$(a)\;8 \pi \in_{0} a R \qquad(b)\;4 \pi \in_{0} a R \qquad(c)\;2 \pi \in_{0} a R \qquad(d)\; \pi \in_{0} a R $

Answer : (b $\;4\pi \in_{0} a R \;$

Explanation :

At any point P(x , y ,z) on the sphere $x^2+y^2+z^2=R^2$

and a unit vector perpendicular to the sphere radially outwards is

$\hat{n} = \large\frac{x}{\sqrt{x^2+y^2+z^2}} \hat{i} + \large\frac{y}{\sqrt{x^2+y^2+z^2}} \hat{j}+ \large\frac{z}{\sqrt{x^2+y^2+z^2}} \hat{k}$

$\hat {n} = \large\frac{x}{R} \hat{i} + \large\frac{y}{R} \hat{j} + \large\frac{z}{R} \hat{k}$

Let us find the electric flux through a small area dS at point P on the sphere

$d \phi_{e} = \overrightarrow{E} . dS \hat {n}$

$=(\large\frac{ax^2}{R(x^2+y^2)}+\large\frac{ay^2}{R(x^2+y^2)})$ dS

$d \phi_{e}=\large\frac{a\;dS}{R}$

$\phi_{e}=\oint\;d \phi_{e} = \large\frac{a}{R} \oint dS =\large\frac{a}{R}\;4 \pi R^2$

$\phi_{e} = 4 \pi a R $

From Gauss's Law

$\large\frac{q_{inc}}{\in_{0}}= \phi_{e}$

$q_{inc}=4 \pi \in_{0} a R$

$=\large\frac{aR}{k}\;.$

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