Question

A spherically symmetric charge distribution is characterized by a charge density having the following variation:
$$p\left(r\right)=p_0\left(1-{\displaystyle \frac{r}{R}}\right)$$ for $$r<R$$
$$p\left(r\right)=0$$ for $$r⩾R$$
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. $${\displaystyle \frac{p_0}{4{\epsilon }_0}}\left({\displaystyle \frac{r}{3}}-{\displaystyle \frac{r^2}{4R}}\right)$$
B. $${\displaystyle \frac{p_0}{{\epsilon }_0}}\left({\displaystyle \frac{r}{3}}-{\displaystyle \frac{r^2}{4R}}\right)$$
C. $${\displaystyle \frac{p_0}{3{\epsilon }_0}}\left({\displaystyle \frac{r}{3}}-{\displaystyle \frac{r^2}{4R}}\right)$$
D. $${\displaystyle \frac{p_0}{12{\epsilon }_0}}\left({\displaystyle \frac{r}{3}}-{\displaystyle \frac{r^2}{4R}}\right)$$

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:
$$E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q}{r^2}}$$
Here, $${\epsilon }_0$$ 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.
$$\int dq=q=\underset{0}{\overset{r}{\int }}p\left(r\right)4\pi r^{2}dr$$
$$\Rightarrow q=4\pi p_0\underset{0}{\overset{r}{\int }}\left(1-{\displaystyle \frac{r}{R}}\right)r^{2}dr$$
$$\Rightarrow q=4\pi p_0\underset{0}{\overset{r}{\int }}\left(r^2-{\displaystyle \frac{r^3}{R}}\right)dr$$
$$\Rightarrow q=4\pi p_0\left({\displaystyle \frac{r^3}{3}}+{\displaystyle \frac{r^4}{4R}}\right)$$ …… (1)
The electric field at a distance r from the centre of the sphere is,
$$E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q}{r^2}}$$
Here, $${\epsilon }_0$$ is the permittivity of the medium.
Substituting equation (1) in the above equation, we get,
$$E={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{1}{r^2}}\left(4\pi p_0\left({\displaystyle \frac{r^3}{3}}+{\displaystyle \frac{r^4}{4R}}\right)\right)$$
$$\Rightarrow E={\displaystyle \frac{p_0}{{\epsilon }_0}}\left({\displaystyle \frac{r}{3}}+{\displaystyle \frac{r^2}{4R}}\right)$$

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,
$$Eds={\displaystyle \frac{q_{enc}}{{\epsilon }_0}}$$
$$\Rightarrow E\left(4\pi r^2\right)={\displaystyle \frac{q_{enc}}{{\epsilon }_0}}$$,
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.