Question

A sphere of radius R carries charge density proportional to the square of the distance from the center\[\rho = A{r^2}\], where \[A\] is a positive constant. At a distance of \[R/2\] from the centre, the magnitude of electric field is:
(A) $ \dfrac {A}{4 \pi \varepsilon_0} $
(B) $ \dfrac {AR^3}{40 \varepsilon_0} $
(C) $ \dfrac {AR^3}{24 \varepsilon_0} $
(D) $ \dfrac {AR^3}{5 \varepsilon_0} $

Answer 1

Hint
To find the magnitude of the electric field we need to know the expression for the charge density caused by the influence of the electric field. First we need to apply the gauss law to get the distance of the electric field at radius R.
In this solution we will be using the following formula,
 $ \Rightarrow \rho = \dfrac{Q}{{\dfrac{4}{3}\pi {r^3}}} $
 $ \Rightarrow E = \dfrac{Q}{{A{\varepsilon _0}}} = \dfrac{{\rho V}}{{A{\varepsilon _0}}} $
Where, $ \rho $ is the charge density, $ r $ is the radius, $ A $ is a constant, $ {\varepsilon _0} $ permittivity of free space, $ Q $ is the charge enclosed, $ V $ is the volume of the sphere, $ E $ is the electric field.

Complete step by step answer
The charge density of a body is given by the formula,
 $ \Rightarrow \rho = \dfrac{{{\text{charge enclosed}}}}{{{\text{volume of the body}}}} $
 $ \Rightarrow \rho = \dfrac{Q}{V} $
When we substitute the expression of the volume in the above formula we get,
 $ \Rightarrow \rho = \dfrac{Q}{{\dfrac{4}{3}\pi {r^3}}} $
Now, the gauss law states that, the net electric flux through a closed surface area is equal to $ \dfrac{1}{{{\varepsilon _0}}} $ times the charge enclosed.
Then, we get,
Electric field $ E $ at distance from the center is $ \dfrac{Q}{{A{\varepsilon _0}}} = \dfrac{{\rho V}}{{A{\varepsilon _0}}} $ ,
As we substituted the value of $ Q $ in the equation we get,
Now, Electric field $ E $ at distance $ \dfrac{R}{2} $ from center is $ \dfrac{{\rho \left( {\dfrac{4}{3}\pi {{\left( {\dfrac{R}{2}} \right)}^3}} \right)}}{{4\pi {{\left( {\dfrac{R}{2}} \right)}^3}{\varepsilon _0}}} $
After simplification, we get the value as,
 $ \Rightarrow \dfrac{{\rho \left( {\dfrac{R}{2}} \right)}}{{3{\varepsilon _0}}} $
As, $ \rho = A{r^2} $ ,
So, $ \rho {\text{ at }}\dfrac{R}{2} = A \times \dfrac{{{R^2}}}{4} $
Thus, Electric field at distance $ \dfrac{R}{2} $ from center
 $ \Rightarrow \dfrac{{\dfrac{AR^2}{4}}{\dfrac{R}{2}}}{3{\varepsilon _0}} = \dfrac {AR^3}{24 \varepsilon_0} $
Hence, we get the correct option as (C).

Note
According to electromagnetism, charge density is a measure of electric charge per unit volume of the space in one, two or three dimensions. The three kinds of charge densities are,
-Linear charge density
-Surface charge density
-Volume charge density