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PHYSICS DEPARTMENT
PHY 2049 2nd Exam February 22, 2002
SOLUTIONS
- A thin straight glass rod of length L is mounted on the x-axis with one end at the origin. See sketch. The rod is charged NON- uniformly with a linear charge density λ = ax. Here a is a constant with dimensions charge/length^2. Calculate the electric potential at the point P a distance D away, setting potential at infinity to be zero.
P
D L
x
(1) ka (L − D ln (1 + L/D)) (2) ka (L + D ln (1 + L/D)) (3) −ka (L − D ln (1 + L/D)) (4) ka (D − L ln (1 + L/D)) (5) −ka (D − L ln (1 + L/D))
SOLUTION:
Line element dx has charge dq = λdx = axdx and is a distance x + D from point P. Therefore
V = k
∫ L
0
dq x + D = ka
∫ L
0
x dx x + D
Change variables using u = x + D and du = dx
V = ka
∫ D+L
D
u − D u du = ka
∫ D+L
D
D
u
du
= ka(u − D ln u)|DD+L = ka(L − D ln(1 + L/D))
NOTE: One could eliminate the wrong answers by applying common sense only: a) At L = 0, V must be zero (no charge left). This eliminates (4) and (5). b) V must be positive. This eliminates (3): e.g., try L = D c) At D = ∞, V = 0. This eliminates (2): ka
L + D ln
1 + DL
≈ ka
L + D LD
≈ 2 kaL 6 = 0.
- A parallel plate capacitor (plate area A, plate separation d) of capacitance C is charged to a potential difference V, creating an electrostatic field E the between plates. The stored energy is U. The charging battery is disconnected and a slab of dielectric (dielectric constant k) is now inserted between the plates without touching them. The electrostatic field between the plates and the stored energy become, respectively:
(1) E/k; U/k (2) kE; kU (3) kE; U/k (4) E/k; U/2k (5) E/k^2 ; U/k^2
SOLUTION:
a) The electrostatic field in material of dielectric constant k is k times smaller than it would be in vacuum. Therefore Ek = Ek b) With dielectric inserted Ck = kC, Uk = 12 Q 2 Ck =^
1 2
Q^2 kC =^
U k
- An automobile battery jumper cable is composed of 9 identical strands of copper wire, twisted together. A length L of this cable having resistance R has the wires untwisted, laid end-to-end, and welded together making a single-strand conductor of length 9L. Neglecting the effects of welding, what is the resistance of the ‘new’ wire?
(1) 81R (2) 9R (3) R (4) 18R (5) R/ 9
SOLUTION:
A parallel combination of 9 strands with resistance R implies that each strand has resistance r = 9R. Therefore, 9 strands laid end-to-end should have resistance of 9r = 81R.
- How many time constants τ must elapse for an initially uncharged capacitor in an RC series circuit to be charged to 99% of its equilibrium charge?
(1) 4. 6 τ (2) 0. 99 τ (3) 0. 03 τ (4) 1. 7 τ (5) 2. 3 τ
SOLUTION:
Charge q across the capacitor is q = q 0 (1 − e−t/τ^ ) Solve 0.99 = 1 − e−t/τ^ to find t/τ = 4.6, or t = 4. 6 τ.
- Find the current in the 5.0Ω resistor in the circuit shown. (^) 6.0 Ω
12 .0 Ω
3 .0 Ω 5 .0 Ω
4 .0 Ω
12 V
(1) 1.5A
(2) 0.42A
(3) 3.0A
(4) 2.4A
(5) 0.67A
SOLUTION:
Resistance of upper branch is the parallel combination of 6.0Ω and 12Ω in series with 4Ω. Calculate Rupper = (^) 6+12^6 ·^12 + 4 = 8Ω. Resistance of lower branch is the series combination 3Ω + 5Ω = 8Ω. The two branches in parallel have a resistance of 4Ω. The current through the combined resistors is (^12) 4ΩV = 3 A. This current splits evenly between the two branches (they have equal resistance); hence 1.5 A per branch or 1.5 A through the 5.0Ω resistor.
- Two small spheres of mass m = 1 kg are charged with q = 1C each and placed at a distance of 1 m from each other. Since they repel each other, they start flying apart. Find the velocity of each of the spheres when they are separated by a distance of 2 m.
(1) 67 km/s (2) 47 km/s (3) 134 km/s (4) 190 km/s (5) 268 km/s
SOLUTION:
Initial energy = Ei = KE|i + P E|i = 0 + k q^2 ri
Final energy = Ef = 2( 12 mv^2 ) + k q^2 rf Set Ei = Ef to get
v = 1
k m
ri
rf
km s
- A cylindrical resistor of radius 5.0 mm and length 2.0 cm is made of material that has a resistivity of 3. 5 × 10 −^5 Ω·m. What is the current density when the energy dissipation rate is 1.0 W?
(1) 1.35 E5 Am−^2 (2) 2.65 E3 A (3) 6.58 E10 Am−^1 (4) 2.41 E6 Am−^2 (5) 1.19 E2 Am−^2
SOLUTION:
Power P = i^2 R = (JA)^2 ρLA = ρLJ^2 A. Solve for J =
P ρLA
= 1. 35 × (^105) mA 2 Using L = 2. 0 cm = 0. 02 m, ρ = 3. 5 × 10 −^5 Ω · m, P = 1 Watt and A = πr^2 with r = 5. 0 mm = 5 × 10 −^3 m.
- What is the current through R 2 in the figure?
R 4
R 2 R 3
R 1
I 2
I 1
R 1 = 100 Ω
R 2 = R 3 = 50 Ω
R 4 = 75 Ω
E = 6. 0 V
E
(1) 0.02 A (2) 1 A (3) 0.05 A (4) 0.015 A (5) 0.25 A
SOLUTION:
R 2 ,^ R 3 ,^ R 4 are parallel. Therefore^ Req =^ R 1 +^
R 2 R 3 R 4
R 2 R 3 + R 2 R 4 + R 3 R 4
= 100Ω + 18.75Ω = 118.75Ω. Voltage across R 1 is
I 1 R 1 = 5.05 V and voltage across the parallel combination is therefore 6 V – 5.05 V = 0.95 V. Current I 2 =^0 .95V R 2
A = 0.02 A.
