Equipotential Surfaces and Potential Gradient Problems and Solutions 1

Q#23.47

A metal sphere with radius $r_a$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$. There is charge +q on the inner sphere and charge $-q$ on the outer spherical shell. 

(a) Calculate the potential V(r) for 

(i) r < $r_a$

(ii) $r_a$ < r < $r_b$

(iii) r > $r_b$

(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 

$V_{ab}=\frac{q}{4\pi \epsilon_0}\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$

(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 

$E(r)=\frac{V_{ab}}{(1/r_a-1/r_b)}\frac{1}{r^2}$

(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 $r>r_b$.

(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.

Answer:

For a solid metal sphere or for a spherical shell, $V=\frac{kq}{r}$ outside the sphere and $V=\frac{kq}{R}$ at all points inside the sphere, where R is the radius of the sphere. 

When the electric field is radial, $E=-\frac{\partial V }{\partial r}$ 

(a) (i) r < $r_a$: This region is inside both spheres.

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

(ii) $r_a$ < r < $r_b$: This region is outside the inner shell and inside the outer shell. 

$V=\frac{kq}{r}-\frac{kq}{r_b}=kq\left(\frac{1}{r}-\frac{1}{r_b}\right)$

(iii) r > $r_b$: This region is outside both spheres and V = 0 since outside a sphere the potential is the same as for a point charge. 

Therefore the potential is the same as for two oppositely charged point charges at the same location. These potentials cancel. 

(b) $V_a=\frac{1}{4\pi \epsilon_0}\left(\frac{q}{r_a}-\frac{q}{r_b}\right)$ and $V_b$ = 0

$V_{ab}=\frac{1}{4\pi \epsilon_0}q\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$

(c) Between the spheres $r_a$ < r < $r_b$ and $V=kq\left(\frac{1}{r}-\frac{1}{r_b}\right)$

$E=-\frac{\partial V }{\partial r}=-\frac{q}{4\pi \epsilon_0}\frac{\partial}{\partial r}\left(\frac{1}{r}-\frac{1}{r_b}\right)$

$E=+\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

$E=\frac{V_{ab}}{\left(\frac{1}{r_1}-\frac{1}{r_b}\right)}\frac{1}{r^2}$

(d) From Eq. $E=-\frac{\partial V }{\partial r}$: E = 0, since V is constant (zero) outside the spheres. 

(e) If the outer charge is different, then outside the outer sphere the potential is no longer zero but is

$V=\frac{1}{4\pi \epsilon_0}\left(\frac{q}{r}-\frac{Q}{r}\right)=\frac{1}{4\pi \epsilon_0}\frac{(q-Q)}{r}$ 

All potentials inside the outer shell are just shifted by an amount 

$V=\frac{1}{4\pi \epsilon_0}\frac{Q}{r_b}$ 

Therefore relative potentials within the shells are not affected. Thus (b) and (c) do not change.

However, now that the potential does vary outside the spheres, there is an electric field there: 

$E=-\frac{\partial V }{\partial r}=-\frac{\partial}{\partial r}\left(\frac{kq}{r}+\frac{-kQ}{r}\right)$

$E=\frac{kq}{r^2}\left(1-\frac{Q}{r}\right)=\frac{kq}{r^2}(q-Q)$

 In part (a) the potential is greater than zero for all r < $r_b$.

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