Question

A spherically symmetric charge distribution is characterized by a charge density having the following variation:
\[p\left( r \right) = {p_0}\left( {1 - \dfrac{r}{R}} \right)\] for \[r < R\]
\[p\left( r \right) = 0\] for \[r \geqslant R\]
Where r is the distance from the centre of the charge distribution and \[{p_0}\] is the constant. The electric field at an internal point r:
A. \[\dfrac{{{p_0}}}{{4{\varepsilon _0}}}\left( {\dfrac{r}{3} - \dfrac{{{r^2}}}{{4R}}} \right)\]
B. \[\dfrac{{{p_0}}}{{{\varepsilon _0}}}\left( {\dfrac{r}{3} - \dfrac{{{r^2}}}{{4R}}} \right)\]
C. \[\dfrac{{{p_0}}}{{3{\varepsilon _0}}}\left( {\dfrac{r}{3} - \dfrac{{{r^2}}}{{4R}}} \right)\]
D. \[\dfrac{{{p_0}}}{{12{\varepsilon _0}}}\left( {\dfrac{r}{3} - \dfrac{{{r^2}}}{{4R}}} \right)\]

Answer 1

Hint: Consider the spherical shell of certain thickness at a distance r from the centre of the sphere and calculate the charge on this shell. Integrating the charge on this shell from 0 to r, you will get the charge enclosed by this region. Recall the expression for the electric field and substitute the expression for the charge.

Formula used:
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\]
Here, \[{\varepsilon _0}\] is the permittivity of the medium, q is the charge and r is the distance of the point from the charge.

Complete step by step answer:
We have given that the charge density outside the sphere is zero. That means the electric field outside the sphere is zero. We consider the spherical shell of radius r and thickness dr inside the sphere at a distance r from the origin of the sphere. We can express the charge on this shell as,
\[dq = p\left( r \right)4\pi {r^2}dr\]
We can calculate the total charge in the region between the origin and distance r by integrating the above equation.
\[\int {dq = q} = \int\limits_0^r {p\left( r \right)4\pi {r^2}dr} \]
\[ \Rightarrow q = 4\pi {p_0}\int\limits_0^r {\left( {1 - \dfrac{r}{R}} \right){r^2}dr} \]
\[ \Rightarrow q = 4\pi {p_0}\int\limits_0^r {\left( {{r^2} - \dfrac{{{r^3}}}{R}} \right)dr} \]
\[ \Rightarrow q = 4\pi {p_0}\left( {\dfrac{{{r^3}}}{3} + \dfrac{{{r^4}}}{{4R}}} \right)\] …… (1)
The electric field at a distance r from the centre of the sphere is,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\]
Here, \[{\varepsilon _0}\] is the permittivity of the medium.
Substituting equation (1) in the above equation, we get,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{1}{{{r^2}}}\left( {4\pi {p_0}\left( {\dfrac{{{r^3}}}{3} + \dfrac{{{r^4}}}{{4R}}} \right)} \right)\]
\[ \Rightarrow E = \dfrac{{{p_0}}}{{{\varepsilon _0}}}\left( {\dfrac{r}{3} + \dfrac{{{r^2}}}{{4R}}} \right)\]

So, the correct answer is option B.

Note: Another way to express the electric field in the region is by using Gauss’s law. According to Gauss’s law,
\[Eds = \dfrac{{{q_{enc}}}}{{{\varepsilon _0}}}\]
\[ \Rightarrow E\left( {4\pi {r^2}} \right) = \dfrac{{{q_{enc}}}}{{{\varepsilon _0}}}\],
where, \[{q_{enc}}\] is the charge enclosed in region between 0 to r. By substituting the expression for the charge enclosed, we can get the value of the electric field at distance r from the centre.