A metal sphere with radius ${\mathit{r}}_{\mathbf{a}}$is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius ${\mathit{r}}_{\mathbf{b}}$. There is charge +q on the inner sphere and charge -q on the outer spherical shell. (a) Calculate the potential V(r) for (i)$\mathit{r}\mathbf{<}{\mathit{r}}_{\mathbf{a}}$; (ii)localid="1665125184425" ${\mathit{r}}_{\mathbf{a}}\mathbf{<}\mathit{r}\mathbf{<}{\mathit{r}}_{\mathbf{b}}$; (iii)$\mathit{r}\mathbf{<}{\mathit{r}}_{\mathbf{b}}$. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take V to be zero when r is infinite.

(b) Show that the potential of the inner sphere with respect to the outer is

${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$

(c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude

$\mathit{E}\left(r\right)\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{a}\mathbf{b}}}{\mathbf{(}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{a}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{r}}_{\mathbf{b}}}\mathbf{)}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}$

(d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance r from the center, where $\mathit{r}\mathbf{>}{\mathit{r}}_{\mathbf{b}}$. (e) Suppose the charge on the outer sphere is not -q but a negative charge of different magnitude, say -Q. Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

For $r<r_a:$

The enclosed charge is equal to zero so the potential difference is constant and is equal to the potential at radius $r_a$but if it is relative to $V_b$then:

$V_a=V_a+V_b=\frac{kq}{r_a}-\frac{kq}{r_b}=kq\left[\frac{1}{r_a}-\frac{1}{r_b}\right]$

For localid="1668176138194" $r_a<r<r_b$:

The enclosed charge is equal to q so the potential difference is inversely proportional to r if it is relative to then:

$V_{rb}=V\left(r\right)+V_b=\frac{kq}{r}-\frac{kq}{r_b}=kq\left[\frac{1}{r}-\frac{1}{r_b}\right]$

For $r>r_b$:

The enclosed charge is equal to zero so the potential difference is constant and is given by:

$V\left(r\right)=C$

To get that constant, finding a valid condition, the condition which is valid for any case is the potential at infinity is equal to zero:

$V\left(\infty \right)=0$

So from (1) and (2) get the following:

$V\left(r\right)=C=0$.

The electric potential with respect to other sphere is the sum of them:

$V_{ab}=\frac{kq}{r_a}-\frac{kq}{r_b}=kq\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$

The electric field is equal to the negative gradient of the potential so the following can be written as,

$$\begin{gathered} E=-∆V=-\frac{\partial V}{\partial r} \\ =-kq\frac{\partial \left({\displaystyle \frac{1}{r}}-\frac{1}{r_b}\right)}{\partial r} \\ =\frac{kq}{r^2}=\frac{1}{r^2}\left(\frac{V_{ab}}{{\displaystyle \frac{1}{r_a}}-\frac{1}{r_b}}\right) \end{gathered}$$

The electric field is the same as steps in part © and is given by

$$\begin{gathered} E=-∆V=-\frac{\partial V}{\partial r} \\ =\frac{dC}{dr}=0 \end{gathered}$$

The electric field for $r<r_b$is the same because the electric enclosed charge is the same so to prove it:

$V_{ab}=-\underset{b}{\overset{a}{\int }}\frac{kq}{r^2}dr=\frac{kq}{r_a}-\frac{kq}{r_b}$

Similarly for part (c).

In part (d) the enclosed charge is not q anymore but the net enclosed charge is $q_n=q-Q$so the field is not zero but it depends on values of charges:

$E=\frac{k{q}_n}{r^2}=\frac{k\left(q-Q\right)}{r^2}$.