Thin Spherical Shell by Direct Integration One of the first examples texts use to
illustrate Gauss’ law is to show that the field due to
a thin spherical shell of charge is zero everywhere
inside the shell and equivalent to the field from a
point charge everywhere outside. Authors also
qualitatively argue that the field should be zero, but
the result is rarely shown with direct integration.
This document shows how to directly integrate
Coulomb’s law to get the result described above.
The diagram at right shows the problem. In
general, the contribution
to the field at the position
r
represented by r0 due to a small amount of charge
r
dq at position r is given by
r
dq r r
dE = k r r 3 ( r0 − r )
r0 − r
(0, 0, z0)
r
r0
θ
r
r
dq=σ dA
The field at some point on the z-axis due to any small section of the shell becomes
r
dq
dq r r
dE = k r r 3 ( r0 − r ) = k r r 3 [(0 − x) xˆ + (0 − y) yˆ + (z0 − z) zˆ ]
r0 − r
r0 − r
Given the symmetry it seems reasonable to assume that the x- and y-components of the
r
r
field are zero (i.e. E = E z zˆ ), so I’ll focus on the z-component of dE :
dq
dE z = k r r 3 ( z 0 − z )
r0 − r
To get the total field to the entire shell, we’ll need to integrate and spherical coordinates
are likely to be easier than other choices. To do the integral, we’ll need:
•
•
•
Definition of charge density: σ ≡ q / A (and consequentially dq = σdA )
Some spherical coordinates details:
o dA = R 2 dφ sinθdθ and A = 4πR 2
o limits of integration: φ ∈ [0,2π ] and θ ∈ [0, π ]
o z = Rcosθ on the surface of the sphere
r r
r r
Law of cosines to express r0 − r for our system: r0 − r = (z02 + R 2 − 2z0 R cosθ )1/ 2
Combining these elements:
Ez =
2π
π
z 0 − R cos θ
dq
k
(
z
−
z
)
=
k
d
φ
σR 2 (sin θ )dθ 2
r r3 0
∫
∫
∫
( z 0 + R 2 − 2 z 0 R cos θ ) 3 / 2
r0 − r
0
shell
0
Use some substitutions ( ε ≡ z 0 /R and u ≡ −cosθ , du = [sin θ ]dθ ) to simplify the
appearance of the integral:
2π
E z = kσ ∫ dφ
0
− cos π
∫
− cos 0
du
ε +u
[1 + ε + 2εu ]3 / 2
2
Thin spherical shell integral (David Abbott 6/1/08) Page 1 Thin Spherical Shell by Direct Integration Do the integration on φ to get:
1
E z = 2πkσ ∫ du
−1
1
⎡1
⎤
ε
u
ε +u
=
+
2
π
k
σ
du
du
⎢∫
2
3/ 2
2
3/ 2
2
3/ 2 ⎥
∫
[1 + ε + 2εu ]
[1 + ε + 2εu ] ⎦
−1
⎣ −1 [1 + ε + 2εu ]
The first term inside the brackets also integrates pretty easily:
1
I 1 ≡ ε ∫ du[1 + ε 2 + 2εu ] −3 / 2 =ε [1 + ε 2 + 2εu ] −1 / 2 (−2)
−1
1
2ε
+1
= − [1 + ε 2 + 2εu ] −1 / 2
−1
+1
−1
The second term requires integration by parts ( ∫ udv = uv − ∫ vdu ). If you choose
dv = [1 + ε 2 + 2εu]−3 / 2 du and u = u , then the integral needed to find v is essentially I1 :
v = ∫ du[1 + ε 2 + 2εu ] −1 / 2 =
I1
ε
1
= − [1 + ε 2 + 2εu ] −1 / 2
ε
And then integration by parts gives
1
I 2 ≡ ∫ u[1 + ε + 2εu ]
2
−3 / 2
u
du = − [1 + ε + 2εu ]
u
I 2 = − [1 + ε 2 + 2εu]−1 / 2
−1 / 2
ε
−1
ε
+1
2
+1
+
−1
−1
1
ε
+
[1 + ε 2 + 2εu ]1 / 2
2
1
ε
+1
∫ [1 + ε
2
+ 2εu ]−1 / 2 du
−1
+1
−1
Combining everything:
+1
1
u
⎡
⎤
E z = 2πkσ ⎢− [1 + ε 2 + 2εu ] −1 / 2 − [1 + ε 2 + 2εu ] −1 / 2 + 2 [1 + ε 2 + 2εu ]1 / 2 ⎥
ε
ε
⎣
⎦ −1
+1
⎡
u 1
⎞⎤
⎛
E z = 2πkσ ⎢[1 + ε 2 + 2εu ] −1 / 2 ⎜ − 1 − + 2 [1 + ε 2 + 2εu ] ⎟⎥
ε ε
⎝
⎠⎦ −1
⎣
+1
⎡
2πkσ
εu + 1
⎛ u 1 ⎞⎤
E z = 2πkσ ⎢[1 + ε 2 + 2εu ] −1 / 2 ⎜ + 2 ⎟⎥ =
2
ε [1 + ε 2 + 2εu ]1 / 2
⎝ ε ε ⎠⎦ −1
⎣
+1
−1
Evaluating the result (and being careful about the signs of the square root) gives
Ez =
2πkσ ⎡ ε + 1 − ε + 1⎤ 2πkσ
−
⎢
⎥=
ε 2 ⎣⎢ ε + 1 ε − 1 ⎦⎥
ε2
⎡ ε +1 ε −1 ⎤
+
⎢
⎥
⎣⎢ ε + 1 ε − 1 ⎦⎥
The last step is to show that this answer reduces to what we expect. Before we do, it
might be helpful to notice that each term inside the brackets is actually quite simple- any
number (except zero) divided by its absolute value is either one (if the number is
positive) or negative one (if the number is negative).
Thin spherical shell integral (David Abbott 6/1/08) Page 2 Thin Spherical Shell by Direct Integration Inside the shell
If we are interested in the field at points inside the shell, z 0 > R , and, since ε ≡ z 0 /R , this
also means that ε < 1 . Under these conditions, (ε + 1) must be positive and (ε − 1) must
be negative, so
Ez =
2πkσ ⎡ ε + 1 ε − 1 ⎤ 2πkσ
[1 + −1] = 0
+
⎢
⎥=
ε 2 ⎣⎢ ε + 1 ε − 1 ⎦⎥
ε2
Outside the shell
Case 1: At points outside the shell on the positive z-axis, z 0 > R and ε > 1 , and
both (ε + 1) and (ε − 1) are positive so
Ez =
2πkσ ⎡ ε + 1 ε − 1 ⎤ 2πkσ
+
[1 + 1] = 4πk2σ
⎢
⎥=
2
2
ε ⎢⎣ ε + 1 ε − 1 ⎥⎦
ε
ε
Substituting ε ≡ z 0 /R and A = 4πR 2 back in, this becomes
Ez =
4πkσ
ε
2
=
4πk
ε
2
1
q
kq
= 2
2
2
4πR ( z 0 / R)
z0
Case 2: At points outside the shell on the negative z-axis, both (ε + 1) and (ε − 1) are
negative and
Ez =
2πkσ ⎡ ε + 1 ε − 1 ⎤ 2πkσ
[− 1 + −1] = − 4πk2σ = − kq2
+
⎢
⎥=
2
2
ε ⎢⎣ ε + 1 ε − 1 ⎥⎦
ε
ε
z0
The bottom line is that Case 1 and Case 2 have the same interpretation: outside the shell,
the electric field points away from the shell with the strength given by Coulomb’s law:
kq
z 02
If the reader accepts the original claim from symmetry that the field points only in the zr
direction ( E = E z zˆ ), we have proven what we set out to prove: that the field inside the
shell is zero and the field outside the shell is identical to the field due a point charge at
the origin.
Ez =
Thin spherical shell integral (David Abbott 6/1/08) Page 3