Question

Two point charges $q_1\left(\sqrt{10}\mu \text{C}\right)$ and $q_2(-25\mu \text{C})$ are placed on the x-axis at $x=1\text{m}$ and $x=4\text{m}$ respectively.The electric field in $\left(\text{V/m}\right)$ at a point $y=3\text{m}$ on $y-$axis is

$[\text{Take}\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}]$

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

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



Electric field at point $y=3$ will be-
$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$
Let ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ are the value of electric field due to charge, $q_1$ and $q_2$ respectivley


From$\Delta ABC$
Value of $r_1=\sqrt{1^2+3^2}=\sqrt{10}$
$\sin {\theta }_1=\frac{3}{\sqrt{10}},\cos {\theta }_1=\frac{1}{\sqrt{10}}$
So,
Magnitude of $E_1=\frac{1}{4\pi {ϵ}_0}\frac{q_1}{r_1^2}$
$E_1=\frac{1}{4\pi {ϵ}_0}\frac{\sqrt{10}\times {10}^{-6}}{(1^2+3^2)}$
$E_1=(9\times {10}^9)\times \sqrt{10}\times {10}^{-7}$
$E_1=9\times {10}^2\sqrt{10}$
${\overrightarrow{E}}_1=E_1\cos {\theta }_1(-\hat{i})+E_1\sin {\theta }_1\hat{j}$
${\overrightarrow{E}}_1=E_1[-\cos {\theta }_1\hat{i}+\sin {\theta }_1\hat{j}]$
${\overrightarrow{E}}_1=9\sqrt{10}\times {10}^2[-\frac{1}{\sqrt{10}}\hat{i}+\frac{3}{\sqrt{10}}\hat{j}]$
${\overrightarrow{E}}_1=9\times {10}^2[-\hat{i}+3\hat{j}]=[-9\hat{i}+27\hat{j}]\times {10}^2$



From$\Delta PQR$
Value of $r_2=\sqrt{4^2+3^2}=5$

$\sin {\theta }_2=\frac{3}{5},\cos {\theta }_2=\frac{4}{5}$
Similarly, $E_2=\frac{1}{4\pi {ϵ}_0}\frac{q_2}{r_2^2}$
$E_2=\frac{9\times {10}^9\times \left(25\right)\times {10}^{-6}}{5^2}\Rightarrow E_2=9\times {10}^3\text{V/m}$
${\overrightarrow{E}}_2=[E_2\cos {\theta }_2\hat{i}+E_2\sin {\theta }_2(-\hat{j}\left)\right]$
${\overrightarrow{E}}_2=9\times {10}^3(\cos {\theta }_2\hat{i}-{\sin }_2\hat{j})$
${\overrightarrow{E}}_2=9\times {10}^3(\frac{4}{5}\hat{i}-\frac{3}{5}\hat{j})=(72\hat{i}-54\hat{j})\times {10}^2$
So, electric field at point $y=3$ will be-
$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=(63\hat{i}-27\hat{j})\times {10}^2\text{V/m}$