Download CCR 20: Electrostatic Energy of a Sphere of Charge

CLASSICAL CONCEPT REVIEW 20
Electrostatic Energy
of a Sphere of Charge
Introductory physics textbooks typically show that the potential energy density u of
the electric field E is given by
u =
1
 E 2
2 0
SC-1
While this result is usually derived for a parallel-plate capacitor, it is actually quite
generally true. In this CCR section we will show how to obtain the electrostatic potential energy U for a ball or sphere of charge with uniform charge density r, such as that
approximated by an atomic nucleus.
Let us assume that the sphere has radius R and ultimately will contain a total
charge Q uniformly distributed throughout its volume. The electrostatic potential
energy U is equal to the work done in assembling the total charge Q within the volume, that is, the work done in bringing Q from infinity to the sphere. We can do this
by bringing a series of very small charges dq from infinity and spreading each over
the surface of a sphere whose infinitesimal thickness is from r to r 1 dr and whose
center coincides with that of the final sphere of radius R. We continue to bring these
small amounts of charge from infinity until we have assembled the total charge Q.
The work dW done in bringing up each increment of charge dq to the radius r is (see
Figure SC-1)
dW =
1 q1r2dq
r
4p0
SC-2
where q(r) is the amount of charge already assembled up to the radius r. That amount
of charge is
4 3
pr r
3
SC-3
dq = 4pr 2r dr
SC-4
q1r2 =
and therefore
Substituting q(r) and dq from Equations SC-3 and SC-4 into Equation SC-2 gives us
dW =
4 p 2 4
1 14pr 3r>32 14pr 2r dr2
=
r r dr
r
4p0
3 0
SC-5
60
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Classical Concept Review 20
61
dr
r
dq
R
SC-1 A small amounts of charge dq is moved
from infinity and spread uniformly over the
surface of the sphere of radius r, which contains
an amount of charge q(r) moved in similar
increments earlier. When r reaches the value R,
the sphere contains a total charge Q.
The total work done in assembling the sphere of charge to radius R is
R
4 p 2 4
4 pr2 5
W = 3
r r dr =
R 
15 0
0 3 0
SC-6
Since the total charge Q is
Q =
4 3
pR r
3
SC-7
and then
W =
=
=
4 pr2 5
R
p
3
4
R * a *
* * b
p
15 0
R
4
3
3 4 3 2 1
1
a pR rb a
ba b
5 3
4p0 R
3 1 Q2
5 4p0 R
Thus, the electrostatic potential energy of the spherical ball of charge is
U =
3 1 Q2
5 4p0 R
SC-8
EXAMPLE CR-1 Determine the electrostatic potential energy of the 56
26Fe nucleus.
TIPLER_CCR.indd 61
R = 1.07 A1>3 fm = 1.071562 1>3 fm = 4.1 * 10-15 m
U 1 56
26Fe2 =
3 1 126 * 1.60 * 10-19 C2 2
5 4p0
4.1 * 10-15 m
-11
U 1 56
J = 140 MeV
26Fe2 = 2.2 * 10
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