PHY2049 Fall 2000 FINAL EXAM. Four charges (two of them have charge +Q, the other Q) are placed in the corners of a square with a side a. Find the magnitude of forces acting on each of the charges (Q = 00 µc, a =3m). Answer: 9 N -Q +Q F F 2 F3 +Q -Q The net force on any charge is the vector sum of the forces due to the three neighboring charges. Because of the symmetry of the charges and their locations, each charge experiences the same force. Consider the lower left charge: The forces are given by Coulomb s law and F = k Q Q a 2 F = F + F 2 + F 3 = F ĵ F 2 cos 45 î F 2 sin 45 ĵ + F 3 î = (F 3 F 2 / 2)î +(F F 2 / 2)ĵ. = kq2 a 2 = (8.99 09 Nm 2 /C 2 )(00 0 6 C) 2 (3 m) 2 =0.0 N F 2 = k Q Q ( 2a) 2 = kq2 2a 2 = (8.99 09 Nm 2 /C 2 )(00 0 6 C) 2 2(3 m) 2 =5.0 N F 3 =. Therefore, k Q Q a 2 = kq2 a 2 = (8.99 09 Nm 2 /C 2 )(00 0 6 C) 2 (3 m) 2 =0.0 N F = (0.0 N 5.0/ 2N)î +(0.0 N 5.0/ 2N)ĵ = 6.46 Nî +6.46 Nĵ F = (6.46 N) 2 +(6.46 N) 2 = 9.4 N. 2. The electric field in three dimensions is E =(ay + b)ĵ + cˆk, witha =3N/Cm,b =2 N/C, and c = N/C. Find the amount of electric charge trapped inside a cube with a side of m and positioned as shown in the Figure. The answers are rounded to two significant figures. Answer: 27 pc
y x z Gauss law states that the enclosed charge is related to the total flux through the cube. Φ = E da = E da + E da 2 + E da 3 + E da 4 + E da 5 + E da 6. 2 3 Let L = m be the length of a cube edge. Assume also that the sides are enumerated as follows: Area Location Normal E da yz plane î 0 2 parallel to yz plane at x = L î 0 3 xy plane ˆk cda 4 parallel to xy plane at z = L ˆk +cda 5 xz plane ĵ (a0+b) da 6 parallel to xz plane at y = L ĵ (al + b) da The total flux is Φ = E d A 3 + 3 = ( c) da + = al da = al da = al(l 2 ) = (3 N/(Cm))( m)( m) 2 = 3 Nm 2 /C. 4 4 E da 4 + E d A 5 + E da 6 5 6 cda+ (a0+b) da + (al + b) da Gauss law implies q enc = ɛ 0 Φ=(8.85 0 2 C 2 /Nm 2 )(3 Nm 2 /C) = 2.66 0 C. 3. One has two lights; both designed to operate at 20 V. One of them is a 00 W bulb; the other one is a 25 W bulb. The two light bulbs are connected in series and plugged in a 5 6
20 V bulb bulb 2 20 V power outlet as shown in figure. What is the total power dissipated in this circuit? Answer: 20 W The power dissipated in a resistor is given by P = V 2 /R. Solving for the resistance, R = V 2 /P.So R = V 2 P = R 2 = V 2 P 2 = (20 V)2 00 W = 44 Ω (20 V)2 = 576 Ω. 25 W The resistor are in series in the new circuit. The total resistance is R = R + R 2 = 720 Ω. The power lost in this circuit is P = V 2 R = (20 V)2 720 Ω =20W. 4. Three identical parallel plate capacitors have capacitance C 0 each. The gaps between capacitor plates are filled with air. All the capacitors are modified as follows: the gap between plates of the capacitor is decreased by a factor of 2, the gap between plates of the capacitor 2 is increased by a factor of 2, the gap between plates of the capacitor 3 is filled with oil whose dielectric constant equals 2. Then the three modified capacitors are connected as shown in the figure. Find the equivalent capacitance of the circuit. Answer: 2.4C 0 () (2) (3) The capacitance of a parallel plate capacitor is related to its geometry, C = ɛ 0 A/d. Therefore, halving the gap will double the capacitance and doubling the gap will halve the capacitance. This means that C =2C 0 and C 2 = C 0 /2. The addition of the oil multiples the original capacitance by the dielectric constant giving C 3 =2C 0. The circuit consists of C and C 2 in series and C 3 in parallel with the combination. For capacitors in series C = C + C 2 = 2C 0 + 2 C 0 = 5 2C 0
which yields, C =0.4C 0. Capacitors in parallel add and the equivalent circuit is C eq = C + C 3 =0.4C 0 +2C 0 =2.4C 0. 5. An infinitely long wire has a loop as shown in the figure. What is the magnitude of the magnetic field at the center point of the loop, if the wire carries 2 Acurrent and R =2 mm? Answer: 0.8 mt R The magnetic field at the center of the circle is the vector sum of the fields due to the circle and the line. By the right hand rule, both fields are directed out of the plane. Taking that direction as positive, B = B c + B l = µ 0I 2R + µ 0I 2πR = µ ( ) 0I 2R π + = (4π 0 7 Tm/A)(2 A) 2(2 0 3 m) = 8.28 0 4 T. ( π + ) 6. Anegatively charged particle moves with velocity v in a uniform magnetic field B as shown in the figure. Both v and B are in the plane of the paper. What is the direction of the force on the particle? Answer: into the paper v B - The velocity has components v = v x î + v y ĵ and the magnetic field has components B = B x î B y ĵ. The charge is negative will be designated as Q. Theforceisgivenby F = q v B = Q( v x î + v y ĵ) ( B x î B y ĵ) = Q(( v x )( B y ) (v y )( B x ))ˆk = Q(v x B y + v y B x )ˆk.
Since the coefficient of ˆk is negative, the force is in the negative z direction and into the paper. (This result can also be found by directly applying the right hand rule for the cross product.) 7. Aray of light enters a glass cube at point Aat an incident angle of 30 and then undergoes total internal reflection at point B. What minimum value of index of refraction n of glass can be inferred from this information? Answer:.2 B A 30 o θ ο 90 - θ Applying Snell s law at A n sin θ = n 2 sin θ 2 sin30 = n sin θ sin θ = 2n. For total internal reflection to occur at B, 90 θ θ c. Taking the sine of both sides, sin(90 θ) sin θ c. The critical angle is given by sin θ c =/n. This leads to sin(90 θ) sin θ c cos θ sin θ c cos 2 θ sin 2 θ c sin 2 θ n 2 (2n) 2 n 2 4n 2 4 n 2 5/4 n 5/4 =.2. 8. An object is placed 3.6 cm in front of a concave mirror. Its image is virtual and four times as tall as the object. Calculate where the same object should be positioned in order that its virtual image would be only twice as large as the object. Answer: 2.4 cm
Since the virtual image from a single mirror is upright, the image height is positive and the magnification, m = h i /h o = +4. The magnification also involves the object and image distances, m = i/p. Solvingfori gives i = mp = 4(3.6 cm)= 4.4 cm. Substituting into the mirror equation, it is possible to find the focal length of the mirror, f = p + i = 3.6 cm + 4.4 cm = 3 4.4 cm, and f =4.8 cm. The object is moved to another position where the virtual image is now 2 times as tall as the object. Therefore, i = 2p. Substituting into the mirror equation f = p + i = p + 2p = 2p p = f 2 = 2.4 cm. 9. What is the smallest object on the Moon that can be resolved with the Hubble Space Telescope whose diameter is 2.4 m. The typical light wavelength is 500 nm. The distance between the Earth and the Moon is 400,000 km. Neglect the fact that the telescope is at 600 km above the Earth. Answer: 00 m The Rayleigh criterion for the limit of resolution is θ R =.22 λ ( 500 0 9 ) d =.22 m =2.54 0 7 rad. 2.4 m This angle is subtended by an object of length L locatedonthemoon,adistancer from the telescope. This tiny angle is well approximated by θ = L/R. Assuming this object can just be resolved, θ = θ R L R = θ R L = θ R R = (2.54 0 7 rad)(400,000 0 3 m) = 02 m. 0. Two coherent radio-frequency point sources separated by d = 2.0 m radiate in phase with a wavelength λ =0.5 m. Adetector moves in a circular path of some radius R around the mid-point between the two sources (see figure) and measures the radio-wave intensity. Find how many maxima in detects. Answer: 6
d R Maxima occur when the path difference for the two signals is equal to an integer number of wavelengths, = mλ. Points on the x-axis are the same distance from the two sources. The path difference on the x-axis is zero and, correspondingly, those points will experience constructive interference. The path difference for points on the y-axis is 2.0 m. Since the wavelength is 0.5 m, this corresponds to m = 4 wavelengths and these points also experience contructive interference. As the detector rotates from the +x-axis to the +y-axis, the path difference must change continuously from 0 to 4 wavelengths. Therefore, somewhere during its trip, the detector must pass though points where the path difference corresponds to m =, 2, and 3 wavelengths. As the rotation continues into the second quadrant, the pattern reverses to 3, 2, and wavelength path differences. In the top half of the diagram, there are 3 points in the first quadrant and 3 points in the second quadrant. This pattern continues in the lower half of the diagram. Counting all the locations for maxima, there are 4(points on the axes) + 3(points in a quadrant) 4(quadrants) = 6(maxima).