Question

Electric field at point (30, 30, 0) due to a point charge of $8\times {10}^{-3}\mu C$ placed at origin will be (coordinates are in cm)

• A. $8000N/C$
• B. $4000(\hat{i}+\hat{j})N/C$
• C. $200\sqrt{2}(\hat{i}+\hat{j})N/C$
• D. $400\sqrt{2}(\hat{i}+\hat{j})N/C$

Answer 1

The correct option is C $200\sqrt{2}(\hat{i}+\hat{j})N/C$

Distance of the charge from the origin=d=

$\begin{array}{c}\sqrt{{0.30}^2+{0.30}^2}=0.30\times \sqrt{2}cm\\ E=\frac{1}{4\pi {\epsilon }_0}\times \frac{q}{d^2}\\ =9\times {10}^9\times \left(-8\times {10}^{-9}\right)\times \frac{1}{\left({0.30}^2\right)\times 2}\frac{N}{C}\\ E=-400N/C\end{array}$

The electric field strength is in the direction of $\theta$ where

$\begin{array}{c}Tan\theta =\frac{30cm}{30cm}=1\\ {45}^{0}withx-axis.\\ \cos \theta =\frac{1}{\sqrt{2}}and\sin \theta \frac{1}{\sqrt{2}}\end{array}$

Since $E$ is negative , the $\theta$ is in the $3th$ quardrant, i.e $\theta =180+45={225}^0$

The vector form of electric field=

$\begin{array}{c}\overrightarrow{E}=E\left[\cos \varphi \hat{i}+\sin \varphi \hat{j}\right]\\ \overrightarrow{E}=-\left(\frac{400}{\sqrt{2}}\right)\left[\hat{i}+\hat{j}\right]\frac{N}{C}\\ \overrightarrow{E}=-\frac{200}{\sqrt{2}}\left(\hat{i}+\hat{j}\right)\frac{N}{C}\end{array}$