Electrostatic Energy of a Sphere of Charge [PDF]

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

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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.



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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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