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Unformatted text preview: Husain, Zeena – Homework 2 – Due: Feb 3 2004, 4:00 am – Inst: Sonia Paban 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points 1) Two uncharged metal balls, X and Y , each stand on a glass rod and are touching. Y X 2) A third ball carrying a positive charge, is brought near the first two. Y X + 3) While the positions of these balls are fixed, ball X is connected to ground. Y X + 4) Then the ground wire is disconnected. Y X + 5) While X and Y remain in touch, the ball carring the positive charge is removed. Y X 6) Then ball X and Y are separated. Y X After these procedures, the signs of the charge q X on X and q Y on Y are 1. q X is negative and q Y is neutral. 2. q X is positive and q Y is negative. 3. q X is positive and q Y is neutral. 4. q X is neutral and q Y is negative. 5. q X is neutral and q Y is neutral. 6. q X is positive and q Y is positive. 7. q X is negative and q Y is positive. 8. q X is neutral and q Y is positive. 9. q X is negative and q Y is negative. cor rect Explanation: When the ball with positive charge is brought nearby, the free charges inside X and Y rearrange themselves. The negative charges are attracted and go to the right ( i.e. move to Y ), whereas the positive charges are repelled and collect in the left hand side of the system XY , i.e., in X . When we ground X , the positive charges which have collected in X are allowed to es cape (they strive to the left), whereas the negative charges in Y are still held enthralled by the positive charge on the third ball. We break the ground. Now we remove the third ball with positive charge. The charge on Y is redistributed in the system XY , i.e. they share the negative charge (equally if identical). Finally we separate X and Y . The signs of the charge on X and that on Y are both negative. Husain, Zeena – Homework 2 – Due: Feb 3 2004, 4:00 am – Inst: Sonia Paban 2 002 (part 1 of 1) 10 points A rod 19 . 5 cm long is uniformly charged and has a total charge of 13 . 2 μ C. Determine the magnitude of the electric field along the axis of the rod at a point 25 . 0685 cm from the center of the rod. Correct answer: 2 . 22428 × 10 6 N / C. Explanation: Given : ‘ = 19 . 5 cm , Q = 13 . 2 μ C , and r = 25 . 0685 cm . For a rod of length ‘ and linear charge density (charge per unit length) λ , the field at a dis tance d from the end of the rod along the axis is E = k e Z d + ‘ d λ x 2 dx = k e λ x fl fl fl fl d + ‘ d = k e λ‘ d ( ‘ + d ) , where dq = λdx . The linear charge density (if the total charge is Q ) is λ = Q ‘ so that E = k e Q ‘ ‘ d ( ‘ + d ) = k e Q d ( ‘ + d ) ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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