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PH504
PROBLEM 1
A solid sphere of radius R = 40cm has a total positive charge of 26
C uniformly distributed throughout its volume. Calculate the
magnitude of the electric field at
(a) 0 cm,
(b) 30 cm, and
(c) 60 cm from the centre of the sphere.
Solution:
(a) From Gauss' law we have E(r) = Q/(40r2) for r > R.
For a volume charge density ,
E(r) = r/(30) for r < R
Increases linearly (in proportion) to r.
At r = 0 we have E(r) = 0.
(b) We have
(r) = 26C/Vsphere, where Vsphere = 4R3/3 = 4(0.4m)3/3.
This yields (r) = 9.7 10-5 C/m3.
We therefore have at r = 0.3 m,
E(r) = r/(30) = (9.710-50.3/(38.8510-12))(N/C)
= 1.1106 N/C.
(c) At r = 0.6 m we are outside the sphere and so have
E(r) = Q/(40r2) = (91092.610-5/(0.6)2) N/C
= 6.5105 N/C.
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PROBLEM 2
A sphere of radius R = 40cm has a total positive charge of 26C
uniformly distributed over its surface. Calculate the magnitude of
the electric field at
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(a) 0 cm,
(b) 30 cm, and
(c) 60 cm from the centre of the sphere.
Solution:
(a) From Gauss' law we have E(r) = Q/(40r2) for r > R, and
E(r) = r/(30) for r < R.
At r = 0 we have E(r) = 0.
(b) We have (r) = 0 inside the sphere, therefore E(r) = 0 inside the sphere.
(c) At r = 0.6 m we have E(r) = Q/(40r2)
= (91092.610-5/(0.6)2) N/C
= 6.5105 N/C.
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PROBLEM 3
The results for a conducting sphere applies whether it's solid or hollow.
Suppose a sphere holds a charge of -3x10-6 C. Let's look at a hollow
sphere, and make it more interesting by adding a point charge at the
centre with +5 x10-6 C.
What does the electric field look like around this charge inside the
hollow sphere? How is the negative charge distributed on the
hollow sphere?
To find the answers, keep these things in mind:

The electric field must be zero inside the solid part of the sphere
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
Outside the solid part of the sphere, you can find the net electric
field by adding, as vectors, the electric field from the point charge
alone and from the sphere alone
We know that the electric field from the point charge is given by
q / 0r2.
Because the charge is positive, the field points away from the charge.
If we took the point charge out of the sphere, the field from the negative
charge on the sphere would be zero inside the sphere, and given by
Q / 0r2
outside the sphere.
The net electric field with the point charge and the charged sphere, then,
is the sum of the fields from the point charge alone and from the sphere
alone (except inside the solid part of the sphere, where the field must be
zero). This is shown in the picture:
Take k = 1/ (0).
How is the charge distributed on the sphere? The electrons must
distribute themselves so the field is zero in the solid part. This means
there must be -5 microcoulombs of charge on the inner surface, to stop all
the field lines from the +5 microcoulomb point charge. There must then
be +2 microcoulombs of charge on the outer surface of the sphere, to give
a net charge of -5+2 = -3 microcoulombs.
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PROBLEM 4
Consider the superposition of the fields from two planar charge distributions
which are separated by some distance d. Place a uniformly charged plane
with charge density  at x = +d/2 and a similar plane with charge density - at
x = -d/2. Describe the electric field strength and direction.
The field due to the upper plane of charge is
E1 = +/20 j, x > d/2, E1 = - /20 j, x < d/2.
The field due to the lower plane of charge is
E2 = /20 j, x < - d/2, E2 = - /20 j, x > -d/2.
The total field in the region x < -d/2 is E = E1+E2 = -/20 j +/20 j = 0.
Similarly, the total field in the region x > d/2 is zero.
The field is ZERO outside!
In the region between -d/2 and +d/2 the total field is is
E = E1+E2 = -/20 j -/20 j = -/0 j.
The fields add to yield a uniform field between the planes, but they precisely
cancel outside the planes to give zero net field outside.
Between the planes the field points from the positive towards the negative
plane. This is the common configuration of a parallel plate capacitor.
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PROBLEM 5
A surface charge density (x,y) is given by (x,y) = 3x2+4y2-xy C
m-2. Calculate the total charge contained within the area bounded
by x=0+a, y=0+a.
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Answer:
25a4/12 Coulombs
This result is obtained by integrating the equation twice - with
respect to x and y (the order doesn't matter) and using the
appropriate limits.
.
PROBLEM 6
A line charge has a constant charge density  Cm-1. It extends along the
y-axis from - to +. Calculate the E-field at a distance a along the
positive x-axis.
Answer: E=/(20a) radially outwards for positive 
Method 1: Split the line charge up into a series a small point charges.
Their contribution to the E-field along the y-axis cancel leaving only a
component along the x-axis. You should end up with an integral which
involves three related variables (distance along the line charge, radial
distance from segment of line charge to point a and an angle). You need
to eliminate two of these by expressing them in terms of the third
variable. The final integral is much easier if expressed in terms of the
angle variable.
Method 2: This question is much easier if solved using Gauss's law.
PROBLEM 7
Two charges of –Q are placed at positions –a and +a on the x-axis. Find the
potential energy of the system when a third charge +Q is placed at a
position x on the x-axis. Find any values of x for which the system is in
equilibrium and identify the type of equilibrium. Are there any equilibrium
configurations corresponding to positions of the +Q charge not on the x-axis?
Potential energy for -a <x < a , x< -a, x > a :
There is one equilibrium position x=0 which is an unstable position.
There are no equilibrium positions away from the x-axis.
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This question has a complication which is easy to miss in that you
have to be very careful concerning the signs of the three terms which
contribute to the P.E. For example in the region -a<x<+a the P.E. is
given by the equation below. However if you try and apply this to the
region x>a then the third term gives the incorrect sign requiring the
factor (a-x) in the denominator to be written (x-a). For points away
from the x-axis there is always a net force towards this axis due to the
two negative charges attracting the positive one. It is not possible to
cancel out this component and hence there are no points of
equilibrium away
from the x-axis.
PROBLEM 8
A parallel plate capacitor has plates of area 5 cm2 and separation
0.1 mm. If the capacitor is connected to a battery of voltage 25V
calculate the charge on the plates of the capacitor and the energy
stored. The capacitor is now disconnected from the battery and a
dielectric of relative permittivity 5 is used to completely fill the
region between the plates. Calculate the voltage between the
plates of the capacitor and the energy stored.
Answers:
Q = 1.1 nC, U = 1.4x10-8 J
V = 5 V, U = 2.8x10-9 J
Here the charge on the capacitor remains constant.
PROBLEM 9
For the capacitor of question 8 the dielectric is inserted with the
capacitor still connected to the battery. Calculate the charge on the
plates of the capacitor and the energy stored. Is any energy lost or
gained by the battery in this process?
Q=5.5nC, U = 6.9x10-8 J
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Energy stored has increased so energy is lost by the battery.
Here the voltage between the plates of the capacitor remains
constant.
PROBLEM 10
A parallel plate capacitor has plates of area 15cm2 and separation
2mm. Calculate its capacitance. A dielectric plate of thickness
1mm and r=4 is inserted between the plates. Find the new
capacitance. (Hint: the capacitor is equivalent to two capacitors in
series. For two capacitors in series the reciprocal of the total
capacitance is equal to the sum of the reciprocals of the individual
capacitances).
Answers:
C=6.6x10-12F, C=1.1x10-11F
You can either treat this as two capacitors in series or find
expressions for the E-field in the two regions of the capacitor and
then integrate E between the plates to find the potential difference
and hence the capacitance.
PROBLEM 11
A polythene sheet of r=2.3 is placed in an electric field of 2x106
V/m. Find E, D and P within the polythene and the surface charge
density.
E=8.7x105 V/m, D = 1.8x10-5 Cm-2, P = 1x10-5 Cm-2=
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