PHY2054 Summer 2011 
Exam 1 (2 June) Solutions

Answer (1) is the correct answer unless otherwise noted. Solutions follow, in italics.

Please refer to your own copy of the exam for clarification of figures, circuit diagrams, and peculiar/weird typos which may have resulted from (1) inevitable mutations as the exam & solutions evolved from 

.doc —>.que—>.pdf —>.txt—>.htm, and (2) this being done in haste on my home computer..

1. Charge A of +5 microC and charge B of  25 microC are placed 10 cm apart. The ratio of the electrostatic force on charge A to the force on charge B is:

(1) 1:1 (2) 5:25 (3) 25:5 (4) zero (5) none of these

The mutual forces are equal and opposite!

2. In the previous question, what is the electric field (in megavolts/m) at a point halfway between charge A and charge B??

(1)108,toward B (2)108,toward A (3)72,toward A (4)72,toward B (5)92,toward B

E=K/r^2 . The field due to each charge is away from the +, toward the - charge, so they add together. Hence E1= k x 5e^-6 C/(5e^-2m)^2;

E2 = k x 25e^-6 C/(5e^-2m)^2. Total E = E1 + E2 = 108 MV/m

3. Three identical charges of 40 microC are placed at the corners of an equilateral

triangle of side length 20 cm. What is the magnitude of the electric field (in

megavolts/m) at the center of the triangle?

(1) zero (2) 144 (3) 25 (4) 1.26 (5) 0.42

The fields of the three charges, directed symmetrically away from one another, cancel at the center, hence E = 0

4. What is the electrostatic potential energy (in J) of the charge configuration described in the previous question?

(1) 216 (2) 144 (3) 540,000 (4) 72 (5) zero

The potential at distance r due to one charge is kq/r = k 40e^-6/20e^-2 = 18e^5 volts.

The potential energy of a second charge at distance r from the first one = [work required to move the charge to that location] and is just W = qV = 40e^-6 x 18e^5 = 72 J = U1

The work required to move a third equal charge up to distance r from both charges is

2 x 72J, = 144J = U2

The total work = U total = = U1 + U2 = 72 + 144 = 216 J

5. In the above diagram, suppose that the lower right-hand charge is made negative, all other parameters remaining

unchanged. The net force on the charge at lower left will then be directed:

(1) downward and to the right

(2) downward and to the left

(3) upward and to the right

(4) upward and to the left

(5) directly toward the right-hand side of the page

Sketch the vectors for the fields. See below.

E1 due to upper charge is down & to left; E2 from - Q is to right .

Resultant field E is down & to the right.

6. Identical 10 microfarad capacitors A and B are charged to 100V and 200V, respectively. They are then connected together, + to – and – to +. After charge equilibrium is reached, what is the potential difference across capacitor B (in volts)?

(1) 50 (2) 100 (3) 150 (4) 200 (5) 300

Charge stored on C1 = 10µF x 100V = 1000µC; on C2, 10µF x 200V = 2000µC

Connecting them - to + and + to - causes the charges to partially cancel:

2000µC - 1000µC = 1000µC shared between the two capacitors that are now

connected in parallel. The parallel combination has C = 10 + 10 = 20µF

The potential difference thus  = 1000 µC/20µF = 50V

7. For the above situation, the total stored energy in both capacitors after charge equilibrium is, with respect to the initial stored energy U,

(1)U/10 (2)U/5 (3)1.8U (4)3U (5)5U

The final stored energy U' = Ω Q∆V = Ω (1000e -6)C x 50V= 25e^-3 J

Initial stored energy U = Ω (1000e^-6) x 100V + Ω (2000e^-6) x 200V = 250e^-3J

U/U' = 250/25 = 10

8. Oops! Answering this question correctly requires info from Ch. 18.

Thus it was not scored.

9. Two parallel conducting plates are mounted one above the other, 5 cm apart. A voltage V is placed across the plates, creating a vertically oriented electric field between them. A 1-gram particle carrying a charge of ¡1 microCoulomb placed halfway between the plates is held stationary by the field. What is V (in volts) and which way (upward U or downward D) is the field directed?

(1) 490, D (2) 490, U (3) 9800, U (4) 9800, D (5) insufficient data for solution

For balance, mg = qE = q ∆V/d ==> ∆V = mgd/q = (5 e^-2 x 9.8 x e^-3)/ e^-6 =

= 490V . The electrostatic force is upward (to balance downward g). But the force is exerted on a negative charge, so the field must be in the opposite, or Down direction.

10. An electron is shot towards a ¡1 nanoCoulomb point charge. How close does the electron come to the ¡1 microCoulomb charge (in m) if its velocity is 3:7 • 106 m/s when it is far away? Assume that the ¡1 microCoulomb charge remains at rest.

(1) 0.23 (2) 0.46 (3) 0.12 (4) 0.34 (5) 0.60

To stop the moving electron,

(work done by the E-field of the point charge) = KE of electron.

Work W = eV and V = kQ/r where r = stopping distance, thus Ω mv^2 = kQ/r ==>

r = 2kQ/(mv^2) = [(2k) (1.6 e^-19)]/[ 9.1e^-31 x (3.7e^6)^2] = 0.23m

11. A charge Q is a distance d from another charge -9Q. Somewhere along the line joining these two charges is a point

where the electric field is zero. How far away from the Q charge is this point?

(1) d/2 (2) 2d (3) 3d (4) 9d (5) d

See sketch, below, of charges Q and -9Q , separated by distance "d".

The only location where the E-field can be zero is at "x", at distance "r" to the left of Q — closer to the smaller charge & farther from the larger. It should be obvious after some thought that this E = 0 point can't be between the two charges. [Why??]

So if the field at x is zero, E(Q) = E(-9Q) ==> |kQ/r^2| = |k(9Q)/((r+d)^2|

==> 9r^2 = (r+d)^2 ==> 9r^2 = r^2 + 2rd + d^2 ==> 8r^2 + 2rd + d^2 = 0 .

Using the quadratic formula and taking the positive root, we obtain r = d/2 .

12. A proton and an electron are each accelerated through a potential difference of 1000 volts. The kinetic energy of the

proton and the electron, in electron volts are, respectively,

(1) 1000, 1000 (2) 1000, 5.4e-4 (3) 5.4e-4, 1000 (4) 1000, 0.54 (5) 0.54, 1000

From the definition of the electron-volt , the KE is the same: 1000 eV.

[Of course, the speed of the proton will be much less because of its greater mass!!]

13. A 10,000-microfarad capacitor is charged to a potential V. It is disconnected from the charging source and connected

across a 1-watt light bulb. If the stored energy is su±cient to operate the light bulb for 2 minutes, what was the potential

di®erence across the capacitor, in volts?

(1) 155 (2) 0.155 (3) 14 (4) 110 (5) 346

Energy delivered = Pt = 1W x 2min = 1 J/s x 120 s = 120J = stored energy

Stored energy U = Ω C (∆V )^2 ==> ∆V = √(2U/C ) = √( 2 x 120)/e^-2) = 154.92 -> 155V

14. Two equipotential surfaces lying near the middle of the space between the plates of a parallel-plate capacitor are 2.0 mm apart and have a potential difference of 0.0012 volt. The area of each plate is 7.5 cm2 . What is the charge on each plate,

in (coulombs x10-15)?

(1) 4 (2) 16 (3) 5310 (4) 6.6

The uniform E-field between the plates is ∆V/d = 12e^-4/ 2e^-3 = 0.6 V/m

For the parallel-plate capacitor, C = e₀ A/d . Q = C∆V = (e₀ A/d) ∆V = e₀ A ∆V/d,

So Q = 8.85 e^-12 x 7.5 e^-4 x 0.6 = 3.98e^-15 ≈ 4 e^-15 C

15. A 3-microfarad capacitor and a 5-microfarad capacitor are connected in series

across a 30-volt battery. A 4-microfarad capacitor is then connected in parallel

across the 3-microfarad capacitor [please refer to the figure]. After charge

equilibrium is reached, the potential diifference in volts across the 4-microfarad

capacitor is

(1) 12.5 (2) 30 (3) 17.5 (4) 4 (5) 26

For the completed circuit we have 5 mF in series with (3mF + 4mF in parallel = 7mF). The equivalent capacitance of the whole combination is 1/C = 1/5 + 1/7 ==> C = 2.92 mF.

This is a series combination so he total charge as well as that on both the 5mF and the lower combination is Q = C∆V = 2.92 x 30 = 87.5mC

The voltage across the lower pair is thus ∆V= Q/C = 87.5/7 = 12.5V

16. The charge [in microcoulombs] on the 5-microfarad capacitor in the above question, before and after the addition of the third capacitor, is:

(1) 56, 88 (2) 56, 56 (3) 240, 240 (4) 125, 175 (5) 16, 10

We already have the 'afterward' charge on the 5 mF capacitor (87.5mC, from the above). The 'before' charge we get from the initial series capacitance 1/C = 1/5 + 1/3 ==> C= 1.875 mF . Q = C∆V = 1.875 x 30 = 56.25 mC. So before, after is 56, 88.

17. A capacitor is hooked up to a battery. When fully charged the capacitor has potential energy of 8 Joules stored in it. The battery is now disconnected and the capacitor is isolated. If the capacitor's surface area were now doubled and the

separation distance between the plates halved, how much energy [in J] remains stored in the capacitor?

(1)2 (2)4 (3)8 (4)16 (5)32

C = e₀ A/d; C' = e₀ (2A/(d/2) = e₀ (4A/d) = 4C. Since U = Q^2/2C and Q is

unchanged, U' = Q^2/ 2(4C) = Q^2/ 8C = U/4 = 8J/4 = 2J

18. Two identical charges Q are placed distance D apart.

The electric potential energy of this system (in J) is:

(1) KQ2=D (2) KQ=D (3) KQ=D2 (4) KQ2=D2 (5) KQ=D

The potential at distance D for the first charge is V = k Q/D.

PE = work to move second charge Q up to distance D = VQ = k Q^2/D

19. Transmission of electrical energy over long distances is usually done at higher voltages because under these circumstances:

(1) There is less energy loss in the transmission lines.

(2) The resistance of the transmission line diminishes.

(3) More current is transmitted.

(4) The line insulation is more e®ective.

(5) The ohmic voltage drop is greater.

Ohmic heating or power loss in the line is I^2R. For a given amount of transmitted power, increasing the voltage reduces the current [P = I∆V].. So energy loss in the line is less at higher voltages.

20. You wish to increase by a factor of two the heat output of an electric (\resistance") heater. Which one of the following options is most likely to work?

(1) double the resistance, keeping the current the same

(2) double the resistance, keeping the potential difference the same

(3) double the potential di®erence, keeping the resistance the same

(4) double both the current and the potential di®erence

(5) double the current, keeping the resistance the same

Power P = I^2 R = ∆V^2/R .

Of the options given, only doubling R with same current will double P.