Electric field of a hollow sphere

University Electromagnetism:
Electric field of a hollow sphere with surface charge

1. E-field of a hollow sphere 1
Electrical Field of a hollow Electrical Field of a hollow
SphereSphere
© Frits F.M. de Mul

2. E-field of a hollow sphere 2
E-field of a hollow sphere
Available:
A hollow sphere, radius R, with
surface charge density [C/m
2
]
Question:
Calculate E-field in arbitrary
points outside the sphere

3. E-field of a hollow sphere 3
E-field of a hollow sphere
•Analysis and symmetry
•Approach to solution
•Calculations
•Conclusions

4. E-field of a hollow sphere 4
Analysis and Symmetry
1. Charge distribution:
(surface charge)
s [C/m
2
]
Z
X
Y
R
R
R
2. Coordinate axes:
Z-axis = polar axis
3. Symmetry: spherical
4. Spherical coordinates:
r, q, j
e
r
e
q
q
e
j
j

5. E-field of a hollow sphere 5
Analysis, field build-up
1. XYZ-axes
X
Y
j
q
R
Z
2. Point P on Z-axis
. P
3. all Q
i
’s at r
i
, q
i

, j
i
contribute E
i
to E
in P
r
Q
i
E
i
e
r
4. E
i,xy
, E
i,z
5. expect: S E
i,xy
= 0,
to be checked !!
6. E = E
z
e
z
only !

6. E-field of a hollow sphere 6
Approach to solution
Z
e
r
q
r
X
Y
. P
e
z
j
dq
dj
dQ at dA
Distributed charge: dQ
dE
)(
4
2
0
zz
r
dQ
eedE
r
·=
pe
dQ = s dA
dA=(R.dq).(R.sinq.dj)
R.sinq
r and (e
r
.e
z
):
see next page

7. E-field of a hollow sphere 7
Calculations (1)Calculations (1)
Z
X
Y
e
r
j
q
. P
r
dQ at dA
dE
dq
dj
e
z
R.sinq
)(
4
2
0
zz
r
dA
eedE
r
·=
pe
s
dA=(R.dq).(R.sinq.dj)
r and (e
r
.e
z)
:
see next page

8. E-field of a hollow sphere 8
Calculations (2)Calculations (2)
Z
X
Y
e
r
j
q
. P
r
dQ at dA
dE
dq
dj
e
z
R.sinq
r
q
R
e
r
e
z
z
P
R.sinq
R.sinq
R.cosq
z
P
- R.cosq
r
2
=(R.sinq)
2
+ (z
P
- R.cosq)
2
(e
r
.e
z
)= (z
P
- R.cosq) / r

9. E-field of a hollow sphere 9
Calculations (3)Calculations (3)
Z
X
Y
e
r
j
q
. P
r
dQ
dE
dq
dj
e
z
R.sinq
)(
4
2
0
zz
r
dA
eedE
r
·=
pe
s
dA=(R.dq).(R.sinq.dj)
r
2
=(R.sinq)
2
+ (z
P
- R.cosq)
2
(e
r .e
z)= (z
P - R.cosq) / r
òò
==
=
p
j
p
q
2
00
zz
dEE

10. E-field of a hollow sphere 10
Calculations (4)Calculations (4)
Z
X
Y
e
r
j
q
. P
r
dQ
dE
dq
dj
e
z
R.sinq
òò
==
=
p
j
p
q
2
00
zz
dEE
result for E in P:
z
P
< R : E = 0
zeE
2
0
2
4
4
Pz
R
pe
ps
=z
P
> R :
zeE
2
04
P
tot
z
Q
pe
=

11. E-field of a hollow sphere 11
ConclusionsConclusions
for homogeneous
charge distribution:
r < R : E = 0
E=0
zeE
2
04r
Q
tot
pe
=r > R :
E
E
total charge seems to
be in center
the endthe end