Today in Physics 217: charged spheres
‰ Finish example from
Wednesday.
‰ Field from a charged
spherical shell,
z
calculated with
Coulomb’s Law.
‰ Same, calculated with
Gauss’ Law.
‰ Analogy with
gravity: Gauss’ Law
for gravity.
x
20 September 2002
dE
dq = σ da′
r
da′ = R 2 sin θ ′dθ ′dφ ′rˆ
θ′
R
Physics 217, Fall 2002
φ′
y
1
Using Coulomb’s Law:
dE
Break the plane into annuli with
radius r and width dr, and break the
annuli into segments of width rdφ .
The charge of each segment is
z
dq = σ rdrdφ
r
α
φ
Horizontal components
dq
of field from segments
at φ and φ+π cancel,
and their vertical components add, so above the plane, we have:
σr
ˆ
dE = 2 2 cos α z = 2 2
z + r2
r
dq
π
∞
(
E = 2σ zzˆ ∫ dφ ∫ r r 2 + z2
0
20 September 2002
0
)
−3 2
z
z2 + r 2
∞
drdφ zˆ
dr = πσ zzˆ ∫ u− 3 2 du = πσ zzˆ
z2
Physics 217, Fall 2002
−1 2 ∞
u
−1 2
= 2πσ zˆ
z2
2
Using Gauss’ Law
E
First note that E must point perpendicular
to, and away from, the plane, since
r
the plane is infinite and there’s
z
no difference between the
view to the right and the
z
view to the left. Then
draw a cylinder,
bisected by the plane. By symmetry, E is perpendicular to the area
element vectors on the cylinder walls, parallel to those on the circular
faces, and constant on those faces, so
∫
E ⋅ da = 2 Eπ r 2 = 4π Qenclosed = 4π 2 r 2σ , or
E = ±2πσ zˆ .
Harder setup (finding and exploiting symmetry), easier math.
20 September 2002
Physics 217, Fall 2002
3
Electric fields from spherically-symmetrical charge
distributions
Today we will prove two important, though perhaps
intuitively obvious, facts about spherical charge distributions:
‰ The field outside a uniformly-charged spherical shell is
the same as that from a point charge of the same
magnitude, the same distance away as the sphere’s center.
‰ The field inside a uniformly-charged spherical shell is
zero.
The proof will serve also as another useful example of the
application of Coulomb’s and Gauss’ laws to the
determination of electric fields from specified charge
distributions.
20 September 2002
Physics 217, Fall 2002
4
Coulomb’s Law example: field from a uniformlycharged spherical shell
dE
Griffiths, problem 2.7:
What is the electric field
a distance z away from
the center of a spherical
shell with radius R and z
uniform surface charge
density σ?
dq = σ da′
r
da′ = R 2 sin θ ′dθ ′dφ ′rˆ
θ′
R
φ′
y
x
20 September 2002
Physics 217, Fall 2002
5
Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
dE
In the plane at azimuth φ, it
can be seen more easily that
2
r 2 = R 2 sin 2 θ ′ + ( z − R cosθ ′ )
dE (φ )
dE (φ + π )
= R 2 sin 2 θ ′ + z2 + R 2 cos 2 θ ′ − 2 Rz cosθ ′
α
= R 2 + z2 − 2 Rz cosθ ′
R sin θ
dq at φ + π
Consider two area elements at
azimuth φ and φ + π: as before,
the horizontal components of
their contribution to E cancel,
and the vertical components
add.
20 September 2002
Physics 217, Fall 2002
R cosθ
θ
r
dq at φ
R
6
Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
So dE = zˆ 2
= zˆ 2
dq
r
2
cos α
σ R 2 sin θ ′dθ ′dφ ′
z − R cosθ ′
R 2 + z 2 − 2 Rz cosθ ′
= zˆ 2σ R 2
E = zˆ 2σ R
2
(
π
R 2 + z2 − 2 Rz cosθ ′
sin θ ′ ( z − R cosθ ′ )
′dφ ′
d
θ
32
2
2
R + z − 2 Rz cosθ ′
)
π
∫ dφ ′∫
0
0
(
sin θ ′ ( z − R cosθ ′ )
R 2 + z2 − 2 Rz cosθ ′
)
32
dθ ′
The first integral is trivial: it just comes out to π.
20 September 2002
Physics 217, Fall 2002
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Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
For the second, substitute
w = cosθ ′, dw = − sin θ ′dθ ′, w = 1 → −1 :
E = zˆ 2πσ R 2
1
∫
−1
(
( z − Rw )
R 2 + z2 − 2 Rzw
)
32
dw
Break this integral in two. For the first one, substitute
u = R 2 + z2 − 2 Rzw , du = −2 Rzdw ,
u = R 2 + z2 + 2 Rz → R 2 + z2 − 2 Rz
1
z
∫
−1
(R
1
2
20 September 2002
2
+ z − 2 Rzw
)
32
1
dw =
2R
R 2 + z2 + 2 Rz − 3 2
u
du
2
2
R + z − 2 Rz
∫
Physics 217, Fall 2002
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Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
1
2R
R 2 + z 2 + 2 Rz − 3 2
u
du
2
2
R + z − 2 Rz
∫
2
2
1 
− 1 2  R + z + 2 Rz
=
−2 u

 R 2 + z2 − 2 Rz
2R

1
1
1
= 
−

R  R 2 + z2 − 2 Rz
R 2 + z2 + 2 Rz 
The second half of the integral needs to be done by parts.
Take
− Rdw
u=w
dv =
−3 2
2
2
R + z − 2 Rzw
(
du = dw v =
20 September 2002
1
z
)
1
R 2 + z2 − 2 Rzw
Physics 217, Fall 2002
(as we just saw)
9
Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
∫ udv = uv C − ∫ vdu
C
1
∫
−1
C
(R
− Rw
2
2
+ z − 2 Rzw
)
32
1
dw =
z
1
w
R 2 + z2 − 2 Rzw
1
−
z
1
∫
−1
−1
dw
R 2 + z2 − 2 Rzw
In the last term, use (again)
u = R 2 + z2 − 2 Rzw , du = −2 Rzdw ,
u = R 2 + z2 + 2 Rz → R 2 + z2 − 2 Rz
20 September 2002
Physics 217, Fall 2002
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Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
and it becomes
1
−
z
1
∫
−1
dw
2
2
R + z − 2 Rzw
=−
1
2 Rz2
R 2 + z2 + 2 Rz − 1 2
u
du
2
2
R + z − 2 Rz
∫
R 2 + z2 + 2 Rz
1
2 u  2 2
 R + z − 2 Rz
2 Rz2 
1
= − 2 R 2 + z2 + 2 Rz − R 2 + z2 − 2 Rz
Rz
=−
(
So, putting all these terms together (and factoring out 1 z2
as we do), we get
20 September 2002
Physics 217, Fall 2002
11
)
Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
2πσ R 2  z2 
1
1
ˆ


E=z
−
2
2
2
 R  R 2 + z2 − 2 Rz
z
R
z
+
+ 2 Rz



1
1

+ z
+
 R 2 + z2 − 2 Rz

2
2
2
R
z
Rz
+
+


1

R 2 + z2 + 2 Rz − R 2 + z2 − 2 Rz 
−
R

(




)
This looks like a mess until you notice that
R + z + 2 Rz =
( z + R)
2
= z+R
R 2 + z 2 − 2 Rz =
( z − R )2
= z−R
2
20 September 2002
2
Physics 217, Fall 2002
Positive, since they
represent the length of r,
which is always positive.
12
Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
2πσ R 2  z2
E = zˆ

2
z
 R
 1
1
−

 z−R z+R



 1
1  1

z
R
z
R
+ z 
+
−
+
−
−
(
)


−
+
z
R
z
R
R


2πσ R 2  z2  1
1 
= zˆ
−
 

2
z
 R  z − R z + R 

+ z 

20 September 2002
2
2
 z2 − R 2
z
R
−
1
1  1
+
−
 − 
z−R z+R  R z−R
z+R

Physics 217, Fall 2002





13
Coulomb’s Law example: field from a uniformlycharged spherical shell (continued)
which gives us, finally,
2πσ R 2  z − R z + R
E = zˆ
+

2
z
 z − R z + R



Two cases: z larger than, or smaller than, R. (P outside, inside)
‰ Larger (outside):
Behaves like a
2
4πσ R
z−R
z+R
Q point charge at
ˆz
=1=
⇒ E = zˆ
=
z−R
z+R
z2
z2 the sphere’s
center.
‰ Smaller (inside): means z − R = R − z , so
z − R z + R z2 − R 2 + R 2 − z2
+
=
=0 ⇒ E=0
2
2
R−z z+R
R −z
20 September 2002
Physics 217, Fall 2002
14
Gauss’ Law example: field from a uniformlycharged spherical shell (continued)
First note that the field must be spherically symmetric as
well, and point radially outward or inward – that is, E is
perpendicular to all sphere’s centered at the same point as the
charged sphere. So draw two Gaussian spheres, one inside
and one outside:
E ⋅ da = 4π Qenclosed
∫
r >R:
( E ) ( 4π r
2
) = 4π ( 4π R σ ) = 4π Q
2
r <R:
( E ) ( 4π r 2 ) = 0
20 September 2002
⇒ E = rˆ
Q
r2
⇒ E=0
Physics 217, Fall 2002
15
Gauss’ Law for gravity
Newton was the first to realize these results, in the context of
the other 1/r2 force, gravity. He convinced himself by use of a
proof similar to our Coulomb’s law demonstration, Gauss
still not having been born by then. We could have saved
Newton a lot of trouble by pointing out the following.
The force of gravity on a mass M from a mass m is
F = rˆGmM r 2
Gravitational forces superpose: the force on M from N
charges is
N
mi
Gm1 M
Gm2 M
rˆ1 +
rˆ2 + … = M G 2 rˆi ≡ Mg ( r )
F (r ) =
2
2
r1
r2
i = 1 ri
∑
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Physics 217, Fall 2002
16
Gauss’ Law for gravity (continued)
For a continuous distribution of mass (density ρ ( r ) ), the
gravitational field g is obtained by letting N → ∞ :
rˆ
ρ ( r ′ ) dτ ′
g (r ) = G
∫ r2
V
Take the divergence of both sides, and carry out the resulting
integral on the RHS, as we did on Wednesday, and we get
— ⋅ g = 4π G ρ ( r )
Now integrate this result over volume, and use the
divergence theorem, as we also did on Wednesday :
g ⋅ da = 4π GMenclosed
∫
20 September 2002
Physics 217, Fall 2002
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