Transcript BilaksPhysiks

A conducting spherical shell with inner radius a and
outer radius b has a positive point charge Q
located at its center. The total charge on the shell is
-3 Q, and it is insulated from its surroundings.
Derive expressions for the electric field magnitude
E in terms of the distance from the center r for the
regions r<a, a<r<b, and r>b.
a
Bold type denotes
vector quantities
b
+Q
-3Q
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Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
1. Which of the following physics principles
should one use to solve this problem?
I. Ampere’s Law
II. Faraday’s Law of Induction
III. Gauss’s Law
IV.Superposition of Electric Fields
•
•
•
•
A:
B:
C:
D:
I
II & IV
III
III & IV
Choice: A
Incorrect
This law deals with magnetic fields
produced by electric current.
Choice: B
Incorrect
Faraday’s law deals with the time rate of change of
magnetic flux, so this is not applicable to our situation.
On the other hand, the principle of superposition of
electric fields is very helpful here. Considering the
contributions to the electric field from each charge will
make this task easier to evaluate.
Choice: C
Incorrect
We can use this law to solve for E as
we exploit symmetry, but another
physics principle will also be helpful.
Choice: D
Correct
Gauss’s law and the principle of superposition of
electric fields are very helpful here. Considering the
contributions to the electric field from the charge in each
region will make this task easier to evaluate.
2. Which statement correctly
describes Gauss’s Law?
• A : The total electric field through a closed surface is
equal to the net charge inside the surface.
• B : The total electric flux through a closed surface is
equal to the total charge inside the surface divided by
the area.
• C : The total electric flux through a closed surface is
equal to the total charge inside the surface divided by o
(permittivity of free space).
Choice: A
Incorrect
E  Qenc
Gauss’s Law relates the electric
flux through
a
surface
to
the
total

charge enclosed by the surface.
Choice: B
Incorrect
According to Gauss’s Law, the total
electric flux through a closed surface is
equal to the total charge inside the
surface divided by
o , not by the area.
Choice: C
Correct
In mathematical form, Gauss’s Law is expressed as:
net

Qenc
  E  dA 
o
3. What is a convenient Gaussian surface
for this system?
• A: circle
• B: cube
• C: sphere
Choice: A
Incorrect
The system is three dimensional.
Choice: B
Incorrect
The magnitude of the electric
field is not the same at all points
on a cubical surface.
Choice: C
Correct
This is a good choice, because
the system shows spherical
symmetry.
Please get out a piece of scratch paper and sketch
the charge configuration of the situation described in the
problem statement. For now, only draw the arrangement
due to the charge +Q inside the cavity. Draw your
Gaussian surface with a radius r<a.
Example: Your sketch should look
something like this:
+
+Q
+
+
-
+
a -
-
-
+Q r
1. A negative charge
of magnitude -Q is
induced uniformly
around the inner
surface of the cavity.
2. The negative
charge is drawn to
the inner surface and
a positive charge
remains on the outer
surface of the
conducting shell.
+
+
-
-Q
-
+
+
-
+
Gaussian surface
4. For r<a, as depicted in our sketch, the field strength is equal in
magnitude everywhere on the surface and is radially outward in
direction. Choose all of the following that are correct expressions for
the total flux in this case.
I)

A: I only
B: II only
C: III only

D: I and II only
E: I and III only

E4r 2
II)
Q
o
III)
Q
4r 2o
Choice: A
Incorrect
This is a correct expression, but it
is not the only one.
Choice: B
Incorrect
This is a correct expression,
but it is not the only one.
Choice: C
Incorrect
Check the units. This is an
expression for electric field.
Choice: D
Correct
Both these expressions
are correct.
Mathematical expression of Gauss’s
Law:
net 
 E  dA 
Qenc
o
The total charge enclosed in our
Gaussian surface is only +Q,
 therefore:
net
Q

o


Since the electric field is
perpendicular to the infinitesimal
area dA, the magnitude of the
electric field E can be pulled out of
the integral.
net 
2
E

dA

E
dA

EA

E4r


Choice: E
Incorrect
One of these expressions is not
correct. Try Again.
5. Which one of the following expressions
describes the electric field E in the region
r<a?
• A:
• B:
Q
2
4r o
• C:
3Q
2
4r o


Q
2
4r o
Choice: A
Incorrect
The total charge enclosed in
the Gaussian sphere is +Q.
Choice: B
Correct
From the previous question we have:
Simple algebra shows that:

Q
E4r 
o
2
Q
E
2
4r o
Choice: C
Incorrect
Since our Gaussian surface has
a radius r less than a, the only
charge that is enclosed is the
point charge +Q.
6. For electrostatic equilibrium, the electric
field inside the conductor (metal) is which
of the following?
• A: uniform but non-zero
• B: zero
• C: non-uniform
Choice: A
Incorrect
This is in violation of the
equilibrium condition.
Choice: B
Correct
This implies equilibrium.
Choice: C
Incorrect
The charges are uniformly
distributed.
• Please make another sketch of the charge
configuration of the spherical shell, this
time only consider the -3Q charge that is
placed on the outer surface of the
conductor.
Example
The charge -3Q distributes
uniformly on the surface of the
conductor, making the field
everywhere inside of the
conductor equal to zero.
-
Notice that there is
no charge enclosed if
the gaussian surface
is placed anywhere
inside of the surface
of the conductor.
-3Q
-
-
-
• Considering a superposition of the electric fields
produced by the point charge +Q and the surface charge
-3Q will help us find expressions for the electric field in
the regions a<r<b and r>b.
• Sketch the charge configuration of the spherical shell
considering all sources of electric field (combine your
two previous drawings).
Superposition of electric fields and charge
summation
+
+
-
+
-
+
-
+Q
-
+Q
-Q
-
+
+
=
-
+
-
+
The red arrows depict electric field lines.
Notice that there are more field lines in the
second drawing pointing inwards than there
are in the first pointing outwards.
-3Q
-+
+Q
-
-
+-
-
-Q
+Q
-
-
-
+
-
-
-
-+
-
7. Add a Gaussian surface to your drawing with a
radius a<r<b.
What is the electric field in this region (inside of
the conductor).
• A:
E=0
• B:
Q
E= 4r 2
o
• C:
Q
E=
4r 2o

Choice: A
Correct
The electric field inside of a
conductor is always equal to zero.
Qenc
E4r 
o
a
Qenc  (Q)  (Q)  0
r
2
E0
b
Choice: B
Incorrect
Remember that the electric field
inside of a conductor is always zero.
Your Gaussian surface should
enclose more than just the point
charge +Q.
Choice: C
Correct
Remember that the electric field
inside of a conductor is always zero.
Your Gaussian surface should
enclose more than just the induced
charge -Q on the inside of the shell.
8. What is the net charge enclosed by a
Gaussian sphere in the r>b region?
•
•
•
•
A: +Q
B: -3Q
C: -Q
D: -2Q
Choice: A
Incorrect
The point charge in the
center, the induced charges
-Q and +Q on the inner
and outer surfaces of the
shell, and the -3Q spread
around the outside of the
shell are all enclosed by
the Gaussian sphere.
Gaussian
surface
r
b
Qenc  (Q)  (Q)  (Q)  (3Q)
Qenc  2Q
Choice: B
Incorrect
The point charge in the
center, the induced charges
-Q and +Q on the inner
and outer surfaces of the
shell, and the -3Q spread
around the outside of the
shell are all enclosed by
the Gaussian sphere.
Gaussian
surface
r
b
Qenc  (Q)  (Q)  (Q)  (3Q)
Qenc  2Q
Choice: C
Incorrect
The point charge in the
center, the induced charges
-Q and +Q on the inner
and outer surfaces of the
shell, and the -3Q spread
around the outside of the
shell are all enclosed by
the Gaussian sphere.
Gaussian
surface
r
b
Qenc  (Q)  (Q)  (Q)  (3Q)
Qenc  2Q
Choice: D
Correct
The point charge in the
center, the induced charges
-Q and +Q on the inner
and outer surfaces of the
shell, and the -3Q spread
around the outside of the
shell are all enclosed by
the Gaussian sphere.
Qenc  (Q)  (Q)  (Q)  (3Q)
Qenc  2Q
Gaussian
surface
r
b
9. Which one of the following is the expression for
the electric field E in the region r>b?
A:
Q
(4r 2o )
(radially outward)
B:
Q
2r 2o
(radially outward)
C:
Q
2r 2o
(radially inward)
D:
2Q
o
(radially inward)




Choice: A
Incorrect
The total enclosed charge is not +Q.
Also, field lines from a negative
source charge are directed inward.
Please check your sketch and try
again.
Choice: B
Incorrect
Field lines from a negative source
charge are directed inward.
Choice: C
Correct
Reasoning:
Q enc
2Q
Q
E


2
2
2
4r 0 4r 0 2r 0
Field
 lines from a negative source
charge are directed inward.
Choice: D
Incorrect
This is the total flux through
the Gaussian sphere.