Question 120 of 311

Two point charges ${q}_{1}\left(\sqrt{10}\mu C\right)$ and ${q}_{2}\left(-25\mu C\right)$ are placed on the $x$ -axis at $x=1m$ and $x=4m$ respectively. The electric field $\left(\text{in}V/m\right)$ at a point $y=3m$ on $y$-axis is,

$\left[\text{Take}\dfrac{1}{4\pi {ϵ}_{0}}=9\times {10}^{9}N{m}^{2}{C}^{-2}\right]$

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

Show answer and solution▾

Answer: (D) $\left(63\hat{i}-27\hat{j}\right)\times {10}^{2}$

Electric field due to $\sqrt{10}\mu C$


${\vec{E}}_{1}=\dfrac{1}{4\pi {\in }_{0}}\dfrac{\sqrt{10}\times {10}^{-6}}{{r}_{1}^{3}}{\vec{r}}_{1}$

$=\dfrac{1}{4\pi {\in }_{0}}\dfrac{\sqrt{10}\times {10}^{-6}}{{\left(\sqrt{10}\right)}^{3}}\left(-\hat{i}+3\hat{j}\right)$

Similarly, electric field due to $-25\mu C$

${\vec{E}}_{2}=\dfrac{1}{4\pi {\in }_{0}}\dfrac{25\times {10}^{-6}}{{\left(5\right)}^{3}}\left(4\hat{i}-3\hat{j}\right)$

Net electric field,
$\vec{E}=9\times {10}^{9}\times {10}^{-6}\left(\dfrac{-\hat{i}}{10}+\dfrac{3\hat{j}}{10}+\dfrac{4}{5}\hat{i}-\dfrac{3}{5}\hat{j}\right)$

$=9\times {10}^{3}\left(\dfrac{7\hat{i}}{10}-\dfrac{3\hat{j}}{10}\right)$

$=\left(63\hat{i}-27\hat{j}\right)\times {10}^{2}\dfrac{N}{C}$