Question

Let there be a spherically symmetric charge distribution with charge density varying as \[\rho (r)={{\rho }_{o}}\left( \dfrac{5}{4}-\dfrac{r}{R} \right)\] 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.
\[A)\dfrac{r{{\rho }_{0}}}{4{{\varepsilon }_{0}}}\left( \dfrac{5}{4}-\dfrac{r}{R} \right)\]
\[B)\dfrac{{{\rho }_{o}}r}{4{{\varepsilon }_{o}}}\left( \dfrac{5}{3}-\dfrac{r}{R} \right)\]
\[C)\dfrac{4{{\rho }_{o}}r}{3{{\varepsilon }_{o}}}\left( \dfrac{5}{4}-\dfrac{r}{R} \right)\]
\[D)\dfrac{{{\rho }_{o}}r}{4{{\varepsilon }_{o}}}\left( \dfrac{5}{4}-\dfrac{r}{R} \right)\]

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:
\[1)dQ=\rho .dv\]
\[2)E=\dfrac{Q}{4\pi {{r}^{2}}{{\varepsilon }_{0}}}\]

Complete answer:
The variation of charge density of the given spherically symmetric charge distribution is given as \[\rho (r)={{\rho }_{o}}\left( \dfrac{5}{4}-\dfrac{r}{R} \right)\] 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:
\[\int dQ=\int \rho .dv\]
\[\Rightarrow Q=\int\limits_{0}^{r}{{{\rho }_{0}}\left[ \dfrac{5}{4}-\dfrac{r}{R} \right]4\pi {{r}^{2}}dr}\]
\[\Rightarrow Q={{\rho }_{0}}4\pi \int\limits_{0}^{r}{\left( \dfrac{5{{r}^{2}}}{4}-\dfrac{{{r}^{3}}}{R} \right)}dr\]
\[\Rightarrow Q=\dfrac{{{\rho }_{0}}4\pi }{4}\left[ \dfrac{5{{r}^{3}}}{3}-\dfrac{{{r}^{4}}}{R} \right]_{{}}^{{}}\]
\[\Rightarrow Q={{\rho }_{o}}\pi \left[ \dfrac{5{{r}^{3}}}{3}-\dfrac{{{r}^{4}}}{R} \right]_{{}}^{{}}\]
Let this be equation 2.
Now, according to Gauss’s law, electric flux or electric field through a Gaussian surface is given by
\[E=\dfrac{Q}{4\pi {{r}^{2}}{{\varepsilon }_{0}}}\]
Let this be equation 3.
Substituting equation 2 in equation 4, we have
\[E=\dfrac{Q}{4\pi {{r}^{2}}{{\varepsilon }_{0}}}=\dfrac{{{\rho }_{o}}\pi }{4\pi {{r}^{2}}{{\varepsilon }_{0}}}\left[ \dfrac{5{{r}^{3}}}{3}-\dfrac{{{r}^{4}}}{R} \right]\Rightarrow E=\dfrac{r{{\rho }_{0}}}{4{{\varepsilon }_{0}}}\left[ \dfrac{5}{3}-\dfrac{r}{R} \right]\]
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 \[\dfrac{{{\rho }_{o}}r}{4{{\varepsilon }_{o}}}\left( \dfrac{5}{3}-\dfrac{r}{R} \right)\]

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.