Question

The electric field strength in a region is given as $\overline{E}=\frac{x\overline{i}+y\overline{j}}{x^2+y^2}$. Find the net charge inside a sphere of radius $a$ with its centre at origin.

• A. $4\pi {\epsilon }_{0}a$
• B. $8\pi {\epsilon }_{0}a$
• C. $4\pi {\epsilon }_{0}Ra$
• D. $2\pi {\epsilon }_{0}a$

Answer 1

The correct option is C $4\pi {\epsilon }_{0}Ra$

$\varphi =\oint \overrightarrow{E}.\overrightarrow{ds}\overrightarrow{E}=\frac{a(x\hat{i}+y\hat{j})}{x^2+y^2}\varphi =\oint \frac{a(x\hat{i}+y\hat{j})}{x^2+y^2}.\frac{(x\hat{i}+y\hat{j}+k\hat{z})}{\sqrt{x^2+y^2+z^2}}ds{\varphi }_{net}=a\int \frac{x^2+y^2}{x^2+y^2}\frac{1}{\sqrt{x^2+y^2+z^2}}ds{\varphi }_{net}=\frac{a}{R}\int ds=\frac{a}{R}4\pi R^2{\varphi }_{net}=4\pi Ra\Rightarrow \frac{q}{{\epsilon }_0}=4\pi Raq=4\pi {\epsilon }_{0}Ra$