May 30, 2008
PHY 2049
Summer 2008, Exam I
Solutions
1. Two identical conducting spheres A and B carry equal charge. They are separated by a distance much
larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first
touched to A, then to B, and then removed. Finally, spheres A and B are brought into contact and then
separated. As a result, the electrostatic force between A and B, which was originally F, becomes:
(1) 25F/64
(2) F/2
(3) F/4
(4) 5F/8
(5) 3F/8
When the balls touch, their potentials become the same. Since they are the same size (same radius R,
their capacitances are equal, C = 4πεR), therefore their charges are the same. In the first touch, balls
(A,B,C) start out with charges (Q, Q, 0) but end up with charges (Q/2, Q, Q/2). After the second touch
involving C and B, the charges are (Q/2, 3Q/4, 3Q/4). After the third touch involving A and B, the
charges are (5Q/8, 5Q/8, 3Q/4). The electrostatic force is then reduced by (5/8)2 = 25/64.
2. A particle with charge 3μC is placed at the origin, an identical particle, with the same charge, is
placed 2m from the origin on the x axis, and a third identical particle, with the same charge, is placed
2m from the origin on the y axis. The magnitude of the force on the particle at the origin is:
(1) 2.9X 10-2 N
(2)1.4 X 10 -2 N
(3)2.0 X 10-2 N
(4)4.0 X 10-2 N
(5)8.1 X 10-2 N
Here we can calculate the electric field at the origin and then calculate the force by F = qE. The x and
y components of the field E are the same, each given by Ex = Ey = kq/r2 = 9x109 x3x10-6/22 = 6.75
kV/m. Then |E| = √[Ex2+Ey2] = 9.55 kV/m and |F| = 3x10-6 x 9.55 x 103 = .0286N.
3. A charged oil drop with a mass of 2x10-4 kg is held suspended by a downward electric field of
300N/C. The charge on the drop (in μC) is:
(1) -6.5
(2) -1.5
(3) +6.5
(4) +1.5
(5) 0
Here the charge has to be negative or else both the gravity and the electrical force will be downward
and there will be no suspension. This rules out answers 3-5. The equation describing balance of
forces here is mg+qE = 0. Recall that both g and E are given to be downwards, taken here as the
positive direction. Solution is q = -mg/E = - 6.54 μC.
4. Positive charge +Q is uniformly distributed on the upper half a semi-circular rod (from North to West
to South) and negative charge -Q is uniformly distributed on the lower half. What is the direction of the
electric field at point P, the center of the semicircle? (North is up)
(1) South
(2) North
(3) West
(4) East
(5) North-East
The electric field at P, due to the positive charge in the upper half of the rod must be pointed southeast. Due to the negative charge it must be pointed south-west. All things being the same, their sum
must be pointed south.
5. Choose the INCORRECT statement:
(1) According to Gauss' law, if a closed surface encloses no charge, then the electric field must vanish
everywhere on the surface.
(2) Gauss' law states that the net number of lines crossing any closed surface in an outward direction is
proportional to the net charge enclosed within the surface.
(3) Coulomb's law can be derived from Gauss' law and symmetry.
(4) Gauss' law applies to a closed surface of any shape.
(5) Gauss' law can be derived from Coulomb's law.
Here the first is the only incorrect statement. Gauss' theorem only says that without charge, the flux
must vanish not the electrical field at all points. As the examples in the book show, the flux is often
zero int eh presence of a uniform electric field.
6. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2.
A particle with charge q is placed at the center of the cavity. The magnitude of the electric field at a
point in the cavity, a distance r from the center, is:
(1) k q/r2
(2) k Q/R12
(3) zero
(4) k(q + Q)/r2
(5) k(q + Q)/(R12 - r2)
Here the only charge in the cavity is q. Inside the conducting spherical shell, the electric field is zero.
Finally outside the shell, the electric field is given by answer 4.
7. The electric potential at points in an xy plane is given by V = (3.0V/m2)x2 – (4.0V/m2)y2 . What is
the electric field at the point (3.0 m, 3.0 m) (Ex, Ey in V/m)?
(1) (-18, 24) (2) (18, -24) (3) (27, -36) (4) (-27, 36) (5) none of these.
Ex = -∂V/∂x = - 6x = -18 V/m
Ey = - ∂V/∂y = 8y = 24 V/m
8. In the rectangle shown, the sides have lengths a = 5 cm and b = 15 cm, q1 = -7.5 C, and q2 = +2.5 C.
How much work (in joules) is required to move a third charge q3 = +2.5C from A to B along a diagonal
of the rectangle?
(1) -3.0 x1012 J
(2) -1.2 x1012 J
(3) -4.2 x1012 J
(4) 9.0 x1012 J (5) none of these
The second sentence in this problem was supposed to read, “ How much work (in joules) is done by the
electrical forces as the charge q3 = 2.5C is moved from A to B.” In that case, answer 1 is correct. As
the problem is stated here, the correct answer is +3 terrajoules, i. e. answer 5, none of these.
The easiest calculation for the problem as stated would note that
Wapp = q3 (VB-VA) = q3 (q1/b +q2/a -q1/a – q2/b) = 3 x 1012 J.
9. An air filled parallel plate capacitor has a capacitance of 4 μF. If the separation between the plates is
decreased by a factor of 2 and a material with a dielectric constant of 12 (like Silicon) is inserted
between the parallel plates, what is the new capacitance, in units of μF?
(1) 96 (2) 48 (3) 4
(4) 24 (5) 0.67
The capacitance of a parallel plate capacitor is C = κεo A/d. Where A is the cross sectional area, d is
the distance between the plates and κ is the dielectric constant of the insert. With the numbers given
here the new capacitance is C' = 12x2x4 = 96 μF.
10. Capacitor C1 is connected alone to a battery and charged until the magnitude of the charge on each
plate is 4.0 x 10-8C. Then it is removed from the battery and connected to two other capacitors C2
and C3, as shown. The charge on the positive plate of C1 is then 1.0 x10-8C. The charges on the positive
plates of C2 and C3 are:
(1) q2 = 3.0 x10-8C and q3 = 3.0 x10-8C
(2) q2 = 2.0 x10-8C and q3 = 2.0 x10-8C
(3) q2 = 5.0 x10-8C and q3 = 1.0 x10-8C
(4) q2 = 3.0 x10-8C and q3 = 1.0 x10-8C
(5) q2 = 1.0 x10-8C and q3 = 3.0 x10-8C
This problem requires no calculation. C1 starts with a charge of 40nC and ends up with only 10 nC.
The series combination of the two capacitors must have the same charge and it must be 30nC on each.
11. Charges are arranged on a square of side d as shown in the diagram. In what direction
is the electric field at the center of the square?
(1) Fourth quadrant (2) First quadrant (3) Second quadrant (4) Third quadrant (5) E = 0
In the diagram, the electric field due to the corner charges cancels. Due to the charges at the North
and South locations, the field adds and is directed south. The effect of charges at the east and west
locations is to provide a field directed east. Hence the net field is directed somewhere in the fourht
quadrant.
12. Refer to the previous problem. What is the potential at the center of the square (in units of kQ/d),
assuming V = 0 at infinity?
(1) -7.7
V = sum kq/r.
(2) +1.3
(3) -9.5
(4) -0.50
(5) +0