b
c
a
vacuum
Solid sphere (+2Q)
conducting spherical shell (-5Q)
A solid spherical conductor having a radius “a” is surrounded by
a concentric spherical conducting shell having an inner radius
“b” and outer radius “c”. The charge on the solid sphere is +2Q
and the total or net charge on the shell is -5Q. The diagram
shows a cross section of the sphere and shell.
qc
qb
+2Q
c
b
a
S
First Objective: Find the charge qb that is located on the inside
surface of the shell at r = b.
(1) We let the charge on the inner and outer surfaces of the shell be qb and qc.
Both the charge and electric field inside this conductor must be zero.
(2) The charge qb can be found by applying Gauss’s Law over a spherical
surface “S” of radius r which is inside the shell (b < r < c).
1
qb
+2Q
c
b
a
S
E = 0 inside
shell
The total or net charge inside the surface “S” is (+2Q + qb).
Using Gauss’s Law and the fact that the field is zero over the
surface “S” gives:
∫ E cosθdA = ε ( 2Q + qb )
1
S
⇒ 0 = 2Q + qb ⇒ qb = −2Q
0
qb
+2Q
c
b
a
qc
Second Objective: Find the charge qc on the outer surface of
the shell
Since there can be no charge inside the shell all of the charge is
on the inner and outer surfaces so:
qb + qc = -5Q and qc = -5Q - (-2Q) = -3Q
2
-2Q
+2Q
-3Q
dA
E
r
n
surface S
Third Objective: Find the electric field outside the shell (r>c)
From the spherical symmetry of all of the charges the electric field can only
be parallel or anti-parallel to the radial direction. Since we want the field
outside the shell (r > c) we construct the spherical surface “S” of radius “r”.
On this surface E is perpendicular to the differential area dA.
Since the net charge inside the surface “S” is negative, the field points
inwards as shown.
-2Q
+2Q
-3Q
dA
E
r
n
surface S
The total charge inside “S” is: +2Q -2Q - 3Q = -3Q. Note that the angle
between the outward normal to dA and the direction of E is 180o. We apply
Gauss’ Law over the surface “S”:
∫ E cos180dA = ε ( − 3Q)
1
S
0
⇒ E ( − 1) ∫ dA =
S
−3Q
ε0
In the above equation E and cos180 are constants and have been placed outside
the integral.
3
-2Q
+2Q
-3Q
dA
E
r
Since the integral over dA is 4πr2 , we find that
the magnitude of the field is (note that the
magnitude of the field must be positive):
If we let r be an outward radial unit vector
the field is:
n
E=
3Q
4πε 0 r 2
r
E=−
3Q
4πε 0 r 2
r$
4