91Ó°ÊÓ

a) A charged particle is held at the center of the shell.

b) The scale of the vertical axis,${\mathrm{E}}_{\mathrm{s}}=10×{10}^7 \mathrm{N}/\mathrm{C} $

Using the concept of the electric field on a body due to a charge, we can get the respective charges on the inner and the outer surfaces of the shell. By subtracting the calculated difference between the charges, we can get the net charge on the shell.

Formula:

The electric field on a point due to a particle charge, $\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$ (1)

The electric field is 0 between the regions $\mathrm{r}=2.5 \mathrm{cm}$to $\mathrm{r}=3.0 \mathrm{cm}$. It shows that the shell is conducting in nature.

Therefore from the graph, the inner radius of the shell is ${\mathrm{r}}_{\mathrm{i}}=2.5 \mathrm{cm}$ and E at this radius is ${\mathrm{E}}_{\mathrm{i}}=2.0×{10}^7\mathrm{N}/\mathrm{C}$,

By using the formula for the Electric Field of equation (1), we get the value of the inner charge of the shell as follows:

$$\begin{gathered} {\mathrm{q}}_i=\left(4{\mathrm{Ï€Î\mu }}_0{{\mathrm{r}}_0}^2\right)×{\mathrm{E}}_{\mathrm{i}} \\ =\frac{{\left(0.025\mathrm{m}\right)}^2\left(2.0×{10}^7\mathrm{N}/\mathrm{C}\right)}{\left(9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)} \\ =1.4×{10}^{-6}\mathrm{C} \end{gathered}$$

Similarly, with the values of the outer radius of the shell, ${\mathrm{r}}_{\mathrm{o}}=3.0 \mathrm{cm}$ and electric field,

${\mathrm{E}}_{\mathrm{o}}=8.0×{10}^7 \mathrm{N}/\mathrm{C}$

Thus, the outer charge of the shell is given using equation (i) as follows:

$$\begin{gathered} {\mathrm{q}}_0=\left(4{\mathrm{Ï€Î\mu }}_0{{\mathrm{r}}_0}^2\right)×\mathrm{E} \\ =\frac{{\left(0.030\mathrm{m}\right)}^2\left(8.0×{10}^7\mathrm{N}/\mathrm{C}\right)}{9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2} \\ =8.0×{10}^{-6}\mathrm{C} \end{gathered}$$

Net charge on the shell using the above values can be given as follows:

$$\begin{gathered} \mathrm{Q}={\mathrm{q}}_0-{\mathrm{q}}_{\mathrm{i}} \\ =\left(8.0×{10}^6\mathrm{C}\right)-\left(1.4×{10}^{-6}\mathrm{C}\right) \\ =6.6×{10}^{-6}\mathrm{C} \end{gathered}$$

Hence, the value of the net charge is $6.6×{10}^{-6}\mathrm{C}$.