Question

A solid sphere of radius R has a charge Q distributed in its volume with a charge density $\rho =k{{r}^{a}}$, 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, find the value of a is

Answer 1

Hint: We are given the radius and charge density of a solenoid. We need to find one of the constants given in the value of the charge density. Since we are given the relation between the electric fields in two different cases, we can find those electric fields and equate them. Thus we will get the solution.
Formula used:
$E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{r}^{2}}}$
\[q=\int\limits_{0}^{r}{\rho dv}\]

Complete answer:
In the question we have a solid sphere with a radius R.
A charge ‘Q’ is distributed in the volume of the sphere and the charge density is given as, $\rho =k{{r}^{a}}$, where ‘k’ and ‘a’ are constants and ‘r’ is the distance from the spheres centre.
Consider the figure of the solid sphere given below.

First let us find the electric field on the surface of the sphere, i.e. when
$r=R$
We know that the equation for electric field is given by,
$E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{r}^{2}}}$, were ‘q’ is the charge and ‘r’ is distance.
Therefore the electric field on the surface of the given solid sphere will be,
${{E}_{1}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}_{1}}}{{{R}^{2}}}$
Here we need to find the charge on the conductor.
We know that total charge is given by the equation,
\[q=\int\limits_{0}^{r}{\rho dv}\]
Therefore the charge on the surface of the conductor for the given solid sphere will be,
${{Q}_{1}}=\int\limits_{0}^{R}{\rho dv}$
We know that volume of a small area on a sphere ‘$dv$’ is,
$dv=4\pi {{r}^{2}}dr$
And we also know the value of charge density $\rho $
 By substituting these values we get,
$\Rightarrow {{Q}_{1}}=\int\limits_{0}^{R}{k{{r}^{a}}\left( 4\pi {{r}^{2}}dr \right)}$
Now we can take the constants out of the integral. Thus,
$\Rightarrow {{Q}_{1}}=4\pi k\int\limits_{0}^{R}{{{r}^{a}}{{r}^{2}}dr}$
$\Rightarrow {{Q}_{1}}=4\pi k\int\limits_{0}^{R}{{{r}^{a+2}}dr}$
Integrating this we will get,
$\Rightarrow {{Q}_{1}}=4\pi k\times \left[ \dfrac{{{r}^{a+2+1}}}{a+2+1} \right]_{0}^{R}$
$\Rightarrow {{Q}_{1}}=4\pi k\times \left[ \dfrac{{{r}^{a+3}}}{a+3} \right]_{0}^{R}$
$\Rightarrow {{Q}_{1}}=4\pi k\times \left( \dfrac{{{R}^{a+3}}}{a+3} \right)$
Therefore we get the electric field on the surface of the given solid sphere as,
\[{{E}_{1}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{4\pi k\left( {{R}^{a+3}} \right)}{\left( a+3 \right){{R}^{2}}}\]
By simplifying the above equation we get
\[\Rightarrow {{E}_{1}}=\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{k{{R}^{a+1}}}{\left( a+3 \right)}\]
This is the electric field when $r=R$.
Now we can find the electric field at $r=\dfrac{R}{2}$.
Similarly, we get
${{E}_{2}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}_{2}}}{{{\left( \dfrac{R}{2} \right)}^{2}}}$
Now we need to find the total charge at $r=\dfrac{R}{2}$.
Let ${{Q}_{2}}$ be the total charge at $r=\dfrac{R}{2}$, then
${{Q}_{2}}=\int\limits_{0}^{{}^{R}/{}_{2}}{\rho dv}$
We know $dv=4\pi {{r}^{2}}dr$ and $\rho =k{{r}^{a}}$. Therefore,
$\Rightarrow {{Q}_{2}}=\int\limits_{0}^{{}^{R}/{}_{2}}{k{{r}^{a}}4\pi {{r}^{2}}dr}$
By solving this,
$\Rightarrow {{Q}_{2}}=4\pi k\int\limits_{0}^{{}^{R}/{}_{2}}{{{r}^{a+2}}dr}$
\[\Rightarrow {{Q}_{2}}=4\pi k\left[ \dfrac{{{r}^{a+3}}}{a+3} \right]_{0}^{{}^{R}/{}_{2}}\]
By applying limits, we get
$\Rightarrow {{Q}_{2}}=4\pi k\left( \dfrac{{{\left( \dfrac{R}{2} \right)}^{a+3}}}{a+3} \right)$
Therefore the electric field at $r=\dfrac{R}{2}$ will be,
${{E}_{2}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{4\pi k{{\left( \dfrac{R}{2} \right)}^{a+3}}}{\left( a+3 \right){{\left( \dfrac{R}{2} \right)}^{2}}}$
By simplifying this we get,
$\Rightarrow {{E}_{2}}=\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{k{{\left( \dfrac{R}{2} \right)}^{a+3}}{{\left( \dfrac{R}{2} \right)}^{-2}}}{\left( a+3 \right)}$
$\Rightarrow {{E}_{2}}=\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{k\times {{R}^{a+1}}}{\left( a+3 \right){{\left( 2 \right)}^{a+1}}}$
This is the electric field when $r=\dfrac{R}{2}$.
In the question it is said that the electric field when $r=\dfrac{R}{2}$ is $\dfrac{1}{8}$ times the electric field when $r=R$, i.e.
${{E}_{2}}=\dfrac{1}{8}{{E}_{1}}$
By substituting the values of ${{E}_{1}}$ and ${{E}_{2}}$, we get
$\Rightarrow \dfrac{1}{{{\varepsilon }_{0}}}\dfrac{k\times {{R}^{a+1}}}{\left( a+3 \right){{\left( 2 \right)}^{a+1}}}=\dfrac{1}{8}\dfrac{1}{{{\varepsilon }_{0}}}\dfrac{k{{R}^{a+1}}}{\left( a+3 \right)}$
$\Rightarrow \dfrac{1}{{{\left( 2 \right)}^{a+1}}}=\dfrac{1}{8}$
$\Rightarrow {{2}^{a+1}}=8$
We know that $8={{2}^{3}}$. Therefore,
$\Rightarrow {{2}^{a+1}}={{2}^{3}}$
From this we can see that,
$\Rightarrow a+1=3$
Therefore,
$\Rightarrow a=3-1=2$
Hence the value of ‘a’ is 2.

Note:
While finding the electric field when $r=\dfrac{R}{2}$, we might eliminate the common terms at the beginning itself. i.e.
${{E}_{2}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}_{2}}}{{{\left( \dfrac{R}{2} \right)}^{2}}}$
If we eliminate the common terms now itself, we will get
$\Rightarrow {{E}_{2}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{4{{Q}_{2}}}{{{R}^{2}}}$
$\Rightarrow {{E}_{2}}=\dfrac{1}{\pi {{\varepsilon }_{0}}}\dfrac{{{Q}_{2}}}{{{R}^{2}}}$
This is a simple form of ${{E}_{2}}$. But if we do so we might feel difficulty while equating this with ${{E}_{1}}$. Hence try not to simplify ${{E}_{2}}$ first itself.