Download Problem 1. (5 points) A number of point charges with values Qi are

Problem 1. (5 points)
A number of point charges with values Qi are fixed at positions ri. If we double the value
of each and every charge, but keep the positions the same, the following quantities also
double (list all that apply):
(A) the total potential energy of system,
(B) the force between particles one and two,
(C) the electric field due the all particles at a point far away,
(D) the electric field due to all particles at a point in the center of the group (assuming
that there is no particle at this point),
(E) the electric flux through a closed surface around all charges
Forces between charges depend on the product of two charges, and hence if all charges
double their value, they increase by a factor of four. The same holds for potential
energies. The electric field at a point is the force per unit charge, and hence scale with the
charges. If all charges double their value, the electric field does as well. Anything related
to an electric field doubles. Hence the answer is C,D,E
Score:
All three correct: 5 points
Two correct: 3 points
One correct: 1 point
Subtract 1 point for each wrong answer, but do not go below zero.
Problem 2. (5 points)
Gauss’ law does NOT apply when we have:
(A) a spherical surface surrounding a group of charges,
(B) a cubic surface dividing a group of charges into two parts (inside and outside),
(C) a plane separating two plates of a capacitor,
(D) a doughnut shaped surface enclosing a circle of charges.
Gauss’ law deals with a CLOSED surface. A plane is open, so the answer is C.
Problem 3. (5 points)
Two charges are placed along the x-axis. A charge of 3.0 mC sits at x=0.0cm and a
charge of 6.0 mC sits at x=9.0cm. Where is the electric field equal to zero? At x=
(A) 4.5 cm,
(B) 3.0 cm,
(C) 6.0 cm,
(D) 3.75 cm,
(E) 5.25 cm.
If I put a test charge right in the middle, the distance to each charge is the same, but the
charge at 9.0 cm is larger, so the test charge will be pushed to smaller values of x. For
any point with a larger value of x the test charge is closer to the larger charge, so it will
always be pushed to smaller values of x. So the answer is either B or D. For answer B the
ratio of the distances to the test charge is equal to the ratio of the charges. The electric
field goes like charge over distance squared, so this is not the correct ratio, which leaves
us with D as the correct answer. Mathematically:
Call the x coordinate of the point with zero field xP, then we have
1 3.0
1
6.0

40 x P2 40 (9  x P ) 2
(9  x P ) 2
 2.0
x P2
9  xP  xP 2
xP 
9.0
 3.73cm
1 2
Problem 4. (10 points)
This problem should be answered on the opposite page.
An electric field has a field strength E=0.34z V/m, and points away from the z=0 plane,
along the z-axis. What is the total flux through a cylinder with axis perpendicular to the
z=0 plane, centered on the z=0 plane, radius 22cm, and length 3.4m?
(a) 3 points: Make a drawing with all information,
(b) 3 points: Give an expression for the flux through every surface of the cylinder,
(c) 4 points: Find the answer.
R=22cm
z
y
E=0.34z V/m
x
L=3.4m
Flux through side surface is zero, because the field is parallel to the surface: Φside=0
Flux through top surface is area times field, because the field is constant over the top
surface: Φtop=AE. The flux through the bottom surface is the same, because both the
direction of the surface normal and field are reversed: Φbottom=AE.
  0  (r 2 )(0.34 z )  (r 2 )(0.34 z )
  2 (0.22) 2 (0.34)(1.7)  0.18Vm
Problem 5. (10 points)
This problem should be answered on the opposite page.
An electric dipole consists of two charges, +Q and –Q, a distance d apart. You are to find
the potential due to this dipole in a plane bisecting the line connecting the two charges,
with this line being normal to the plane. Consider a point P a distance s from the point
where the connecting line intersects the plane.
(a) 3 points: Make a drawing with all information,
(b) 3 points: Set up the (simple) summation needed,
(c) 4 points: Find the answer.
r
d/2
d/2
P
s
The summation is VP=V++V- , with V  
1
Q
40 r
Because the point P is on the bisecting line, the distance to the positive charge from P is
the same as the distance to the negative charge from P. Hence we have V+=-V- and this
means that V=0 V.
Problem 6. (20 points)
E
This problem should be answered on the opposite page.
1
Q
40 d d 2   L 2
2
The electric field due to a line segment of uniform charge of length L at a point in the
plane that bisects the line segment and a distance d from the line segment has a field
strength given in the formula above, and points in the direction from the center of the
segment. Two such segments are parallel to the z-axis, with distance D between them
along the y-axis. Find the electric field in a point P along a line in the x-direction in this
bisectional plane, going through the center between the two segments.
D
L
P
(a) (5 points) Determine the direction of the total electric field in P using symmetry,
(b) (5 points) Make a drawing in the bisectional plane with all information needed to
solve the problem,
(c) (10 points) Evaluate the result.
There is as much charge above the plane as below the plane, and the distribution is
symmetrical. Hence if we put a test charge in P the force in the z direction down due to
the top part of the charges is equal and opposite to the force due to the bottom parts.
Hence the net force is in the plane. The same reasoning holds for the forces to the left and
right of the line, they cancel. Hence the net force is along the x-direction, along the line.

E1
r
D/2
xP
P
D/2

E2


1
Q
E | E1 || E2 |
40 r r 2  ( L2 )2
E1x  E 2 x  E
Etot, x  2 E
xp
r
xP
xP
1

r
20 r 2 r 2  ( L ) 2
2
r 2  xP2  ( D2 ) 2
Etot, x 
1
xP
20 ( xP2  ( D2 )2 ) xP2  ( D2 )2  ( L2 )2
Problem 7. (25 points)
This problem should be answered on the opposite page.
Two charged small spheres are hanging from the ceiling, attached by strings to the same
point on the ceiling. Each sphere has a charge of +321nC. The mass of each sphere is
17g. Find the angle θ that the strings make with each other. Assume the distance between
the charges is 12 cm.
θ
Q=+321nC, m=17g
Q=+321nC, m=17g
(a) (5 points) Draw a force diagram for the sphere on the left,
(b) (5 points) Give an expression relating all forces on the sphere on the left,
(c) (5 points) Find equations for all components,
(d) (10 points) Find the answer.
 /2

FT

FC

FG


 
FG  FC  FT  0
Q2
For the x-direction: 
 T sin(  / 2)  0
40 d 2
Where T is the magnitude of the tension force.
1
For the y direction:  mg  T cos( / 2)  0
Solving the second equation for T and substitution in the first gives
mg
1 Q2
sin(  / 2) 
cos( / 2)
40 d 2
tan( / 2) 
Q2
40 d 2 mg
1
tan(θ/2)=(9x109)(321x10-9)2/((0.12)2(0.017)(9.8))=0.39
θ=2x0.37 rad =42 degrees
Problem 8. (20 points)
This problem should be answered on the opposite page.
The centripetal acceleration of a particle moving with a uniform speed v in a circle of
radius r is v2/r. A small sphere with mass m and negative charge -q moves with uniform
speed v in a circular orbit of radius r around a positive charge Q fixed at the origin.
(a) (5 points) Give an expression for the Coulomb force acting on mass m,
(b) (5 points) Give an expression relating the centripetal force and the Coulomb force for
mass m,
(c) (5 points) Derive an expression that relates the kinetic energy of mass m and the
electric potential energy,
(d) (5 points) Calculate (you should get a numeric result) the fraction of the total energy
of the mass m that goes into kinetic energy.
(a) Magnitude FC 
(b) FC  Fcent
(c) E K 
(d)
1
Qq
, direction towards the center
4 0 r 2
mv 2
1 Qq

r
40 r 2
1 2 1 1 Qq
1
mv 
  EC
2
2 40 r
2
EK
EK
EK


 1
Etot E K  EC E K  2 E K
The kinetic energy is equal but opposite to the total energy.