A non-conducting spherical shell, with an inner radius of $\mathbf{4}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$and an outer radius of $\mathbf{6}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$, has charge spread non-uniformly through its volume between its inner and outer surfaces. The volume charge density$\mathbf{\text{ρ}}$ is the charge per unit volume, with the unit coulomb per cubic meter. For this shell $\mathbf{\rho }\mathbf{=}\mathbf{b}\mathbf{/}\mathbf{r}$, where$\mathit{r}$ is the distance in meters from the center of the shell and$\mathbf{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.0}\mathbf{\text{\hspace{0.17em}μC}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$ .What is the net charge in the shell?

Inner radius of the non-conducting spherical shell,${\mathrm{r}}_1=4\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.04\mathrm{m}$

Outer radius of the shell${\mathrm{r}}_2=6\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.06\mathrm{m}$

The volume charge density$\mathrm{\rho }=\frac{\mathrm{b}}{\mathrm{r}}$

where,$\mathrm{r}$is the distance from the shell and$\mathrm{b}=3.0\mathrm{\mu C}/{\mathrm{m}}^2$

Using the concept of charge density which is given as charge per volume, we can get the desired result. This can also be given in the form of area and distance change by differentiation. Thus, by integrating the charge equation, we can get the net charge on the shell.

Formulae:

The charge density of a distribution can be given as:

$\begin{array}{l}\mathrm{\rho }=\mathrm{dq}/\mathrm{dV}\\ \mathrm{dq}=\mathrm{\rho Adr}\end{array}$ (1)

Area of a spherical shell,$\mathrm{A}=4{\mathrm{\pi r}}^2$ (2)

The charge$\mathrm{dq}$within a thin shell of thickness$\mathrm{dr}$is given by integrating equation (1) by substituting the value of area from equation (2) and the given value of charge density$\mathrm{\rho }=\frac{\mathrm{b}}{\mathrm{r}}$as follows:

$\begin{array}{c}\int \mathrm{dq}=4\mathrm{\pi b}\underset{\mathrm{r}1}{\overset{\mathrm{r}2}{}}\mathrm{r}\mathrm{dr}\\ \mathrm{q}=2\mathrm{\pi b}({\mathrm{r}}_2^2-{\mathrm{r}}_1^2)\\ \mathrm{q}=2\times \mathrm{\pi }\times 3.0\mathrm{\mu C}/{\mathrm{m}}^2\{{\left(0.06\right)}^2-{\left(0.04\right)}^2\}\\ \mathrm{q}=0.038\mathrm{\mu C}\end{array}$

Hence, the net charge on the shell is $0.038\mathrm{\mu C}$