Question

A charge of ${10}^{-10}C$ is palced at the origin. The eletric field(in N/C) at $(1,1)$ cm due to it, is:

• A. $(\overline{i}+\overline{j})\times {10}^3$
• B. $\frac{4.5}{\sqrt{2}}(\overline{i}+\overline{j})\times {10}^3$
• C. $20\overline{j}\times {10}^3$
• D. $4.5\sqrt{2}(\overline{i}+\overline{j})\times {10}^3$

Answer 1

The correct option is B $\frac{4.5}{\sqrt{2}}(\overline{i}+\overline{j})\times {10}^3$

$\overrightarrow{E}=\frac{k\left({10}^{-10}\right)}{(\sqrt{2}\times {10}^{-2}{)}^2}(\frac{\hat{x}}{\sqrt{2}}+\frac{\hat{y}}{\sqrt{2}})$

$=\frac{9\times {10}^9\times {10}^{-10}}{{10}^{-4}\times 2}\left(\frac{\hat{x}}{\sqrt{2}}+\frac{\hat{y}}{\sqrt{2}}\right)\to$ unit vector in the above shown direction

$\overrightarrow{E}=\frac{4.5\times {10}^3}{\sqrt{2}}(\hat{x}+\hat{y})$