Physics 2049 Exam 1 Spring 2002

Transcription

1 Physics 49 Exam 1 Spring q r1 q1 r13 q3 1. In the figure q 1 = 3µC, q = 4µC, q 3 = 5µC, r 1 = 9m, r 13 = 1m. Compute the magnitude of the total force on q 3. F 3NET = F 31 + F 3 = kq 3q 1 î + kq ( 3q ) cos θî sin θĵ r13 r1 + r13 with θ = arctan( r 1 ) = r 13 F 3NET = 15.8Nî 4.8Nĵ F 3NET = (15.8) + (4.8) N = 16.5N 1

2 q Q q q. Three identical charges q are placed at the corners of the equilateral triangle of side length L. The fourth charge Q is placed in the center of the triangle. Find the value of Q in terms of q that would put the whole system of four charges in equilibrium. Calculating the net force on the top q and setting it to zero to enforce equilibrium we get, F qnet = k qq ĵ + kq sin(6 )ĵ = x L where x is the distance between q and Q. In terms of L, x = (L/) cos(3 ) = L/ 3 (1) so that 3k qq 3kq = L L or Q = q/ 3 (must have opposite sign for equilibrium) e E

3 3. An electron is placed in an electric field as shown. Find the direction of the electrostatic force on the electron. F = qe = ee Since F and E are antiparallel the force is directed to the left,. 4. A particle of mass 1 mg, and charge 6µC traveling with constant velocity v = 1m/sî enters a region with a constant electric field E = 1.N/Cĵ. What is the speed seconds after it enters the region with the electric field? The velocity of the charge only changes in the ĵ direction, F = qe = ma y a y = qe/m = 7.m/s v y (t) = a y t v y (s) = 14.4m/s v(s) = (1) + (14.4) )m/s = 17.53m/s 5. A spherical nonconducting balloon has a positive charge distributed uniformly over its surface. If the balloon is expanded (no extra charge is added) the magnitude of the electric field inside the balloon, and the magnitude of the electric field near the outer surface. If a point P is outside the balloon before expansion, and inside after expansion, the magnitude of the E field at that point. Use Gauss s law to get the electric field, E d a = E(4πr ) = q in S ɛ E = kq in r Since all the charge is uniformly distributed on the surface of the balloon, the electric field inside is zero. Therefore, the field after expansion will remain zero. A point just outside the surface of the balloon the electric field will decrease since the electric field drops off as 1/r. It should also be evident that since the electric field is positive outside the balloon, any point that starts outside and ends inside where the field is zero will experience a decrease. 3

4 y x x=cm x=1cm z 6. A cube with sides of length 1 cm rests on the y = plane, with one edge along the line x = cm and another edge along x = 1cm, as shown in the above figure. A nonuniform field (in N/C) pierces the the cube and is described by: where A =7.E+6 N/(C m). cube? E(x) = Axî How much charge (in C) is inside the Electric field flux only occurs at the x = cm and x = 1cm faces which we will call left and right, respectively. Gauss s law tells us, S E d a = Q ɛ = right E d a + left E d a The differential area elements on these surfaces are d a right = dydzî and d a left = dydzî. ( ) Q = ɛ E(x) L E(x) L x=1cm x=cm where L is the side length. Therefore, Q = ɛ 7Nm /C = 6.E 8C 4

5 7. The electric field at the surface of a sphere of radius R is measured. It is found to be radially outward and to have the same magnitude all over the sphere. Call that magnitude E 1. A second sphere of radius R is found to have a radially inward electric field that is uniform also but with magnitue 4E 1. What is the ratio of the charge in the second sphere to the charge in the first sphere (Q /Q 1 )? If we place Gaussian surfaces on the surfaces of both spheres they would tell us the magnitudes of the electric field are, E 1 = k Q 1, E R1 = k Q R E 1 = R Q 1 E R1 Q = 1 4 = 4 Q 1 Q Since the fields point in opposite directions, the signs of Q 1 and Q are opposite and we have Q Q 1 = A clever physicist produces a very long, cylindrical distribution of electrical charge. The diameter of the cyclindrical distribution is D. Inside the volume charge density varies radially as, ( ) ρ(r) = ρ 1 r D For r > D/, ρ(r) =. Here r is the radial distance from the cylindrical axis and ρ is a negative constant ρ <. Calculate the electric field generated by rho for all r > D/. That is, at any point outside of the region containing the charge. Again we use Gauss s law, but this time for cylindrical symmetry, S E d a = E(πrL) = q in /ɛ π L q in = dv ρ(v ) = dφ dz D/ = πl ρ dr r 5 ( 1 r D D/ ) dr rρ(r)

6 = πl ρ ( D 8 D 1 = πl ρ D 1 E = ρ D 4ɛ r inward Now by the way of the 5 possible answers, only had the electric field directed inward, allowing you to eliminate 3 of the possible choices. Of the remaining two, only the one derived above has the correct dimensions of of an electric field. Therefore, one could have answered this question without having to even know what an integral is! 9. The electric field in megan/c at a point halfway between a point charge of +µc at the origin and a point charge of -3µC at x = +1 cm is: Between the two charges, the electric field is pointed to the right and the components add, E(x) = k q 1 k q + x (.1m x) k(e-6c+ 3E-6C) E(.5m) = = 1.8E7 N/C = 18 megan/c (.5m) 1. An electric dipole is composed of equal and opposite charges each of magnitude.1µc, placed at y = 1 cm, respectively. The magnitude of the electrostatic field due to this dipole (in N/C) at the point x = +3m is: The magnitude of the electric field along the z axis due to the two charges along the y axis comes from adding the y components and cancelling the z components to give, E(z) = kq (d/) + z sin θ = ) kqd ((d/) + z ) 3/ d/ where sin θ =, and d is the distance between the charges. (d/) +z The dipole field comes from approximating the electric field for z d, so we can neglect the d/ in the denominator and we are left with, E(z) kqd z 3 = k (1E-7C)(.m) (3m) =.67 N/C 6