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A sphere of radius R carries charge density proportional to the square of the distance from the centerρ=Ar2, where A is a positive constant. At a distance of R/2 from the centre, the magnitude of electric field is:
(A) A4πε0
(B) AR340ε0
(C) AR324ε0
(D) AR35ε0

Answer
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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,
 ρ=Q43πr3
 E=QAε0=ρVAε0
Where, ρ is the charge density, r is the radius, A is a constant, ε0 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,
 ρ=charge enclosedvolume of the body
 ρ=QV
When we substitute the expression of the volume in the above formula we get,
 ρ=Q43πr3
Now, the gauss law states that, the net electric flux through a closed surface area is equal to 1ε0 times the charge enclosed.
Then, we get,
Electric field E at distance from the center is QAε0=ρVAε0 ,
As we substituted the value of Q in the equation we get,
Now, Electric field E at distance R2 from center is ρ(43π(R2)3)4π(R2)3ε0
After simplification, we get the value as,
 ρ(R2)3ε0
As, ρ=Ar2 ,
So, ρ at R2=A×R24
Thus, Electric field at distance R2 from center
 AR24R23ε0=AR324ε0
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

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