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Prof. Yasu Takano Prof. Paul Avery Dec. 11, 2006
Exam 4 Solutions
( First answer is correct )
- ( Exam 1 ) Consider a cylinder of radius 6 m and height 15 m. At each point on the cylindrical surface, an electric field of 10 V/m points out radially. There is no electric field on the top and bottom surfaces of the cylinder. What is the charge enclosed by the cylinder in coulombs?
(1) 5.0 × 10 −^8
(2) 2.5 × 10 −^8
(3) 1.7 × 10 −^8
(4) 8.3 × 10 −^9
(5) 3.3 × 10 −^8
We use Gauss’ law, and note that only the E field perpendicular to the curved surface contrib-
utes. Using the area of a cylinder, Gauss’ law yields E ( 2 π rh ) = q enc / ε 0 , where r is the cylinder
radius and h is its height. Solving for the enclosed charge yields q enc (^) = 5.0 × 10 −^8 C.
- ( WebAssign 24.5 ) Two large, parallel, conducting plates are 1.2 mm apart and have equal and opposite charges on their facing surfaces. An electrostatic force of 4 × 10 −^13 N acts on an elec- tron placed anywhere between the two plates. What is the potential difference in volts between the plates?
(1) 3000 (2) 360 (3) 1.88 × 106 (4) 1880 (5) 2250
The force on an electron is F = eE. The potential difference between two plates is Ed. Thus the potential difference can be written V = Fd/e = 3000 V.
- ( WebAssign 25.12 ) In the figure shown, the battery has a potential difference of 36 V. If C 1 = C 4 = C 5 = 5 μF and C 2 = C 3 = 10 μF, what is the charge on C 5 in μC?
(1) 120 (2) 240 (3) 180 (4) 360 (5) 60
The total capacitance in the right hand branch is C 2345 = 3.33 μ F. The charge on this equivalent
capacitance is the same as the charge on C 5 , because C 5 is in series with the rest of the branch
C 234 (as discussed in lecture and the textbook). Thus q 5 = 3.33 × 36 = 120 μ C.
- ( Exam 2 ) A current of 15 A flows through a copper wire whose radius is 1.25 mm. Assuming that the current is uniform over the wire cross section, calculate the electron drift speed in m/s. Take the free electron density to be 7 × 1028 electrons/m^3.
(1) 2.7 × 10 −^4
(2) 7.1 × 10 −^3
(3) 1.2 × 10 −^5
(4) 2.5 × 10 −^2
(5) 1.7 × 10 −^7
The total current I = JA is given by I = JA = en v Ae d , where J is the current density, e is the
electronic charge, ne is the free electron density, vd is the drift velocity and A = π R^2 is the area.
With the numbers provided, we obtain vd = 2.7 × 10 −^4 m/s.
- ( Exam 2 ) An electron moves at a speed of v = 4 × 107 m/s parallel to and a distance d = 15 cm from a conducting wire carrying a current of I = 60 A. If the directions of the electron and cur- rent are as shown in the figure, what is the magnitude (in N) and direction of the force experi- enced by the electron?
(1) 5.1 × 10 −^16 toward the wire (2) 7.7 × 10 −^16 away from the wire (3) 1.8 × 10 −^16 away from the wire (4) 6.1 × 10 −^17 toward the wire (5) 2.8 × 10 −^16 toward the wire
The B field a distance d from the wire is B = μ 0 I / 2π d, so the force on the electron is thus
F = evB = ev μ 0 I / 2π d. Plugging in the numbers yields
F = 5.1 10× −^16 N, and the right hand
rule (accounting for negative charge) shows the force is towards the wire (as expected since we effectively have two currents in the same direction).
C 2 C 3
C 4
C 5
C 1
I
e −
d
- ( Webassign 31.43 ) Consider an RLC circuit with a driving emf of amplitude Em = 15 V, R =
25 Ω, L = 0.08 H and C = 3.0μF. What is the amplitude of the voltage (in volts) across the induc- tor at resonance?
(1) 98 (2) 220 (3) 66 (4) 345 (5) 178
At resonance only R contributes to the impedance, so the current amplitude is Im = 15 / 25 =0.
A. The voltage amplitude across the inductor can be found from the inductive reactance to be
VL = X L I m = ω LIm. Using ω = 1/ LC at resonance and substituting values yields 98 V.
- ( Webassign 32.32 ) An electron is placed in a magnetic field B that is directed along a z axis. The energy difference between parallel and antiparallel alignments of the z component of the electron's spin magnetic moment with B is 2.60 × 10 −^25 J. What is the magnitude of B in Tesla?
(1) 0. (2) 0. (3) 0. (4) 1.7 × 10 −^4 (5) 3.3 × 10 −^6
The difference in potential energies of an electron in a magnetic field B is Δ E = 2 μ BB, where μ B
is the Bohr magneton. The above values yield B = 0.014 T.
- ( Exam 3 ) A ray of light from a source placed at the bottom of a pool of water ( n = 1.333) impinges on the water surface with an angle of incidence of 52°. Which of the following state- ment is true?
(1) The ray is totally reflected at the surface. (2) The angle of refraction is 36.2°. (3) The angle of refraction is 90°. (4) The angle of reflection is 48.2°. (5) The angle of reflection depends on the wavelength of the light.
The angle of incidence is larger than the critical angle of 48.2 ° , therefore the ray is totally re- flected.
- ( WebAssign 34.52 ) An observer views an object through a diverging lens, whose focal length is 5 cm. If the object is located 10 cm from the lens, what is the lateral magnification of the ob- ject?
(1) 0. (2) 0. (3) 1. (4) 2. (5) 0
We use the lens equation 1/ p + 1/ q = 1/ f, with f = -5. This yields q = -3.33, and the magnifica-
tion is given by m = − q / p = 3.33 /10 ≅0.
- ( Chapter 34, checkpoint 3, in class ) A person standing between two vertical parallel mirrors A and B looks into mirror A. The distance between the two mirrors is 3 m, and a lit candle is po- sitioned 1 m from mirror A. How deep behind this mirror does the second image of the candle appear, in meters?
(1) 5 (2) 2 (3) 3 (4) 4 (5) 6
The first image in A is 1 m behind A, which makes an image in B 4 m behind B, which in turn makes an image in A 7 m behind A. However, the first image of the candle in B is 2 m behind B, which makes an image in A 5 m behind A.
- ( Chapter 34 new ) Let p be the distance between an object and a spherical mirror or a lens, whose focal length is f. Which of the following situation produces an inverted image of the ob- ject?
(1) The spherical mirror is concave, and p > | f |. (2) The spherical mirror is convex, and p > | f |. (3) The lens is a converging lens, and p < | f |. (4) The lens is a diverging lens, and p < | f |. (5) The lens is a diverging lens, and p > | f |.
This can be seen by inspecting the other answers, which are all false.
sheets is given by the expression L n 1 1 − L n 2 2 = 4 ×1.5 − 3.5 ×1.6 = 0.4 μm , which is 0.67 wave-
lengths. If you report the phase after both waves travel the longer distance of 4 μ m (which is
NOT what was asked in the problem, but the WebAssign problem made the same mistaken as-
sumption), then the shorter wave travels an additional 0.5 μ m, so the total optical length differ-
ence is 0.1 μ m or 0.167 wavelengths. Because of possible confusion, we did not count this prob-
lem, giving everyone credit for it.
- ( Chapter 35, Sample problem 35-6, in class ) An anti-reflection coating of a lens consists of a thin layer of MgF 2 , whose index of refraction is larger than that of air but smaller than that of the lens glass. What is the best thickness of the coating that minimizes the reflection? Assume that the light is perpendicular to the lens surface.
(1) 1/4 of the wavelength of the light in MgF 2 (2) 1/16 of the wavelength of the light in MgF 2 (3) 1/8 of the wavelength of the light in MgF 2 (4) 1/2 of the wavelength of the light in MgF 2 (5) exactly the same as the wavelength of the light in MgF 2
The coating must be 1/4 wave thick to make the two reflected waves a total of 1/2 wavelength out
of phase. The extra phase of 1/2 λ from reflection is common to both waves and cancels in the
difference.
- ( Webassign 36.40 ) Some tropical beetles are colored by optical interference that is due to scales whose alignment forms a diffraction grating. When the incident light rays are perpendicu- lar to the grating, the angle between the first-order maxima (on opposite sides of the zeroth-order maximum) is 26° in light with a wavelength of 550 nm. What is the grating spacing of the beetle in μm?
(1) 2. (2) 1. (3) 0. (4) 1. (5) 0.
The angle between the first maximum from the center is 13 °. The spacing d can be calculated
from d sin θ = λ , which gives λ = 2.4 μ m.
- ( Webassign 36.50 ) A line in the spectrum of a certain element is a doublet with wavelengths 581.6 and 583.0 nm. Calculate the minimum number of lines needed in a grating that will resolve this doublet in the second-order spectrum.
(1) 208 (2) 166 (3) 350 (4) 83 (5) 450
The resolving power is R ≡ λ ave /Δλ = Nm, where N is the number of slits and m is the order.
Using m = 2, we solve for N to get N = 582.3 / 2.8 = 208 slits.
- ( Chapter 36 Sample problem 36.3 ) A person observes a pointillistic painting (made of col- ored dots spaced by an average distance of 2 mm). As she steps backwards, at approximately what distance (in m) from the painting will she see the red dots in the painting (λ = 650 nm) merge together? Assume that her vision is limited by diffraction and that her pupil diameter is equal to 2.5 mm.
(1) 6. (2) 8. (3) 10. (4) 3. (5) 4.
The dots will merge when their angular spacing θ = d / L (d is spacing and L is distance) as seen
by the woman is approximately equal to the minimum angle her eyes are able to resolve, which is
given by the Rayleigh angular distance 1.22 λ / D, where λ is the wavelength of light and D is the
diameter of her pupils. Solving for L yields L = 6.3 m.
