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

Hint: 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.