Question

Two charged concentric spherical shells have radii 10.0cmand 15.0cm . The charge on the inner shell is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{C}$ , and that on the outer shell is$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{C}$. Find the electric field (a) at$\mathit{r}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$and (b) at$\mathit{r}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$.

Short answer

a) The value of the electric field at $\mathrm{r}=12\mathrm{cm}\mathrm{is}2.50\times {10}^4\mathrm{N}/\mathrm{C}$b) The value of the electric field at $\mathrm{r}=20\mathrm{cm}\mathrm{is}1.35\times {10}^4\mathrm{N}/\mathrm{C}$.

Step-by-step solution

The given data

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, ${\mathrm{q}}_{\mathrm{i}}=4\times {10}^{-8}\mathrm{C}$

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

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

Understanding the concept of the electric field

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,$\mathrm{E}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}$ (1)

a) Calculation of the electric field at r = 12 cm

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:

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

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

b) Calculation of the electric field at r = 20 cm

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:

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

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