Question

The intensity of an electric field depends only on the coordinates $x$ and $y$ as follows:
$E=\frac{a(x\hat{i}+y\hat{j})}{x^2+y^2}$
where, $a$ is a constant and $\hat{i}$ and $\hat{j}$ 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.

Answer 1

We know that equation of sphere with origin as centre is $x^2+y^2+z^2=R^2$

At any point $P(x,y,z)$ on the sphere a unit vector perpendicular to the sphere radially outwards is
$\hat{n}=\frac{x}{\sqrt{x^2+y^2+z^2}}\hat{i}+\frac{y}{\sqrt{x^2+y^2+z^2}}\hat{j}+\frac{z}{\sqrt{x^2+y^2+z^2}}\hat{k}=\frac{x}{R}\hat{i}+\frac{y}{R}\hat{j}+\frac{z}{R}\hat{k}$

Let us find the electric flux passing through a small area $dS$ at point $P$ on the sphere,
$d\varphi =E.\hat{n}dS=\{\frac{a{x}^2}{R(x^2+y^2)}+\frac{a{y}^2}{R(x^2+y^2)}\}=\left(\frac{a}{R}\right)dS$

Here, we note that $d\varphi$ is independent of the coordinates $x,y$ and $z$. Therefore, total flux passing through the sphere is
$\varphi =\int d\varphi =\frac{a}{R}\int dS=\left(\frac{a}{R}\right)\left(4\pi R^2\right)$
$\Rightarrow \varphi =4\pi aR$
From Gauss's law,
$\varphi =\frac{q_{\text{in}}}{{ϵ}_0}$
$\left(4\pi aR\right)=\frac{q_{\text{in}}}{{ϵ}_0}$
$\therefore q_{\text{in}}=4\pi {ϵ}_{0}aR$