Question

The electric field in a certain region is acting radially outward and is given by$E=Ar$. A charge contained in a sphere of radius $'a'$ centred at the origin of the field, will be given by:
$A.4\pi {{\varepsilon }_{o}}A{{a}^{3}}$
$B.{{\varepsilon }_{o}}A{{a}^{3}}$
$C.4\pi {{\varepsilon }_{o}}A{{a}^{2}}$
$D.A{{\varepsilon }_{o}}{{a}^{2}}$

Answer 1

We have to use the concept of Gauss’ law to solve this problem. Gauss’s law states that the net electric flux through a closed surface is equal to $\dfrac{1}{{{\varepsilon }_{o}}}$ times the net charge enclosed within the surface in vacuum. Formula of electric flux will also be applied to get the required result.

Formula used:
We will use the following mentioned formulae to solve the given problem:-
$\phi =\dfrac{{{Q}_{in}}}{{{\varepsilon }_{o}}}$ and $E=Ar$.

Complete step-by-step answer:
From the question given above we have $E=Ar$ ……………. $(i)$
Where $A$ is constant and $r$ is the radial distance
Using Gauss’ law we have,
$\phi =\dfrac{{{Q}_{in}}}{{{\varepsilon }_{o}}}$ …………………. $(ii)$ Where $\phi$ denotes electric flux through the closed surface, ${{Q}_{in}}$ is charge enclosed in the surface and ${{\varepsilon }_{o}}$ is permittivity in free space(constant)
Charge is contained by the sphere of radius, $a$.
$r=a$ …………… $(iii)$
We also know that $\varphi =EA$ ……………… $(iv)$
As electric flux is also defined as the dot product of electric field $(E)$ and area $(A)$
Putting $(iv)$ in $(ii)$ we get,
$EA=\dfrac{{{Q}_{in}}}{{{\varepsilon }_{o}}}$ ………………….. $(v)$
We know that $A=4\pi {{r}^{2}}$ for this case it can be written as
$A=4\pi {{a}^{2}}$ …………………….. $(vi)$
From $(i)$ and $(iii)$ we have,
$E=Aa$ ………………….. $(vii)$
Putting values of $(vi)$ and $(vii)$ in $(v)$ we get,
$Aa(4\pi {{a}^{2}})=\dfrac{{{Q}_{in}}}{{{\varepsilon }_{o}}}$
Therefore, ${{Q}_{in}}=4\pi {{\varepsilon }_{o}}A{{a}^{3}}$
Hence, option $(A)$ is the correct one among the given options.

So, the correct answer is “Option A”.

Note: In solving these problems correct use of Gauss’ law is required. Correct identification of geometry of the Gaussian surface is also a very important task. It should also be noted that if the flux emerging out of a Gaussian surface is zero then it is not necessary that the intensity of the electric field is also zero. We should also be very careful in ${{Q}_{in}}$, it only represents the total charges inside the Gaussian surface. It does not include the charges outside the Gaussian surface.