The intensity of an electric field depends only on the coordinates x and y as follows $\overrightarrow{\mathrm{E}}=\frac{\mathrm{a}\left(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}\right)}{{\mathrm{x}}^2+{\mathrm{y}}^2}$where a is a constant and $\hat{\mathrm{i}}$ and $\hat{\mathrm{j}}$ are the unit vectors of the x and y-axes. Find the charge within a sphere of radius R with the centre at the origin

A unit vector perpendicular to the sphere radially outwards of any point of the sphere is given by

$\begin{array}{l}\widehat{\mathrm{n}}=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}}\widehat{\mathrm{i}}+\frac{\mathrm{y}}{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}}\widehat{\mathrm{j}}\\ =\frac{\mathrm{x}}{\mathrm{R}}\widehat{\mathrm{i}}+\frac{\mathrm{y}}{\mathrm{R}}\widehat{\mathrm{j}}+\frac{\mathrm{z}}{\mathrm{R}}\widehat{\mathrm{k}}\left(\text{ as }{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{R}}^2\right)\end{array}$

The electric flux through a differential area dA at point P on the sphere, $\mathrm{d}{\varphi }_{\mathrm{E}}=\overrightarrow{\mathrm{E}}\cdot \widehat{n}d\mathrm{A}$

$=\left[\frac{{\mathrm{ax}}^2}{\mathrm{R}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}+\frac{{\mathrm{ay}}^2}{\mathrm{R}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}\right]\mathrm{dA}=\left(\frac{\mathrm{a}}{\mathrm{R}}\right)\mathrm{dA}$

Note that  ${\mathrm{d\varphi }}_{\mathrm{E}}$ is independent of the coordinates x, y and z. Therefore, total flux passing through the sphere,

${\mathrm{\varphi }}_{\mathrm{E}}=\int {\mathrm{d\varphi }}_{\mathrm{E}}=\frac{\mathrm{a}}{\mathrm{R}}\int \mathrm{dS}=\left(\frac{\mathrm{a}}{\mathrm{R}}\right)\left(4{\mathrm{\pi R}}^2\right)=4\mathrm{\pi aR}$

From Gasuss’s law, we have ${\mathrm{\varphi }}_{\mathbb{E}}=\frac{{\mathrm{q}}_{\text{enclosed }}}{{\mathrm{\epsilon }}_0}$ 

$\Rightarrow \left(4\mathrm{\pi aR}\right)=\frac{{\mathrm{q}}_{\text{enclosed }}}{{\mathrm{\epsilon }}_0}\text{⇒}{\mathrm{q}}_{\text{enclosed }}=4{\mathrm{\pi \epsilon }}_0\mathrm{aR}$