Physics 9 Fall 2011 Homework 4 - Solutions Friday September 16 ...

3. Calculate the potential inside and outside a sphere of radius R and charge Q, in which
the the charge is distributed uniformly throughout the sphere. (Hint: recall the electric
fields inside and outside a uniformly charged sphere from Gauss’s law, and don’t forget
that the potentials must be continuous at the surface of the sphere, where r = R.)
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Solution
The electric field outside a uniformly charged sphere is just the ordinary Coulomb law,
Eout = 1
4πɛ0
Q
,
r2 while inside, as we have seen from Gauss’s law, the field is
Ein = 1
4πɛ0
Q
r.
R3 Now, the potential outside the sphere is simply the electrostatic potential of a point
charge,
Vout = 1 Q
4πɛ0 r .
The potential inside the sphere is a bit harder. We can find it by integrating the
electric field, looking for the total work done in bringing in a unit charge from infinity
and placing it at some position inside the sphere. Thus,
r
Vin(r) = −
∞
R r
E · ds = − Eout · ds −
∞
R
Ein · ds,
where we have split up the integral because the electric field is different in the two
regions. So, plugging in the fields and integrating over radial distance, r, gives
Vin(r) = − Q
4πɛ0
R
∞
dr Q
−
r2 4πɛ0R3 r
r dr =
R
1 Q 1
−
4πɛ0 R 4πɛ0
Q
2R3 2 2
r − R .
Simplifying and combining the terms gives our final result for the potential,
Vin(r) = Q
8πɛ0R
Note that Vin(R) = Vout(R), as we require.
4
3 − r2
R 2
.