Group Problems Week 6 Feb 26
Problem 1 E-field of a continuous distribution of charge
and potential
A point charge Q = 1µC is located on the axis of a uniformly
charged disk of net charge Q’ = 2 µC and radius R = 2 cm at
a distance b = 3 cm from the plane of the disk. Calculate:
1) The electric field at the charge position due to the disk
(show the full calculation).
3) The potential of the disk at the point where the charge is
located due to the disk. After you find the potential calculate
the electric field at that point and compare it to the answer
you gave to 1).
2) The work the electric field has to do to bring the charge
from x=b to infinity along the x axis.
x
Solution
a) The E-field for any point of the axis is along the axis for
symmetry reasons. The charge density of the disk is:
"=
!
Q
dq
=
= 1.6mC /m 2 $ dq = " 2#ada
2
#R
dS
Each elementary ring that forms the disk contributes an
electric field:
dq
b
b
dE x = dE cos" = k 2 cos " = k# 2$ada 3 = k# 2$ada
3/2
r
r
(b2 + a2 )
!
So integrating over the radius from 0 to the disk radius R:
R
bdq
# 2$ada
dE x = k
"
E
=
kb
%
x
3/2
2
2 3/2
0 (b + a )
(b2 + a2 )
!
and substituting the variable u = b + a " du = 2ada
R
2
2
b 2 +R 2 du
2ada
%1/ 2 b +R
E x = k"#b $
=
k
"#
b
=
k
"#
b
%2u
[
]b 2
$ b 2 u3 / 2
2
2 3/2
0 (b + a )
2
!
2
+
.
0 !
1
0
E x = 2k"# -1$
= 1.5 110 7 N /C
20
% R(
1+ ' * 0
-,
& b ) 0/
The field is in the positive direction along the x axis.
!
2) The potential at the point on the axis at a distance b due to
the charge element of the disk dq = " 2#ada is:
dV = k
dq
" 2#ada
=k
$
2
2
r
b +a
! b +R
R
2ada
V (b) = k"#
$
0
2
b +a
2
= k"#
$
2
b2
2
2
2
du
1/ 2 b +R
=
k
"#
2u
[ ]b 2 = 2k"#
u1/ 2
(
)
b 2 + R 2 % b = 5.5 &10 5 V
The potential at any point x along the axis is:
!
(
x 2 + R2 $ x
)
!
V (x) = 2k"#
!
The field is along the x axis hence the relationship with the
potential is:
%
(
dV
d
x
Ex = "
="
2k#$ x 2 + R 2 " x = "2k#$ ' 2
"1
*
dx
dx
& x + R2 )
and at the point x = b:
!
[
(
)]
+
.
0
1
0
E x = 2k"# -1$
= 1.5 110 7 N /C
20
% R(
1+ ' * 0
-,
& b ) 0/
which is the same result as in 1).
!
3)
W = "#U = "Q#V = "Q(V ($) " V (b)) = Q2k%&
!
In fact the potential goes to zero at infinity.
Same result can be obtained if we calculate the work from
the electric force Fx = QEx.
"'
*
"
x
W = # QE x dx = 2k$Q% # )1&
,dx = 0.5J
b
2
2
(
+
x
+
R
b
(
)
b 2 + R 2 " b = 0.5J
Problem 2 Gauss’ Law
!
Consider a uniformly charged insulating sphere of net charge
+3Q = 3.0 nC and radius a = 1.0cm. Concentric with the
sphere is a metal shell with inner radius b = 1.5 cm and
outer radius c = 1.7 cm having a net charge –Q.
1) Find the electric field in all regions of space determined
by the 3 surfaces and plot it.
2) Find the charge density on the surfaces of the metal
shell.
Solution
1) r > c "
r
r
2
# E $ dA = 4%r E =
meters.
Qenclosed 3Q ' Q
2Q
18
=
"E=
= 2 N /C with r in
2
&0
&0
4 %&0 r
r
b < r < c " E = 0 since this is the region inside a conductor in
!
!
!
equilibrium.
a<r<b"
meters.
r<a"
r
#
r r
Q
3Q
27
E $ dA = 4 %r 2 E = enclosed " E =
= 2 N /C with r in
2
&0
4 %&0 r
r
r
2
# E $ dA = 4%r E =
Qenclosed
&0
Since the sphere has uniform charge:
!
!
3Q
Q
3Qr 3
= enclosed $ Qenclosed = 3
4 3
4 3
a
#a
#r
3
3
Q
3Qr
" E = enclosed2 =
= 2.7 %10 7 r(in m) N/C
4 #$0 r
4 #$0 a 3
" = const =
At the surfaces we have:
E(a) =
!
3Qa
3Q
=
= 2.7 $10 5 N /C
3
2
4 "#0 a
4 "#0 a
and the field is continuous.
3Q
= 1.2 $10 5 N /C
2
4 "#0b
2Q
E(c) =
= 6.2 $10 4 N /C
4 "#0c 2
E(b) =
!
!
So the field has a discontinuity across the metal surface as
we expected since the field is zero inside a conductor in
equilibrium.
2) Since the field is zero for b < r < c then if we have a
Gaussian surface with radius b < r < c the enclosed charge
must be zero for Gauss’ law. Hence Qenclosed = +3Q + q(b) =
0, where q(b) is the charge on the surface with radius b and
q(b) = -3Q so the surface charge density is:
" (b) =
!
If there is a charge -3Q in the inner surface of the metal
shell, the charge on the outer surface is q(c) = +3Q – Q =
+2Q in agreement with the value we calculated fr the field in
the region r > c. Hence:
" (c) =
!
#3Q
= #1.1µC /m 2
2
4 $b
2Q
= 0.53µC /m 2
2
4 #c