Unformatted text preview: 45. To ﬁnd an expression for the electric ﬁeld inside the shell in terms of A and the distance from the center
of the shell, select A so the ﬁeld does not depend on the distance. We use a Gaussian surface in the form
of a sphere with radius rg , concentric with the spherical shell and within it (a < rg < b). Gauss’ law will
be used to ﬁnd the magnitude of the electric ﬁeld a distance rg from the shell center. The charge that
is both in the shell and within the Gaussian sphere is given by the integral qs = ρ dV over the portion
of the shell within the Gaussian surface. Since the charge distribution has spherical symmetry, we may
take dV to be the volume of a spherical shell with radius r and inﬁnitesimal thickness dr : dV = 4πr2 dr .
Thus,
r
r
r
g
g
g
A2
2
qs = 4π
ρr2 dr = 4π
r dr = 2πA(rg − a2 ) .
r dr = 4πA
r
a
a
a
2
The total charge inside the Gaussian surface is q + qs = q + 2πA(rg − a2 ). The electric ﬁeld is radial, so
2
the ﬂux through the Gaussian surface is Φ = 4πrg E , where E is the magnitude of the ﬁeld. Gauss’ law
yields
2
2
4πε0 Erg = q + 2πA(rg − a2 ) . We solve for E :
E= 2πAa2
1
q
+ 2πA −
2
2
4πε0 rg
rg . For the ﬁeld to be uniform, the ﬁrst and last terms in the brackets must cancel. They do if q − 2πAa2 = 0
or A = q/2πa2 . ...
View
Full
Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details