91Ó°ÊÓ

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

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

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

where,$\mathrm{r}$is the distance from the shell and$\mathrm{b}=3.0\text{ }\mathrm{μ°ä}/{\mathrm{m}}^2$

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:

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

Area of a spherical shell,$\mathrm{A}=4{\mathrm{Ï€°ù}}^2$ (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$\mathrm{ÒÏ}=\frac{\mathrm{b}}{\mathrm{r}}$as follows:

$\begin{array}{c}∫\mathrm{dq}=4\mathrm{Ï€²ú}\underset{\mathrm{r}1}{\overset{\mathrm{r}2}{}}\mathrm{r}\mathrm{dr}\\ \mathrm{q}=2\mathrm{Ï€²ú}({\mathrm{r}}_2^2−{\mathrm{r}}_1^2)\\ \mathrm{q}=2×\mathrm{Ï€}×3.0\text{ }\mathrm{μ°ä}/{\mathrm{m}}^2\{{\left(0.06\right)}^2−{\left(0.04\right)}^2\}\\ \mathrm{q}=0.038\text{ }\mathrm{μ°ä}\end{array}$

Hence, the net charge on the shell is $0.038\text{ }\mathrm{μ°ä}$