A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho (r)\) given as follows:

\(\rho (r) = {\rho _0}\left( {1 - \frac{{4r}}{{3R}}} \right)\) for \(r \le R\)

\(\rho (r) = 0\) for \(r \ge R\)

Where \({\rho _0}\) is a positive constant. (a) Find the total charge contained in the charge distribution. Obtain an expression for the electric field in the region (b) \(r \ge R\) ; (c) \(r \le R\). (d) Graph the electric-field magnitude \(E\) as a function of \(r\). (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho (r)\)

It states that the line integral of the electric field over a closed surface is equal to\(\frac{1}{{{\varepsilon _0}}}\)times the total charge enclosed by the closed surface. It is expressed as,

\(\int {\vec E \cdot d\vec s} = \frac{{{q_{{\rm{enc}}}}}}{{{\varepsilon _0}}}\) …(i)

Where, \({\varepsilon _0}\) is permittivity of free space, \({q_{{\rm{enc}}}}\) is the charge enclose by the Gaussian surface, \(\vec E\) is the electric field vector and \(d\vec s\) is the surface area vector

Volume of the spherically symmetric surface is as follow:

\(V = \frac{4}{3}\pi {R^3}\)

The relation between the charge density and the volume charge density for a small region is as follow:

\(dq = \rho dV\)

Where, \(\rho \) is the volume charge density and \(dV\) is the volume in the small region of space

Total charge enclosed is determined by,

\(\begin{aligned}dq &= \rho dV\\q &= \int\limits_0^R {\rho dV} \\ &= \int\limits_0^R {\rho \left( {4\pi {r^2}} \right)dr} \end{aligned}\)

Substitute \(\rho = {\rho _0}\left( {1 - \frac{{4r}}{{3R}}} \right)\)

\(\begin{aligned}q &= \int\limits_0^R {\rho \left( {4\pi {r^2}} \right)dr} \\ &= \int\limits_0^R {{\rho _0}\left( {1 - \frac{{4r}}{{3R}}} \right)\left( {4\pi {r^2}} \right)dr} \\ &= {\rho _0}4\pi \int\limits_0^R {\left( {1 - \frac{{4{r^3}}}{{3R}}} \right)dr} \\ &= {\rho _0}4\pi \left( {\frac{{{r^3}}}{3} - \frac{{4{r^4}}}{{3R \times 4}}} \right)_0^R\\ &= {\rho _0}4\pi \left( {\frac{{{R^3}}}{3} - \frac{{{R^3}}}{3}} \right)\\ &= 0\,{\rm{C}}\end{aligned}\)

Hence the total charge contained in the charge distribution is \(0\,{\rm{C}}\)