Exam 1 (02 June 2009) solution notes

· Answer (1) on this cut-and-pasted exam copy is the correct answer, with the following exceptions:

#14 - (3), #16 - (3), #17 - (4)

· The need to put this material online without further delay, coupled with my self-taught typing skills (?) led to my NOT plugging in the numbers on several of the more complex calculations. Work these out for yourself — you'll probably benefit from the practice!

· Also, my at-home editing software refuses to copy & paste the diagrams and messes up exponentials from the .pdf file of the original exam, so please use your exam copy to refer to these.

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1. It is estimated that an ‘average’ lightning bolt might consist of about 25 coulombs of electric charge moving through a potential difference of 50 megavolts. If the bolt duration is 100 milliseconds, what is the power liberated, in megawatts?

(1) 12.5 billion (2) 5 million (3) 125,000 (4) 20 million (5) none of these

P = VI; I = Q/t => P = QV/t = (25C x 50 e6)/ (100e-3) = 12.5 e9 W

[A transcription error on the printed exam asked for the answer in megawatts; none of the listed answers fit. This was corrected to watts during the exam.]

2. 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 sufficient to operate the light bulb for 2 minutes, what was the potential

difference across the capacitor, in volts?

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

P = VI = VQ/t; W = Pt = ½ CV**2.

Thus V = sqrt[2Pt/C]

3. 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 ×10− 15 )?

(1) 4 (2) 16 (3) 5310 (4) 6.6 (5) 3319

For a uniform field like this, E = DV/x .

Also, for the capacitor's parallel plates, E = DV/d ==> DV = Ed

Since C = e_o A/d and Q = C DV, Q =[e_o A/d] (Ed) = e_o AE

4. The figure at right shows the electric field E between two charged metal

plates. Choose the correct statement:

(1) A positive charge placed at point X experiences the same force that it would at point Y

(2) The upper plate is positive and the lower plate is negative

(3) A positive charge at X experiences a greater force than it would at point Z

(4) A positive charge at X experiences less force than it would at Z

(5) A negative charge at X could have its weight balanced by the electrical force due to E

In a uniform E-field [denoted here by the parallel field lines], F=qE everywhere!

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

(1) None of these

(2) The resistance of the transmission line diminishes

(3) More current is transmitted

(4) The line insulation is more effective

(5) The ohmic voltage drop is greater

Higher transmission voltage reduces the current required to deliver the requisite

power (VI). Lower current ==> lower I²R loss in the transmission line. But that's not in the listed 'answers', so "none of these.

6. An electric lantern is powered by a 6-volt battery. During operation, the current drawn by the bulb is measured to be

300 milliamperes. How much power [in W] is the battery delivering?

(1) 1.8 (2) 120 (3) 20 (4) 0.5 (5) Insufficient data

P = VI = 300 e-3 x 6v = 1.8 W

7. In the electric lantern just described, what is the operating resistance of the filament, in ohms?

(1) 20 (2) 1.8 (3) 120 (4) 0.5 (5) Insufficient data

P = I²R ==> R = P/I² = 1.8/ (3e-3)² = 20 W

8. A +1.0μC point charge is moved from point A to B in the uniform

electric field shown in the figure. Which of the following statements is

necessarily true concerning the potential energy of the point charge?

(1) It decreases by 9.0 × 10− 6 J.

(2) It increases by 6.0 × 10− 6 J.

(3) It decreases by 6.0 × 10− 6 J.

(4) It increases by 10.8 × 10− 6 J.

(5) It decreases by 10.8 × 10− 6 J.

Moving 3m in E= 3V/m ==> DV = 3 x 3 = 9V = 9 J/C; work done by E = q DV,

so 9 J/C x 10e-6C = 9 e-6 J.

9. In the above figure, if the point charge is now moved from point B to point C, the location 2.0 m directly below it, how much work is done on it, in micro Joules?

(1) zero (2) +6 (3) −6 (4) +9 ( 5) −9

Moving from B to C is motion perpendicular to the field lines which is thus along

an equipotential line, thus zero work is done.

10. An electron is released from rest in a uniform electric field of strength 1000 N/C. How long (in ns) does the electron take to move through a distance of 1 cm?

(1) 11 (2) 17 (3) 23 (4) 29 (5) 35

Ft = mv ==> t= mv/F; W = eV = mv²/2 and F = eE ;

thus v = sqrt(2eV/m)

so t = [m sqrt(2eV/m)]/eE

11. A parallel-plate capacitor consists of two square plates of side 3 cm separated by 2 mm. Given that the air between the

plates will ionize and break down (spark) for an electric field of 100,000 V/cm, find the maximum charge this capacitor

can hold, in nC.

(1) 80 (2) 20 (3) 50 (4) 110 (5) 140

. For parallel plates, C = e_o A/d and Q = C DV so

Q = (e_o A DV)/d

12. Wires A and B are the same length and are made of the same material. Wire A, however, has twice the diameter of

wire B. The resistance of wire A is 0.08 Ω. What is the resistance of wire B, in ohms?

(1) 0.32 (2) 0.16 (3) 0.08 (4) 0.04 (5) 0.02

R = rl/A; A = p(d/2)² = pd²/4. For the larger wire, d is doubled,

so its area is p(2d/2) ² =pd² which is 4 times the area of wire B. So B has

4R resistance = 4 x 0.08 = 0.32W

13. A spherical surface surrounds a point charge it its center. If the charge is doubled and if the radius of the surface is also doubled, what happens to the electric flux ΦE out of the surface and the magnitude E of the electric field at the surface as a result of these doublings?

(1) Φ and E do not change.

(2) Φ increases and E remains the same.

(3) Φ increases and E decreases.

(4) Φ increases and E increases.

(5) None of these

The flux F = EA when E is perpendicular to A [always true for a sphere].

For the sphere, A = 4 pr² ; for the larger sphere A = 4 p(2r)² = 16 pr²

Initially, E = K Q/r² . , so F = EA = K Q/r² (4 pr²) = 4 pKQ. If we double r, then

E = KQ/(2r)² and thus F = KQ/(2r)² [16 pr² ] = K Q/r² (4 pr²) which is the same.

14. Four charges are at the corners of a square, with B and C on opposite

corners. Charges A and D, on the other two corners, have equal charge,

while both B and C have a charge of +1.0 C. What is the charge on A

so that the force on B is zero?

(1) −1.0 C (2) −0.50 C **(3) −0.35 C (4) −0.71 C (5) None of these

If d = side length of the square, then the diagonal is √2 d.

The repulsive force at B from C is then K Q²/(√2 d)² = KQ²/2d².

The charge q at A must be negative to counteract the repulsive force from C,

and the force must equal the horizontal component of the diagonal force.

So, K qQ/d² = KQ²/2d² sin 45 => q = (Q sin 45)/2 = -0.35Q = -0.35C

15. Three capacitors have capacitances C1 < C2 < C3 . If these capacitors are connected in series, which of the following is true for the resulting equivalent capacitance?

(1) Ceq < C1 (2) Ceq > C3 (3) Ceq = (C1 + C2 + C3 )/3 (4) None of these (5) Insufficient data

For series-connected capacitors, C_eq is always less than the smallest C!

[That's because we combine their reciprocals].

16. A solid conducting sphere of 10 cm radius has a net charge of 20 nC. If the potential at “infinity” is taken as zero, what is the potential at the center of the sphere?

(1) 36μV (2) 360μV **(3) 1.8 × 10 3 V (4) > 1.8 × 10 4 V

(5) None of these

The potential at the center is the same as at the surface of the sphere, which is the same as that of a point charge: V = Kq/r

17. If C1 = 25μF, C2 = 20μF, C3 = 10μF, and ∆V0 = 21 V, determine

the energy stored by C2 .

(1) 0.72 mJ (2) 0.32 mJ (3) 0.40 mJ **(4) 0.91 mJ (5) None of these

C2 and C3 are in parallel, and this combination is in series with C1. The whole

system's capacitance is 1/[1/C1 + 1/(C2+C3) = 1/[1/25 + 1/(20 + 10)] = 1/[0.04 +0.033] = 1/0.073 =13.64 mF.

The system charge is C DV = 13.64mF x 21V = 286 mC; since they're in series, this is also the charge on C1 and on (C2 + C3).

For the latter, DV = Q/(C2 + C3) = 286 mC/30mF = 9.55 V

The energy stored by C2 = ½ C(DV)² = 0.5 x 20 x (9.55 )² = 911mJ = 0.91mJ

18. A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length.

A voltage source is connected to the wires and a current is passed through the wires. If it takes time T for the average

conduction electron to traverse the 1-mm wire, how long does it take for such an electron to traverse the 2-mm wire?

(1) 8T (2) T (3) 4T (4) T /4 (5) None of these

R = rl/A ; larger wire has length 2l and 4 times the area of the smaller.

The current I = nqvA, ==> v = I/nqA where v is the drift velocity.

I, n and q are the same for both wires (same material!).

So in the larger wire the charge carriers travel twice as far at ¼ the speed; thus the time to traverse this wire is 2x4 = 8T

19. If a certain resistor obeys Ohm’s law, its resistance R is the proportionality constant for:

(1) current and potential difference.

(2) current and length.

(3) current and electric field.

(4) potential difference and electric field.

(5) current and cross-sectional area.

See Ohm's Law!

20. Three electrons are placed at the corners of an equilateral triangle of side 1 nm. Find the electrical potential energy in J of this configuration.

(1) +6.9 × 10− 19

(2) −6.9 × 10− 19

(3) +2.3 × 10− 19

(4) −2.3 × 10− 19

(5) zero

. W = qV = eV. For one electron, V = Ke/r.

A two-electron system thus has energy W = e (Ke/r) = Ke²/r .

Bringing a third electron up to form the third corner of the triangle requires

W = Ke²/r + Ke²/r. Adding this to the initial work gives W = 3 Ke²/r