91Ó°ÊÓ

a) Radii of the two concentric shells are${\mathrm{r}}_{\mathrm{i}}=10\mathrm{cm}\mathrm{and}{\mathrm{r}}_{\mathrm{f}}=15\mathrm{cm}$

b) Charge on the inner shell, ${\mathrm{q}}_{\mathrm{i}}=4×{10}^{-8}\mathrm{C}$

c) Charge on the outer shell, ${\mathrm{q}}_0=4×{10}^{-8}\mathrm{C}$

d) Distances $\mathrm{r}=12\mathrm{cm}\mathrm{and}\mathrm{r}=20\mathrm{cm}$.

Using the concept of the electric field, we can get the value of the individual fields at the respective radial distances. But, in the case of the concentric shells, the net charge acting on the point depends on the distance at which the point lies. For a point within the shell boundaries, the charge on the inner shell acts on the point while for a point outside the outer shell, the charges on both the inner and outer shells act on the point.

Formula:

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

Since ${\mathrm{r}}_1=10.0\mathrm{cm}<\mathrm{r}=12.0\mathrm{cm}<\mathrm{r}15.0\mathrm{cm}$the electric field acting on a point is given using equation (1) as:

$$\begin{gathered} \mathrm{E}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{q}}_1}{{\mathrm{r}}^2} \\ =\frac{(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2)(4.0×{10}^{-8}\mathrm{C})}{{(0.120m)}^2} \\ =2.50×{10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $=2.50×{10}^4\mathrm{N}/\mathrm{C}$.

Since${\mathrm{r}}_1<{\mathrm{r}}_2<\mathrm{r}=20.0\mathrm{cm}$ the electric field acting on a point is given using equation (1) as:

$$\begin{gathered} \mathrm{E}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{q}}_1+{\mathrm{r}}_2}{{\mathrm{r}}^2} \\ =\frac{(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2)(6.0×{10}^{-8}\mathrm{C})}{{(0.20m)}^2} \\ =1.35×{10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is role="math" localid="1657353357117" $1.35×{10}^4\mathrm{N}/\mathrm{C}$.