Physics 2212 GH Quiz #2 Solutions Spring 2015

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1 Physics 2212 GH uiz #2 Solutions Spring 2015 Fundamental Charge e = C Mass of an Electron m e = kg Coulomb constant K = N m 2 /C 2 Vacuum Permittivity ϵ 0 = C 2 /N m 2 Unless otherwise directed, friction and drag should be neglected. Any integrals in free-response problems must be evaluated. uestions about magnitudes will state so explicitly. I. (18 points) A system of two charged particles 3.0 cm apart has an electric potential energy of 93.5 µj, with respect to zero at infinite separation. The total charge in the system is 37 nc. What is the charge of each particle? The electric potential energy of a system of two particles with charges q 1 and q 2 is U = K q 1q 2 r with respect to zero at infinite separation. Let the total charge of the system be = q 1 + q 2. So U = K q 1 ( q 1 ) r which is a quadratic in q 1. So Ur K = q 1 q1 2 q1 2 q 1 + Ur K = 0 q 1 = B ± B 2 4AC 2A where A = 1 B = = C C = Ur ( K = J ) ( m ) N m 2 /C 2 = C 2 Therefore q 1 = ( C ) + ( C) 2 4 (1) ( C 2 ) = C 2 (1) or q 1 = ( C ) ( C) 2 4 (1) ( C 2 ) = C 2 (1) Since it doesn t matter which charge is called q 1 and which is called q 2, the charges of the two particles are 24 nc and 13 nc uiz #2 Solutions Page 1 of 7

2 II. (16 points) An infinite insulating slab of charge has thickness 2t, extending from t to +t in the z direction. In the x and y directions, it is infinite in extent. It has a non-uniform volume charge density ρ that depends on position z according to ρ = ρ 0 z 2 where ρ 0 is a positive constant. What is the magnitude of the electric field at a position +t/2 inside the slab on the z axis? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants. Use Gauss Law ϵ 0 Φ = q in ϵ 0 E d A = q in Choose a Gaussian Surface with the symmetry of the charge distribution, such as a right circular cylinder, extending from z = t/2 to z = +t/2 and having end-cap area A. The flux through the curved sides is zero, and the flux through each end-cap is EA, so Φ = E da = 2EA The charge density varies within this Gaussian Surface, so q in = dq in = ρ dv where the differential volume element dv must be small in the direction of the variation (i.e., in the z direction). A thin disk, of area A and height dz would be such a volume element. To find all the charge within the Gaussian Surface, the charge elements must be integrated from z = t/2 to z = +t/2. q in = ρ dv = = ρ 0A [ z 3 3 +t/2 t/2 ] +t/2 t/2 z [ρ 2 ] 0 = ρ 0A 3 A dz = ρ 0A [ ( ) 3 t 2 +t/2 t/2 z 2 dz ( ) ] 3 t = ρ 0A 2 3 [ ] 2t 3 8 = ρ 0At 12 Putting these together, ϵ 0 2EA = ρ 0At 12 E = ρ 0t 24ϵ 0 1. (5 points) Outside the slab in the problem above, how does the magnitude of the electric field E depend on the distance z from the xy plane? Outside the slab, the field must be the same as that of an infinite sheet of charge. E is constant, but not zero, for z > t. uiz #2 Solutions Page 2 of 7

3 III. (16 points) A thin insulating rod of length L lies on the +x axis, with one end a distance d from the origin, as shown. This rod has a non-uniform linear charge density, λ, that varies with position x according to λ = λ 0 x L e x/l where λ 0 is a positive constant. What is the magnitude of the electric potential at the origin, with respect to zero at infinity? Express your answer in terms of parameters defined in the problem, and physical or mathematical constants Since the rod is thin, it can be divided into point-like elements of charge dq, each of which contributes an element of potential dv = K dq r at the origin. The element of charge can be related to the linear charge density by λ = dq dx dq = λ dx = Each element of charge is a distance r = x from the origin, so [ λ 0 x L e x/l] dx V = dv = K dq r = K = Kλ 0 [ e x/l] d+l d d+l [ x λ0 d L e x/l] dx x = Kλ 0 d+l d e x/l dx L = Kλ 0 [e (d+l)/l e d/l] = Kλ 0 e d/l [1 1/e] 2. (5 points) If it can be determined in the problem above, what is the direction of the electric potential at the origin? Electric potential is a scalar! The potential at the origin doesn t have a direction, regardless of its magnitude. uiz #2 Solutions Page 3 of 7

4 3. (5 points) The figure shows two locations inside an ideal parallel plate capacitor. If the potential is zero at the negative plate, what is the ratio of the potential at poin to that at point 1? The electric potential inside a parallel-plate capacitor varies linearly with position. Poin is four times as far from zero as point 1 is, so V 2 /V 1 = 4 4. (5 points) The figure shows two locations inside an ideal parallel plate capacitor. If the potential is zero at the positive plate, what is the ratio of the potential at poin to that at point 1? The electric potential inside a parallel-plate capacitor varies linearly with position. Poin is one-fourth times as far from zero as point 1 is, so V 2 /V 1 = 1/4 uiz #2 Solutions Page 4 of 7

5 5. (5 points) A proton is given an initial velocity directly toward a fixed positive charge. If the proton reaches a turning point, what is true about the potential energy of the system of two charges at that instant, and what is true about the potential at the location of the proton at that instant? Let the zero of potential energy be at infinite separation, and the zero of potential be infinitely far from the fixed charge. With no external forces doing work on the system, and no non-conservative forces within it, the total electric and kinetic energy is constant. When the proton stops, the kinetic energy of the system is zero, so the potential energy must be maximum. The potential at the location of the proton depends only on its distance from the fixed charge, V = Kq/r (with respect to zero at infinite separation). The separation is small at the turning point, and the fixed charge is positive, so the potential has a large positive value. The potential energy is at a maximum, and the proton is at maximum potential. 6. (5 points) An electron is given an initial velocity directly toward a fixed negative charge. If the electron reaches a turning point, what is true about the potential energy of the system of two charges at that instant, and what is true about the potential at the location of the electron at that instant? Let the zero of potential energy be at infinite separation, and the zero of potential be infinitely far from the fixed charge. With no external forces doing work on the system, and no non-conservative forces within it, the total electric and kinetic energy is constant. When the electron stops, the kinetic energy of the system is zero, so the potential energy must be maximum. The potential at the location of the electron depends only on its distance from the fixed charge, V = Kq/r (with respect to zero at infinite separation). The separation is small at the turning point, and the fixed charge is negative, so the potential has a large negative value. The potential energy is at a maximum, and the electron is at minimum potential. uiz #2 Solutions Page 5 of 7

6 7. (5 points) Consider three separate situations, in which a positively charged particle is near a cylinder. In situation i, the particle is on the perpendicular bisector of the cylinder axis. In situation ii, the particle is in the plane containing the top of the cylinder. In situation iii, the particle is on the cylinder axis. In which situation or situations, if any, is the net flux through the curved side surface of the cylinder positive? Sketch field lines from the charged particle. As it contains no charge, the net flux through the cylinder must be zero. In situation i, the top and bottom of the cylinder have positive flux through them, so the flux through the curved side surface must be negative. In situation ii, there is zero flux through the top of the cylinder, but positive flux through the bottom, so the flux through the curved side surface must be negative. In situation iii there is a large negative flux through the top of the cylinder, and a smaller positive flux though the bottom. The flux through the curved side surface must be positive. So the net flux through the curved side surface of the cylinder is positive Only in situation iii. 8. (5 points) The triangular faces of a triangular prism are isosceles right triangles with legs of length L. These triangular faces are a distance 3L/2 apart, as shown. Note that the angle θ in the figure must be 45. A uniform electric field E points directly up the page, perpendicular to the lower rectangular face with edges L and 3L/2. What is the magnitude of the electric flux through the shaded face, with edges 3L/2 and L 2? Every field line that passes through the shaded surface, also passes through the bottom surface. Those two surfaces must have the same flux magnitude. It is simpler to calculate the flux through the bottom surface. Φ = E A = EA cos (180 ) = E L ( ) 3L 2 Φ = E 3L 2 /2 uiz #2 Solutions Page 6 of 7

7 9. (5 points) A small hollow conducting sphere has inner radius R and outer radius 2R. It has positive charge +, and lies at the center of a larger hollow conducting sphere. This larger hollow conducting sphere has inner radius 3R and outer radius 4R. It has positive charge +3. At which surface or surfaces will the electric field of greatest magnitude be found? The field magnitude at each surface will be the same as that of an infinite sheet with the same area charge density E = η E = q ϵ 0 ϵ 0 A = q ϵ 0 4πr 2 q r 2 Using Gauss Law and knowing that the field within a conductor at equilibrium is zero, the inner sphere s charge + must lie entirely on the outer surface of the sphere. E(2R) (2R) 2 = 1 4 R 2 Now consider a Gaussian Surface within the thickness of the outer sphere. The field within the thickness of the sphere is zero, so the flux is zero. Therefore, the charge within that Gaussian Surface must be zero. The inner sphere carries a charge of +, so the inner surface of the outer sphere must carry a charge +. So, E(3R) (3R) 2 = 1 9 Charge must be conserved on the outer sphere. The total charge is +3, so with on the inner surface, there must be +4 on the outer surface. E(4R) = 4 (4R) 2 = 4 16 R 2 R 2 = 1 4 R 2 Therefore The surfaces of radius 2R and 4R have the same field of greatest magnitude. 10. (5 points) The field outside a uniformly charged solid cylinder of radius R and infinite length is identical to that of an infinite line of charge having the same linear charge density: E = λ/2πϵ 0 r for r greater than or equal to R. What is the electric field like inside the cylinder? From symmetry considerations, the field must be zero on the cylinder axis (r = 0). From Gauss Law, which is always valid, it must be non-zero everywhere else inside the cylinder. The only option provided which is consistent with these requirements, is The field increases linearly from zero at the center to λ/2πϵ 0 R at the surface. uiz #2 Solutions Page 7 of 7