Question

Let there be a spherically symmetric charge distribution with charge density varying as $$\rho (r)={\rho }_o({\displaystyle \frac{5}{4}}-{\displaystyle \frac{r}{R}})$$ up to $r=R$, and $\rho (r)=0$ for $\text{r }$ > $R$, where $r$ is the distance from the origin. Find the electric field at a distance $r$, where $\text{r }$ < $R$, from the origin.
$$A){\displaystyle \frac{r{\rho }_0}{4{\epsilon }_0}}({\displaystyle \frac{5}{4}}-{\displaystyle \frac{r}{R}})$$
$$B){\displaystyle \frac{{\rho }_{o}r}{4{\epsilon }_o}}({\displaystyle \frac{5}{3}}-{\displaystyle \frac{r}{R}})$$
$$C){\displaystyle \frac{4{\rho }_{o}r}{3{\epsilon }_o}}({\displaystyle \frac{5}{4}}-{\displaystyle \frac{r}{R}})$$
$$D){\displaystyle \frac{{\rho }_{o}r}{4{\epsilon }_o}}({\displaystyle \frac{5}{4}}-{\displaystyle \frac{r}{R}})$$

Answer 1

Hint: To solve this problem, we need to consider a Gaussian surface, which is a three dimensional and closed surface. Such a surface is nothing but a surface through which the flux (usually gravitational field, magnetic field or electric field), is calculated. Here, in this problem, the Gaussian surface is considered to be having a radius of $r$, which is greater than the radius of the given spherical charge distribution.

 Formula used:
$$1)dQ=\rho .dv$$
$$2)E={\displaystyle \frac{Q}{4\pi r^2{\epsilon }_0}}$$

Complete answer:
The variation of charge density of the given spherically symmetric charge distribution is given as $$\rho (r)={\rho }_o({\displaystyle \frac{5}{4}}-{\displaystyle \frac{r}{R}})$$ up to $r=R$, and $\rho (r)=0$ for $r$ > $R$, where $r$ is the distance from the origin. We are required to determine the electric field at a distance $r(r$ < $R)$ from the origin. Clearly, $R$ is the radius of the spherical charge distribution.
Now, let’s consider a Gaussian surface of radius $r(r$ < $R)$ in the given charge distribution. We know that the total charge enclosed inside the Gaussian surface is given by,
$$dQ=\rho .dv$$
where
$dQ$ is the total charge enclosed by the Gaussian surface
$\rho$ is the charge density inside this surface
$dv$is the volume of this surface element
Let this be equation 1.
Integrating equation 1 from $0$(origin) to $r$ and applying the given conditions, we can determine the total charge enclosed in the required charge distribution as follows:
$$\int dQ=\int \rho .dv$$
$$\Rightarrow Q=\underset{0}{\overset{r}{\int }}{\rho }_0[{\displaystyle \frac{5}{4}}-{\displaystyle \frac{r}{R}}]4\pi r^{2}dr$$
$$\Rightarrow Q={\rho }_{0}4\pi \underset{0}{\overset{r}{\int }}({\displaystyle \frac{5r^2}{4}}-{\displaystyle \frac{r^3}{R}})dr$$
$$\Rightarrow Q={\displaystyle \frac{{\rho }_{0}4\pi }{4}}{[{\displaystyle \frac{5r^3}{3}}-{\displaystyle \frac{r^4}{R}}]}_{}^{}$$
$$\Rightarrow Q={\rho }_o\pi {[{\displaystyle \frac{5r^3}{3}}-{\displaystyle \frac{r^4}{R}}]}_{}^{}$$
Let this be equation 2.
Now, according to Gauss’s law, electric flux or electric field through a Gaussian surface is given by
$$E={\displaystyle \frac{Q}{4\pi r^2{\epsilon }_0}}$$
Let this be equation 3.
Substituting equation 2 in equation 4, we have
$$E={\displaystyle \frac{Q}{4\pi r^2{\epsilon }_0}}={\displaystyle \frac{{\rho }_o\pi }{4\pi r^2{\epsilon }_0}}[{\displaystyle \frac{5r^3}{3}}-{\displaystyle \frac{r^4}{R}}]\Rightarrow E={\displaystyle \frac{r{\rho }_0}{4{\epsilon }_0}}[{\displaystyle \frac{5}{3}}-{\displaystyle \frac{r}{R}}]$$
Let this be equation 4.
Therefore, from equation 4, we can conclude that the required electric field at a distance $r$ from the origin is equal to $${\displaystyle \frac{{\rho }_{o}r}{4{\epsilon }_o}}({\displaystyle \frac{5}{3}}-{\displaystyle \frac{r}{R}})$$

Hence, the correct answer is option $B$.

Note:
Students need to be thorough with the formula for volume charge density, which uses a volume element $dv$. At the same time, the formula for electric flux through a Gaussian surface also needs to be remembered. Whenever we encounter problems related to electric/magnetic flux through surfaces, we can utilize the concept of Gaussian surfaces, as utilized in the above solution.