Phys 2102 Spring 2002 Exam 1

Transcription

1 Phys 2102 Spring 2002 Exam 1 February 19, When a positively charged conductor touches a neutral conductor, the neutral conductor will: (a) Lose protons (b) Gain electrons (c) Stay neutral (d) Lose electrons (e) Gain protons Correct answer: (d) When the conductors touch, they are then both charged positively. Only electrons can move from one conductor to another, so if the neutral conductor ended up with a positive charge it means it lost electrons. 2. Circle T if the statement is True and F if the statement is False: (a) T F The direction of the electric field at a point gives the direction of the force on a negative particle placed at that point. Correct: The direction of the electric field at a point gives the direction of the force on a positive particle placed at that point. (b) T F Electric field lines extend toward positive charges where they terminate. Correct: Electric field lines extend toward negative charges where they terminate. (c) T F Where field lines are close together, E is large; and where they are far apart, E is small. (d) T F Electric field lines never cross. (e) T Electric field lines are always perpendicular to equipotential surfaces. 1

2 3. Two electrons and a proton are located on the x-axis as shown, where a=20nm. Find the electrostatic force, F, on the electron at the origin due to the other two charges. Give both the magnitude and direction of F. Solution: There are two forces on the electron, one due to the electron on the left, F e, and another due to the proton on the right, F p. The force due to the electron is repulsive, so it points to the right, and has magnitude given by Coulomb s law: F e = ke2 The force due to the proton is repulsive, so it points to the right too, and has the same magnitude given as F e because the proton and electron have charges of the same magnitude: F e = ke2 The total force is then F = F e + F p = 2 ke2 = Nm 2 / ( C) 2 ( m) 2 = N î 4. Charge is uniformly distributed on one side of an infinite non-conducting sheet. The plane of charge is perpendicular to the x-axis, and is located at x=0. The electric field at x=2.0m is given by E=50 N/C î. Find the surface charge density σ of the plane of the charge. Solution We know the field will be perpendicular to the plane, pointing outwards (to the right on the right side of the plane, to the left on the left side of the plane). We want to use Gauss law to calculate the field as a function of the charge on the plane. We choose for our Gaussian surface a small cylinder of radius R (a pillbox ) across the plane, which will then have three surfaces. We need to calculate the flux through those surfaces. The field will be tangent to the curved side of the cylinder, and thus the flux through that surface will be zero. The field will be perpendicular to the right and left cap sides of the cylinder, pointing in the same direction as the normal vector. So, for each of the caps the flux is EA, where A is the area of each cap. The flux through the cylinder is then Φ = 2EA. According to Gauss law, the flux is equal to the charge enclosed within the pillbox, divided by ɛ 0. If σ is the charge per unit area on the plane, the charge enclosed is q enc = σa. So, Gauss law tells us: 2

3 Φ = q enc ɛ 0 2EA = σa ɛ 0 σ = 2ɛ 0 E = /Nm 2 50N/C = C/m 2 [[This is the first paragraph in section 24.8 of the textbook. You did not need to write all of this down if you remembered the formula E = σ/2ɛ 0 correctly, of course]]. 5. Charges q 1, q 2 and q 3 are located at a distance L from the origin as shown. Let q 1 = C, q 2 = C and q 3 = C, and L = m. (a) Find the potential V at the origin. Solution: The potential of each charge at the origin is V i = kq i /L. The total potential at the origin is the sum of the potentials produced by each charge: V = V 1 + V 2 + V 3 = k q 1 L + k q 2 L + k q 3 L = k q 1 + q 2 + q 3 L C m = 27.0 Nm 109 C = V (b) Find the potential energy U of this system of three charges. Solution: To find the potential energy, we need to calculate the work needed to build the system of charges, bringing them into place one by one. To bring the first charge q 1, we don t need to do any work, since there are no electric forces anywhere yet: W 1 = 0. To bring the second charge q 2 into place, we are working against the electric field of the first charge. The work to bring charge q 2 is W 2 = q 2 V, where V is the potential produced by q 1 at the point where we bring q 2. The charge q 1 produces a potential V 1 = kq 1 /r, so at the position where q 2 is left, a distance away, the potential is V 1 = kq 1 /, and the work needed to bring q 2 is then W 2 = q 2 V 1 = q 2 k q 1 3

4 In order to bring the third charge, we need to work against the fields produced by q 1 and q 2. Each of the two charges already in place will produce a potential at the point where we want to bring q 3, which is at equal distances from Q 1 and q 2 : V 1 = kq 1 / and V 2 = kq 2 /. Again, the work will be W 3 = q 3 V, where the potential V is th e superposition of V 1 and V 2, so W 3 = q 3 k q 1 + q 2. The total work needed to bring the charges, which is equal to the potential energy stored in the system, is equal to U = W = (W 1 + W 2 + W 3 ) = 0 + k q 1q 2 + k q 3(q 1 + q 2 ) = Nm C C m Nm C C m = Nm Nm = Nm = J Notice that the work we need to bring the charges is positive, so the energy stored in the system is positive. The work done by the electric field is negative. 6. A spherical metal shell with inner radius R 1 = 0.10m and outer radius R 2 = 0.12m has a net charge Q 1 = 5µC. A point charge Q 2 = 2µC is located at the center of the spherical shell as shown. (a) Use Gauss law to find the electric field E for r > R 2. Solution: Define a spherical Gaussian surface of radius r, centered at the point charge Q 2. In order to use Gauss law ɛ 0 Φ = q enc, we need to calculate the electric flux through the Gaussian sphere Φ = E da and the charge enclosed by the Gaussian surface qenc. The field at all points on this surface will have equal magnitude and point perpendicular to the surface, outwards (since the charges inside are positive). The flux, then, is easy to calculate: E da = EdA cos(0) = E da = EA, where A is the surface of the Gaussian sphere: A = 4π. So, we have Φ = E4π. The charge enclosed by the Gaussian sphere is easy: ALL of the charge is inside, so q enc = Q 1 + Q 2. So, using Gauss law: 4

5 ɛ 0 Φ = q enc ɛ 0 E4π = Q 1 + Q 2 E = Q 1 + Q 2 4πɛ 0 = k Q 1 + Q C = Nm C (b) Use Gauss law to find the electric field E for r < R 1. Solution: Define a spherical Gaussian surface of radius r < R 1, centered at the point charge Q 2. Again, the flux is easy to calculate: E d A = EA = E4π. The charge enclosed by the Gaussian sphere is now JUST the point charge Q 2, since the charges on the shell are outside the Gaussian surface. Then, again using Gauss law: ɛ 0 Φ = q enc ɛ 0 E4π = Q 2 E = Q 2 4πɛ 0 = k Q C = Nm C (c) The electric field is zero inside the conductor (E = 0 for R 2 > r > R 1 ). Use this information and Gauss law to determine the charge Q I on the inner surface of the conducting shell. Solution: Choose a Gaussian sphere inside the thick shell, with a radius larger than the radius of the inner surface, but smaller than the radius of the outer surface. The field at all points on this Gaussian surface is zero, since these points are inside the conductor. Therefore, the flux through the Gaussian surface is zero: Φ = E d A = 0 because E = 0. From Gauss law, then ɛ 0 Φ = q enc, we deduce that the enclosed charge must be zero. The enclosed charge is Q 2 and all of the charge on the inner surface Q I. Since Q 2 + Q I = 0, we have Q I = Q 2 = 2.0µC. This was not asked, but since the total charge on the shell is Q 1 = 5.0µC, and now we know that the charge on the inner surface is Q I = 2.0µC, we know that the charge on the outer surface must be Q E = +7.0µC. 5