Question

The intensity of an electric field depends only on the coordinate x, y and z as follows $\overrightarrow{E}=\dfrac{a({{x}_{i}}+{{y}_{j}}+{{z}_{k}})}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{\dfrac{3}{2}}}}$ Unit. The electrostatic energy stored between two imaginary concentric spherical shells of radii R and 2R with center at origin is
A)$\dfrac{4\pi {{\varepsilon }_{0}}{{a}^{2}}}{R}$
B) $\dfrac{2\pi {{\varepsilon }_{0}}{{a}^{2}}}{R}$
C) $\dfrac{\pi {{\varepsilon }_{0}}{{a}^{2}}}{R}$
D) $\dfrac{aR}{k}$

Answer 1

Hint: The solution to this problem is obtained by gauss’s diversion theorem. Field is a region where each point has a corresponding value of some physical function. For example, an electric field in a region has some specific direction of $\overrightarrow{E}$ component at different points. Electric field(E) is a vector quantity.

Complete answer:
In scalar units if the value of physical functions at each point of a field is a scalar quantity then it is a scalar field like temperature of atmosphere, depth of sea water from surfaces etc.
In a vector field if the value of function at each point of a field is vector quantity then it is called vector field like wind velocity of atmosphere ,the force of gravity on a mass in space ,forces on a charged body placed in an electric field etc. Electric field which has both magnitude and direction.
On the sphere the points P(X, Y, Z)
${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{R}^{2}}$
A unit vector which is perpendicular to the sphere radially outwards is given by:
$\begin{align}
  & \widehat{N}=\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\widehat{i}+\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\widehat{j}+\dfrac{Z}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\widehat{k} \\
 & \widehat{N}=\dfrac{x}{R}\widehat{i}+\dfrac{y}{R}\widehat{j}+\dfrac{z}{R}\widehat{k} \\
\end{align}$
On the sphere at point P the electric flux through small area dS is given by:
$\begin{align}
  & d{{\phi }_{e}}=\overrightarrow{E}.dS\widehat{N} \\
 & d{{\phi }_{e}}=(\dfrac{a{{x}^{2}}}{R({{x}^{2}}+{{y}^{2}})}+\dfrac{a{{y}^{2}}}{R({{x}^{2}}+{{y}^{2}})})dS \\
 & d{{\phi }_{e}}=\dfrac{adS}{R} \\
 & {{\phi }_{e}}={{\oint{d\phi }}_{e}}=\oint{\dfrac{adS}{R}} \\
 & {{\phi }_{e}}=4\pi aR \\
\end{align}$
By applying gauss law
$\begin{align}
  & {{\phi }_{e}}=\dfrac{{{q}_{inc}}}{{{\varepsilon }_{0}}} \\
 & {{q}_{inc}}=4\pi {{\varepsilon }_{0}}aR \\
 & K=\dfrac{1}{4\pi {{\varepsilon }_{0}}} \\
 & {{q}_{inc}}=\dfrac{aR}{K} \\
\end{align}$
The electrostatic energy stored is equal to $\dfrac{aR}{K}$

So option D is correct

Note:
Students unit vector is the ratio of vector itself by its magnitude and Gauss’s diversion theorem which states that volume integral of the divergence of vector field $\overrightarrow{A}$.taken over any volume V is equal to surface integral of $\overrightarrow{A}$ taken over the closed surface that bonds the volume V $\int_{V}{(\nabla }\cdot \overrightarrow{A}).dV=\oint\limits_{S}{\overrightarrow{A}\cdot \overrightarrow{dS}}$.