Question

A solid sphere has a charge Q of radius R is dispensed in its volume with a charge density of, $\rho = kra$ where k and a are constants and r is the distance from its centre. If the electric field at r =$\dfrac{R}{2}$ is $\dfrac{1}{8}$​ times that at r = R, the value of a is,
A. 3
B. 5
C. 2
D. 7

Answer 1

Hint: To solve this problem, firstly we have to understand the concepts of Gauss’s law, distribution of charge in solid sphere and charge density, and then by using the formulas required and the information given in our question, we will find the values (asked in the question) and hence we will easily approach our answer.

Step-By-Step answer:
Gauss’s Law: Another name for this is Gauss’s flux theorem. We can relate this law from distribution of electric charge to resulting electric field. This law states that as electric flux increases in a closed surface electric charge enclosed in a closed surface also increases.
Electric field inside a closed sphere or conductor: As we all know that conductors conduct electricity. And we all know that conductors have freely moving charges. So, charge will come on the surface and density of charge will remain the same throughout the surface.

Charge Density: Defined as total charge present in unit length, volume or density. Here we had made use of volume charge density. It is symbolized by Greek letter (ρ).

As by now we have studied Gauss's law, now we can come to result:
We can also write that,
$\Rightarrow \,\,E \times 4\pi {R^2} = \left( {\dfrac{{4\pi k}}{{{\varepsilon _0}}}} \right)\dfrac{{{R^{\left( {a + 1} \right)}}}}{{{\varepsilon _0}\left( {a + 3} \right)}}$
For
$\Rightarrow \,\,r = \dfrac{R}{{{2_{{E_2}}}}}{\dfrac{{k\left[ {\dfrac{R}{2}} \right]}}{{{\varepsilon _0}\left( {a + 3} \right)}}^{\left( {a + 1} \right)}}$
Given,​​
${E_2} = \dfrac{E}{8}$
Or we can write,
$\Rightarrow \,\,k\dfrac{{{{\left[ {\dfrac{R}{2}} \right]}^{\left( {a + 1} \right)}}}}{{{\varepsilon _0}\left( {a + 3} \right)}} = \dfrac{1}{8}\dfrac{{k{R^{\left( {a + 1} \right)}}}}{{{\varepsilon _0}\left( {a + 3} \right)}}$
​$\Rightarrow \,\,\dfrac{1}{{{2^{a + 1}}}}\dfrac{1}{8}$
We get, a=2

Hence the correct option is C) 2.

Note - In this problem we came across many concepts and studied them well. Electric fields of a sphere of equal charge density throughout the surface and total charge Q can be obtained by applying Gauss' law. Let us take a Gaussian surface in the form of a sphere at radius r > R.
The electric field has equal value at every point of the surface and is toward the outside. Electric flux is then the electric field times the area of the spherical surface.