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Unformatted text preview: Solution for Homework 5 Gauss Law Solution to Homework Problem 5.1(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 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 5.2(Tent Flux) Problem: While sleeping in a tent, you are nearly struck by lightning and receive a charge of 1 2 10- 5 C . Compute the electric flux out of the tent. Select One of the Following: (a) (b) 4 10- 17 Nm 2 C (c) 7 10- 4 Nm 2 C 1 (d-Answer) 6 10 5 Nm 2 C (e) Cannot be calculated because the tent does not have a symmetric shape. Solution By Gausss Law, the total electric flux out of the tent is proportional to the charge it encloses, or = Q enclosed = 1 2 10- 5 C 8 . 85 10- 12 C 2 Nm 2 = 6 10 5 Nm 2 C Total Points for Problem: 3 Points Solution to Homework Problem 5.3(Integration circumvention) Problem: In a spherically symmetric Gauss Law calculation, the following step is performed: integraldisplay S E ( r ) dA = E ( r ) integraldisplay S dA Why can this be done? Select One of the Following: (a) E is constant everywhere. (b) The surface normal of a sphere is r . (c) It is an approximation that makes the integral easier. (d-Answer) The electric field has the same magnitude at all points on S . (e) Because the moon is in the 7th house. Solution The electric field has the same magnitude at all points a distance r from the origin, and so is constant on S ....
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