Question

The intensity of an electric field depends on the co-ordinates x , y and z as follows $E=\frac{a(xi+yj+zk)}{(x^2+y^2+z^2{)}^{3/2}}$ unit The electrostatic energy stored between two imaginary concentric spherical shells of radii r and 2 R with center at origin is

• A. $\frac{4\pi {\epsilon }_0{a}^2}{R}$
• B. $\frac{2\pi {\epsilon }_0{a}^2}{R}$
• C. $\frac{\pi {\epsilon }_0{a}^2}{R}$
• D. $\frac{aR}{k}$

Answer 1

The correct option is D $\frac{aR}{k}$

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

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

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

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

$d{\varphi }_e=\overrightarrow{E}.dS\hat{n}$

$=(\frac{a{x}^2}{R(x^2+y^2)}+\frac{a{y}^2}{R(x^2+y^2)})$ dS

$d{\varphi }_e=\frac{adS}{R}$

${\varphi }_e=\oint d{\varphi }_e=\frac{a}{R}\oint dS=\frac{a}{R}4\pi R^2$

${\varphi }_e=4\pi aR$

From Gauss's Law

$\frac{q_{inc}}{{\in }_0}={\varphi }_e$

$q_{inc}=4\pi {\in }_{0}aR$

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