This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Solution for Homework 5 Gauss Law Solution to Homework Problem 5.1(Net Charge Enclosed By a Gaussian Surface) Problem: The arrows in the figure represent electric field lines of equal magnitude. The Gaussian cube in the figure encloses Select One of the Following: (a) a positive charge. (b) a negative charge. (c) zero charge. (d) a net positive charge. (e-Answer) a net negative charge. (f) no net charge. Solution Four lines point into the box, and only two lines point out; more lines point into the box. The number of field lines is proportional to the flux. The flux out of the surface is proportional to the charge enclosed. The correct choice is (e). Total Points for Problem: 3 Points Solution to Homework Problem 5.2(Balancing Forces) Problem: A pith ball (a small sphere of tree bark) has mass m = 60mg = 6 10- 5 kg and receives a charge of +1nC = 1 10- 9 C when zapped with the electrophorous. Our electrostatic generator (the Van de Graaff) has a spherical surface with radius 12 . 5cm . When in operation a substantial surface charge is transferred to the sphere. If the pith ball is placed directly on top of the Van de Graaff generator (without transferring charge), how much total charge must be placed on the Van de Graaff to lift the pith ball against the force of gravity? Select One of the Following: (a) 20nC (b) . 6 C (c-Answer) 1 C (d) 5 C (e) 150 C Solution 1 If we take upward to be the positive y direction, the force of gravity on the pith ball is vector F g =- mg y , where g = 9 . 81 m s 2 . The electric force of the Van der Graaf on the pith ball is vector F e = vector F 12 = kq 1 q 2 r 2 12 r 12 In order for the pith ball to hover, the forces must cancel, or vector F g + vector F e = 0 . We know q 2 = 1 10- 9 C , m = 6 10- 5 kg , and vector r 12 = 0 . 125my , so solve for q 1 , and plug in the numbers. vector F g =- vector F e- mg y =- kq 1 q 2 r 2 12 y = mg = kq 1 q 2 r 2 12 q 1 = mgr 2 12 kq 2 = (6 10- 5 kg)(9 . 8 m s 2 )(0 . 125m) 2 (8 . 99 10 9 Nm 2 C 2 )(1 10- 9 C) = 1 . 02 10- 6 C q 1 = 1 C Total Points for Problem: 3 Points Solution to Homework Problem 5.3(Compute the Charge in a Box) Problem: A 10 . 0cm box is centered at the origin with sides parallel to the coordinate planes. The box is in an electric field which changes with x , vector E = ( Dx ) x where D = 10 . 0N / Cm is a constant. What charge is contained in the box? Select One of the Following: (a) 1 . 77 10- 13 C (b) 0C (c) 9 . 11 10- 9 C (d-Answer) 8 . 85 10- 14 C (e) It is impossible to tell with the information given. Solution x (cm) y (cm) 5 5-5-5 E n l n r Definitions n l Outward Normal of Left Side of Box n r Outward Normal of Right Side of Box vector E = Dx x Electric Field Q Charge in the box A Area of a side of the box e,l Electric Flux through left side e,r Electric Flux through right side x l Location of Left side of box x r Location of Right side of box...
View Full Document