theta }_1\hat{j}]$
$\therefore tan{\theta }_1=3$
$E_1=9\times \sqrt{10}\times {10}^2\left[\frac{1}{\sqrt{10}}\right(-\hat{i})+\frac{3}{\sqrt{10}}\hat{j}]$
$E_1=9\times {10}^2[-\hat{i}+3\hat{j}]=[-9\hat{i}+27\hat{j}]{10}^2$
$therefore\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}=(63\hat{i}-27\hat{j})\times {10}^{2}V/m$
$\therefore$ correct answer is (3)