February 12, 2010
PHY2054
Solutions Exam I
1-3. Particle 1 has a mass of 1.0 g and charge 6.0 µC, and particle 2 has a mass of 2.0 g and
charge (3.0| 5.0| 8) µC. While particle 2 is held in place, particle 1 is released from rest at a
distance of 2.0 mm from particle 2. After a long time, what is the speed of particle 1 [in m/s]?
Answer: 400|520|660
Energy conservation determines that Ef = ½ mv2 = Ei = kq1q2/d. Solve for v to get the answers.
4-6. Four point charges are placed at the four corners of a 3.0 cm × 3.0 cm square, in two
different configurations as shown in the figure. The magnitude Q of the charges is (3.0|2.0|1.0)
µC. Which configuration has higher electrical potential energy and what is its value [in J]?
Answer: Configuration 1 (-3.8|-1.7|-0.42)
The potential energy
located a distance d apart.
, where the sum is over all pairs of charges (q1, q2) that are
You can tell the answer for the highest energy configuration is 1 since here the like charges are
closer by and like charges always make positive contribution to the potential energy, increasing
it. The terms involving near pairs cancel in the U sum. Only the diagonals survive and lead to
the answer.
7-9. Three point charges (Q1 = -5.0 µC, Q2 = 4.0 µC, and Q3 = 6.0 µC) are located at the corners
of an equilateral triangle, as shown. The electric potential due to these charges is (4.0| 8.0| 16) V
at the center of the triangle. If Q2 is increased to 8.0 µC and Q3 is decreased to 4.0 µC, what will
be the electric potential at the center [in V]?
Answer: ( 5.6| 11.2| 22)
The potential is given by the sum
over all the charges which happen to be located at
the same distance from the center of the triangle. Thus the sum is over only the charges
(alternatively, the value given for one set of charges determines the size of the triangle).
Therefore V2 being the potential with new charges and V1 the potential for original charges,
10-12. In the electric circuit shown, capacitors 1-4 have the same capacitance of 2.0 µF
each, and the amount of charge stored in capacitor 2 is 3.0 µC. What is the total
energy stored in the four capacitors [in µJ]?
Answer: (27| 48| 32)
Since E = ½ CV2, we need the equivalent capacitance as well as the potential difference.
The equivalent capacitance across the series part C = [1/2 +1/2+1/2] -1 = 2/3 µF, which in
parallel with 2 µF gives C = 8/3 µF.
The charge across capacitance 2 is 3µC. It must be the same charge across the net series
capacitance, C = 2/3 µF and therefore the potential difference across the series link is Q/C = 9/2
= 4.5V. The energy stored in all capacitors is ½ CV2 = ½ ×8/3 × 10-6×4.52 = 27 µJ.
13-15. In the electric circuit shown, C1 = 4.0 µF, C2 = 8.0 µF, and C3 = (3.0| 5.0| 7.0) µF. The
amount of charge stored in capacitor Cx is exactly half the amount of charge stored in C2. How
large is Cx [in µF]?
Answer: (2.4| 1.8| 1.6)
The charge is the same in series. Thus the charge on C2 is the same as the one on the series
combination at left and likewise, the charge on Cx is the same as the one on the series
combination at right. Equivalent capacitances on each side should be in the ratio of the charges
on Cx and C2.
Which can be solved for Cx = 2.4µF.
16-18. In the circuit shown, all resistance values are in ohms. Find the equivalent resistance
between points a and b [in Ω].
Answer: (9| 6| 12)
A matter of redrawing the circuits.
19-21
22–24. You bought two light bulbs, one rated (60|150|200) W/120 V and the other 100 W/120 V.
Then, you connected them in series and plugged the two-bulb circuit into a 120 V power outlet
as shown. Approximately, how much power is going to be dissipated by the circuit?
Answer: (40|60|70) W
Let the power ratings of the two light bulbs be P1 and P2, and their resistances be R1 and R2.
Then, P1=V2/R1 and P2=V2/R2, where V is the outlet voltage 120 V. When the light bulbs are
connected in series as shown in figure, the power dissipated by each light bulb will be less than
the rated value because the potential drop across it will be less than V. (Equivalently, the
decrease in the dissipated power occurs because the current through either of the two light bulbs
decreases due to the resistance of the other light bulb in series.) The power dissipated by the two
light bulbs is P=V2/(R1+R2) because their total resistance is R1+R2. This gives
P=V2/(V2/P1+V2/P2)=1/(1/P1+1/P2).
25–27. An electric engineer designs a capacitor that would have two parallel plates of 1 cm2 area
each separated by a 100 µm thick dielectric ceramic sheet with a dielectric constant κ= (5|4|2)
and resistivity ρ = (1018|1017|1016) Ωm. If such a capacitor is charged to 1 µC, estimate how long
it will take for a half of the charge to leak through its slightly conductive ceramic sheet?
Answer: (1 year|1 month|1 day)
The capacitance of the capacitor is C=κε0A/d, where A is the area of the plate and d the
distance between the two plates. At the same time, the resistance of the ceramic sheet is
R=ρd/A. Therefore, the time constant of the charge decrease, τ = RC, is ρκε0. The time it takes
for the charge to decrease to ½ of its initial value is (ln2) τ , which in this case is (ln2) ρκε0. The
unit is sec. (1day = 86,400 sec)
28–29. You have a piece of wire with the total resistance R. You interconnect the two wire ends
to make a square loop and measure the resistance between points A and B as
shown in the figure. Approximately, what is this resistance in terms of R?
Answer: (0.2R|0.25R)
In Problem 28, the resistance of the straight section between points A and B is R/4, whereas the
resistance of the C-shaped section between the two points is 3R/4. These two resistances in
parallel give the net resistance between the two points: 3R/16. Similarly, the two parallel
resistances in Problem 29 are both R/2, resulting in the net resistance of R/4.
30. You have a piece of wire with the total resistance R. You interconnect the two wire ends to
make an equilateral triangular loop and measure the resistance between points A and B as shown
in the figure. Approximately, what is this resistance in terms of R?
Answer: 0.22R
In this problem, the two parallel resistances are R/3 and 2R/3, resulting in the net resistance of
2R/9.
31–33. Three identical charges (q = 5|2|4 µC) lie along a circle of radius 2 m at angles of 40o,
140o and 270o (as usual with respect to the x axis). What is the resultant electric field (in
kilonewton/coulomb) at the center of the circle?
Answer: –(3.2|1.3|2.6) ŷ
Let the radius of the circle be r. At the center of the circle, the net electric field produced by the
charges at 40o and 140o is 2sin(40o)kE q/r2 in the negative y direction, since the x components of
the individual fields due to the two charges cancel out exactly. The field produced by the charge
at 270o is kE q/r2 in the positive y direction. The total field is therefore [2sin(40o)-1]kE q/r2 in the
negative y direction.
34–36. A proton (+e) shoots through tiny holes in two parallel metal plates across which
(40|100|2) V of electric potential difference is applied, as shown in the figure. The speed of the
proton is originally v0 = 2.0×105 m/s as it goes through the hole in the first plate. Find its speed
as it goes through the hole in the second plate.
Answer: (1.80|1.44|1.99)× 105 m/s
Let the applied electrical potential difference be ∆V. Then the potential energy of the proton, as
it goes through the hole in the second plate, is higher by e(∆V) than its potential energy when it
went through the hole in the first plate. Therefore, the kinetic energy of the particle, as it goes
through the hole in the second plate, is mv2/2 =mv02/2 – e(∆V). Solve this for v.
37–39. Three equal charges each of magnitude (7|5|3) µC are arranged at the corners of an
equilateral triangle, with sides of length 10 cm. Calculate the magnitude of the force acting on
each of these charges.
Answer: (76|39|14) N
In this geometry, the magnitudes of the forces acting on the three charges are the same. For
convenience, let us position the triangle so that one charge is on the y axis and the two other
charges are on the x axis, and consider the force on the first charge. This force is 2sin(60o)]kE
q2/r2 in the y direction, where q is the magnitude of each charge and r the length of the sides of
the triangle.
40–42. The electric field required to suspend a proton|electron|He++ ion against the force of
gravity is:
Answer: 102 nanovolts/m directed upward|56 picovolts/m directed downward|205 nanovolts/m
directed upward
Since charge of the electron is negative, the electric field has to point down in order to exert an
upward force against the force of gravity. For other particles, the direction of the field has to be
upward. Let the mass of the particle be m and the magnitude of its charge q. The force of gravity
on the particle is mg and the electrical force qE, where E is the magnitude of the electric field.
Therefore, E=mg/q.
43–45. Three identical (size and shape) conductive balls, labeled A, B and C, carry charge 2q|2q|
q , q| q|–2q and –2q|–q|2q respectively. They are placed equidistant from each other. First A is
brought to touch B and then separated back to the original position.
Next A and C touch each other and are restored to the old locations. Finally, B touches C and
then returns to its place. What is the magnitude of the final force between A and B in proportion
to the original force (Ff=Fi) between them?
Answer: 5/64|7/64|3/64
As an example, let us consider the first version of this problem, in which the initial charges on
balls A, B, and C are 2q, q, and –2q. In the first step, in which balls A and B touch each other
and separate, the total charge 3q will be equally divided, resulting in 3q/2 on each ball. In the
second step, the total charge will be 3q/2–2q= – q/2, which will be equally divided, resulting in –
q/4 on each ball. In the last step, the total charge 3q/2–q/4 = 5q/4 will be equally divided into
two parts, 5q/8 each. After these three steps, the charges on A and B will be respectively –q/4
and 5q/8.
Originally, the force between A and B was kE(2q)q/r2, where r is the distance between the two
balls. After the redistribution of the charges, the force between A and B will be
kE(–q/4) (5q/8)/r2, whose magnitude is 5/64 of the original.