Question

A ball of radius \[R\] carries a positive charge whose volume charge density depends only on the distance \[r\] from the ball's centre as: \[\rho = {\rho _0}(1 - \dfrac{r}{R})\] where \[{\rho _0}\] is constant. Assume \[\varepsilon \] as the permittivity of the ball. Then the magnitude of the electric field as a function of the distance \[r\] outside the ball is given by :
A. \[E = \dfrac{{{\rho _0}{R^3}}}{{8\varepsilon {r^2}}}\]
B. \[E = \dfrac{{{\rho _0}{R^3}}}{{12\varepsilon {r^2}}}\]
C. \[E = \dfrac{{{\rho _0}{R^3}}}{{16\varepsilon {r^2}}}\]
D. \[E = \dfrac{{{\rho _0}{R^3}}}{{24\varepsilon {r^2}}}\]

Answer 1

Hint: The electric field due to a charge at a point is the force acting on a unit positive charge when it is placed at that point. Electric field due to continuous charge distribution is the integration over the whole charge distribution. Gauss’s law states that the total flux due to a volume is proportional to the charge enclosed.

Formula used:
Gauss’s law is given by,
\[\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{q}{{{\varepsilon _0}}}\]
where, \[\vec E\] is the electric field, \[S\] is the closed surface placed in the electric field.
Total charge of continuous charge distribution,
\[q = \int {\rho dv} \]
where, \[\rho \] is the volume charge density and \[dv\]is the volume element.

Complete step by step answer:
We know that the electric field at a point due to charge is nothing but the force acting on a unit positive charge due to the charge at that point. For multiple charges since the electric field follows the superposition principle it is the sum of the electric field due to each charge. For a continuous charge distribution it is the integration over the source points. Here, we have a ball of radius \[R\]which carries a charge distribution \[\rho = {\rho _0}(1 - \dfrac{r}{R})\].

Now, Gauss’s law states that the total electric flux due to a closed surface is equal to the \[\dfrac{1}{\varepsilon }\] times charge enclosed by the surface. Mathematically,
\[\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{q}{\varepsilon }\]
where, \[\vec E\] is the electric field, \[S\] is the closed surface placed in the electric field.
Here, the total charge is the integration over the whole distribution. So, from gauss law we can have,
\[\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{q}{\varepsilon }\]
\[\Rightarrow \oint\limits_s {\vec E} .d\vec S = \dfrac{{\int {\rho dv} }}{\varepsilon }\]
Since the charge distribution is symmetric the electric is also uniform.
\[E\oint\limits_s {dS} = \dfrac{1}{\varepsilon }\int {{\rho _0}(1 - \dfrac{r}{R})4\pi {r^2}dr} \]
\[ \Rightarrow E \cdot 4\pi {r^2} = \dfrac{1}{\varepsilon }\int\limits_0^R {{\rho _0}(1 - \dfrac{r}{R})4\pi {r^2}dr} \]
\[ \Rightarrow E \cdot 4\pi {r^2} = \dfrac{{4\pi }}{\varepsilon }\int\limits_0^R {{\rho _0}{r^2}dr - \int\limits_0^R {{\rho _0}\dfrac{{{r^3}}}{R}dr} } \]

Integrating and putting the limits we have,
\[E \cdot 4\pi {r^2} = \dfrac{{4\pi }}{\varepsilon }\left( {\dfrac{{{\rho _0}{R^3}}}{3} - {\rho _0}\dfrac{{{R^4}}}{{4R}}} \right)\]
\[\Rightarrow E \cdot {r^2} = \dfrac{1}{\varepsilon }\dfrac{{{\rho _0}{R^3}}}{{12}}\]
\[\therefore E = \dfrac{{{\rho _0}{R^3}}}{{12\varepsilon {r^2}}}\]
Hence, the electric field outside the ball is given by, \[E = \dfrac{{{\rho _0}{R^3}}}{{12\varepsilon {r^2}}}\]

Hence, option B is the correct answer.

Note: Gauss’s law can be applied to find any electric field when the charge distribution is symmetric. If the distribution is symmetric the electric field due to the distribution is uniform. For a uniform electric field it becomes a constant of integration and we can easily perform the integration to find the electric field.