# The volume charge density \rho_E within a sphere of radius r_0 is distributed in accordance with...

## Question:

The volume charge density {eq}\rho_E{/eq} within a sphere of radius {eq}r_0{/eq} is distributed in accordance with the following spherically symmetric relation {eq}\rho_E(r)=\rho_0[1-r^2/r_0^2]{/eq}, where *r* is measured from the center of the sphere and {eq}\rho_0{/eq} is a constant.

For a point P inside the sphere (r<{eq}r_0{/eq}), determine the electric potential V. Let V=0 at infinity. Give your answer in terms of *r*, {eq}r_0, \rho_0{/eq}, and electric constant {eq}\epsilon_0{/eq}. (Answer should be in V{eq}_r{/eq}=....)

## Electric Field:

An electric field is the effect of a charged particle in a certain region. It varies according to the charge and the distance of the particle. The region most commonly referred to as the electric field is also the same as the number of field lines surrounding a particle.

## Answer and Explanation:

We are given the following data:

- The radius of the sphere is {eq}{r_0}{/eq}.

- The distance of the point P is
*r*. - The volume charge density of the sphere is {eq}{\rho _E} = {\rho _0}\left( {1 - \dfrac{{{r^2}}}{{r_0^2}}} \right){/eq}.

The expression to calculate the volume of the sphere is:

{eq}V = \dfrac{4}{3}\pi r_0^3 {/eq}

The expression to calculate the charge of the sphere is:

{eq}Q = {\rho _E}V {/eq}

Substituting all the values into the above expression gives us:

{eq}\begin{align*} Q &= \left[ {{\rho _0}\left( {1 - \dfrac{{{r^2}}}{{r_0^2}}} \right)} \right]\left( {\dfrac{4}{3}\pi r_0^3} \right)\\ &= \dfrac{4}{3}\pi {\rho _0}\left[ {r_0^3 - {r^2}{r_0}} \right] \end{align*} {/eq}

The expression to calculate the surface area of the sphere is:

{eq}A = 4\pi r_0^2 {/eq}

Applying Gauss' law to calculate the electric field of the sphere results in:

{eq}EA = \dfrac{Q}{{{\varepsilon _0}}}{/eq}, where:

- {eq}{\varepsilon _0}{/eq} is the permittivity of the free space.

Substituting all the values into the above expression, we get:

{eq}\begin{align*} E\left( {4\pi r_0^2} \right) &= \dfrac{{\dfrac{4}{3}\pi {\rho _0}\left[ {r_0^3 - {r^2}{r_0}} \right]}}{{{\varepsilon _0}}}\\ E &= \dfrac{{{\rho _0}}}{{3{\varepsilon _0}}}\left( {{r_0} - \dfrac{{{r^2}}}{{{r_0}}}} \right) \end{align*} {/eq}

The expression to calculate the electric potential at the point P inside the sphere is:

{eq}{V_r} = E \cdot r {/eq}

Substituting all the values into the above expression gives us:

{eq}\begin{align*} {V_r} &= \left[ {\dfrac{{{\rho _0}}}{{3{\varepsilon _0}}}\left( {{r_0} - \dfrac{{{r^2}}}{{{r_0}}}} \right)} \right]r\\ &= \boxed{\dfrac{{{\rho _0}r}}{{3{\varepsilon _0}}}\left( {{r_0} - \dfrac{{{r^2}}}{{{r_0}}}} \right)} \end{align*} {/eq}

Thus, **the electric potential at the point P inside the sphere is** {eq}\mathbf{\dfrac{{{\rho _0}r}}{{3{\varepsilon _0}}}\left( {{r_0} - \dfrac{{{r^2}}}{{{r_0}}}} \right)} {/eq}.