MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2014
Exam Two Solutions
Problem 1 (25 points)
NOTE: YOU MUST SHOW WORK in order to get any credit for this problem.
A positively charged ion with charge q and mass m , initially at rest, is accelerated across a gap
by an electric potential difference with magnitude ΔV , so that it emerges at the origin with a
velocity in the positive y -direction, as shown below. For y > 0 , the ion enters a region of
uniform magnetic field B1 , as shown below, and follows a semi-circular trajectory of radius R .
Then for y < 0 , the ion enters a region of uniform magnetic field B2 , as shown below, and
follows a semi-circular trajectory of radius R / 3 . This pattern repeats, as shown below.
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a) Determine a vector expression for the magnetic field B1 in terms of q , m , R , and/or
ΔV , as needed. Be sure specify the direction of the magnetic field.
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b) Determine a vector expression for the magnetic field B 2 in terms of q , m , R , and/or
ΔV , as needed. Be sure to specify the direction of the magnetic field.
c) How long does it take the charge to reach point P in the diagram, in terms of q , m , R ,
and/or ΔV , as needed?
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Problem 2 (25 points): answers without work shown will not be given any credit.
NOTE: YOU MUST SHOW WORK in order to get any credit for this problem.
Two dielectric rectangular slabs both have length L and width w , but the first has thickness d1 and
dielectric constant κ 1 , and the second has thickness d2 and dielectric constant κ 2 . The two slabs
are inserted into a parallel-plate capacitor consisting of two conducting plates of width w , length L ,
and thickness d = d1 + d2 . The capacitor is charged up such that the total free charge on the entire
top and bottom plates is +Q and −Q respectively. The charging battery is then removed from the
circuit. Neglect all edge effects.
a) Determine an expression for the magnitude of the electric field E1 in dielectric slab 1 in
terms of Q , κ1 , κ 2 , d1 , d 2 , w , L , ε 0 , and/or d , as needed. Describe the direction of
the electric field.
b) Determine an expression for the magnitude of the electric field E2 in dielectric slab 2.
Express your answer in terms of Q , κ1 , κ 2 , d1 , d 2 , w , L , ε 0 , and/or d , as needed.
Describe the direction of the electric field.
c) Using your expressions for the magnitudes of the electric fields from parts a) and b),
determine the magnitude of potential difference, ΔV , between the top and bottom plates
of the capacitor. Express your answer in terms of Q , κ1 , κ 2 , d1 , d 2 , w , L , ε 0 , and/or
d , as needed.
d) Determine the capacitance of this system of conducting plates and dielectric slabs.
Express your answer in terms of κ1 , κ 2 , d1 , d 2 , w , L , ε 0 , and/or d , as needed. Do not
use Q in your answer.
e) Determine the bound surface charge density on the upper surface of dielectric slab 1.
Express your answer in terms of Q , κ1 , κ 2 , d1 , d 2 , w , L , ε 0 , and/or d , as needed.
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Problem 3 (25 points): answers without work shown will not be given any
credit.
The electric potential, V (x) , for an infinite uniformly charged slab with volume charge density
ρ and an infinite uniformly charged plane with area charge density σ is given by:
⎧(1 V ⋅ m −1 )x − (1 V);
⎪
V (x) = ⎨−(1 V ⋅ m −2 )x 2 + (1 V ⋅ m −1 )x;
⎪(1 V) − (1 V ⋅ m −1 )x;
⎩
x ≤ −1 m
−1m ≤ x ≤1m
x ≥1m
A plot of the electric potential as a function of x is shown in the figure below.
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a) Determine a vector expression for the electric field E(x) for this problem in the following
regions:
(i) For x ≤ −1 m , (ii) −1 m ≤ x ≤ 1 m , and (iii) x ≥ 1 m . Be sure to include the correct units
in your answer.
b) Determine the point(s) where the electric field vanishes.
c) Make a plot of the x -component of the electric field, Ex (x) , as a function of x on the graph
below. Be sure to clearly label your vertical and horizontal axis with the correct scales and
units for Ex (x) and x .
d) Describe the exact location of the charged slab and the charged plane that are the sources of
this electric field.
e) Determine the volume charge density ρ . You may express your answer in terms of ε 0 .
[Hint: Using Gauss’ Law]. Be sure to include the correct units in your answer.
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Problem 4 of 4: (25 points) Concept Questions (Parts A through E)
Part A (5 points):
Three pairs of charged conducting spheres are connected with wires and switches. The spheres
are all very far apart. The large spheres have twice the radius of the small spheres. Each
sphere on the left has a charge of +20 nC and each sphere on the right has a charge of +70 nC
before the switches are closed.
All of the switches are closed and then, after a long time, opened again. Which of the below
rankings of the electric potentials on four of these spheres (A, C, D, E) is correct?
1. VA < VD < VC < VE
2. VE < VC = VD < VA
3. VA = VD < VC = VE
4. VC = VE < VA = VD
5. VA < VC = VD < VE
6. VE < VC < VD < VA
7. The electric potential is the same for all four spheres (A, C, D & E)
8. The ranking of the electric potential cannot be determined or is none of the above
Answer 5. Charge on B and C is each equal to half total charge, hence VA = VB. Charge on
D is more than half of total charge. Charge on E and F are half total charge. Because
radius of E and F are smaller than radius of A and B , VA < VE. Potential on C and D are
equal. Because charge on small sphere C is less than charge on small spheres E and F, VC =
VD < VE. Finally charge on B is less than charge on D so VA = VB < VC = VD. Therefore VA <
VC = VD < VE .
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Part B (5 points):
Consider two hollow concentric conducting shells. The outer conductor carries a total charge
−2Q , and the inner conductor also carries a total charge −2Q . Place a point-like object with
positive charge +Q at the center of the shells. Let V1 be the potential at the inner conductor and
V2 be the potential at the outer conductor.
Which of the following statements is true?
1. The surface charge at O2 is positive and V1 > V2 .
2. The surface charge at O2 is positive and V1 < V2 .
3. The surface charge at O2 is positive and V1 = V2 .
4. The surface charge at O2 is negative and V1 > V2 .
5. The surface charge at O2 is negative and V1 < V2 .
6. The surface charge at O2 is negative and V1 = V2
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Part C (5 points):
A parallel plate capacitor, initially with vacuum between the plates, is connected to a battery and
charged to +Q on the upper plate and −Q on the lower plate, respectively. The battery is then
disconnected. A slab of material with dielectric constant κ is then inserted between the plates.
Which of the following statements is true?
1. The charge stored in the capacitor increases and the energy stored increases.
2. The charge stored in the capacitor remains the same and the energy stored increases.
3. The charge stored in the capacitor decreases and the energy stored increases.
4. The charge stored in the capacitor increases and the energy stored remains the same.
5. The charge stored in the capacitor remains the same and the energy stored remains the same.
6. The charge stored in the capacitor decreases and the energy stored remains the same.
7. The charge stored in the capacitor increases and the energy stored decreases.
8. The charge stored in the capacitor remains the same and the energy stored decreases.
9. The charge stored in the capacitor decreases and the energy stored decreases.
Answer 8. Because the free charge does not change, the capacitance increases when the
dielectric is inserted hence the stored energy Q 2 / 2C decreases.
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Part D (5 points)
Consider the following loop of wire carrying a current I in the counterclockwise direction as
shown in the figure below.
Which of the following statements is true for the magnetic field at the point P located at the
origin?
1. The magnetic field at the point P is zero.
2. The magnetic field at point P is directed in positive y -direction.
3. The magnetic field at point P is directed in negative y -direction.
4. The magnetic field at point P is directed in positive x -direction.
5. The magnetic field at point P is directed in negative x -direction.
6. The magnetic field at point P is directed into the plane of the figure.
7. The magnetic field at point P is directed out of the plane of the figure.
Answer 6. By Biot-Savart Law, the straight legs do not contribute a magnetic field at point
P, the magnetic field of the inner circle points into the plane of the figure at point P, and
the magnetic field of the outer circle points out of the plane of the figure at point P. The
inner circle is closer so the magnetic field is stronger hence the magnetic field of the current
loop points into the plane of the figure at point P.
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Part E (5 Points)
Wire 1 carries a current I from point A to point B . Wire 2 carries an identical current I also
from point A to point B . There is a uniform magnetic field pointing in the positive x -direction.
You may neglect any other non-magnetic forces acting on the wires.
Which of the following statements is true?
1. The magnetic force on wire 1 is greater in magnitude than the magnetic force on wire 2.
2. The magnetic force on wire 1 is smaller in magnitude than the magnetic force on wire 2.
3. The magnetic force on wire 1 is equal in magnitude to the magnetic force on wire 2.
4. It is impossible to determine which wire has a magnetic force of greater magnitude acting on
it without knowing the specific shape of wire 1.
Answer 3. The magnetic force on a current carrying wire loop in a uniform magnetic field
is zero. Therefore the magnetic force on leg 1 is equal to the magnetic force on leg 2.
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