# hw2sol - Flores Michael – Homework 2 – Due Feb 3 2004...

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Unformatted text preview: Flores, Michael – Homework 2 – Due: Feb 3 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 24 questions. Multiple-choice 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, Y and X , each stand on a glass rod and are touching. X Y 2) A third ball carrying a negative charge, is brought near the first two. X Y- 3) While the positions of these balls are fixed, ball Y is connected to ground. X Y- 4) Then the ground wire is disconnected. X Y- 5) While Y and X remain in touch, the ball carring the negative charge is removed. X Y 6) Then ball Y and X are separated. X Y After these procedures, the signs of the charge q Y on Y and q X on X are 1. q Y is negative and q X is positive. 2. q Y is negative and q X is negative. 3. q Y is neutral and q X is neutral. 4. q Y is positive and q X is neutral. 5. q Y is positive and q X is positive. cor- rect 6. q Y is neutral and q X is negative. 7. q Y is neutral and q X is positive. 8. q Y is positive and q X is negative. 9. q Y is negative and q X is neutral. Explanation: When the ball with negative charge is brought nearby, the free charges inside Y and X rearrange themselves. The positive charges are attracted and go to the right ( i.e. move to X ), whereas the negative charges are repelled and collect in the left hand side of the system Y X , i.e., in Y . When we ground Y , the negative charges which have collected in Y are allowed to es- cape (they strive to the left), whereas the positive charges in X are still held enthralled by the negative charge on the third ball. We break the ground. Now we remove the third ball with negative charge. The charge on X is redistributed in the system Y X , i.e. they share the positive charge (equally if identical). Finally we separate Y and X . The signs of the charge on Y and that on X are both positive. Flores, Michael – Homework 2 – Due: Feb 3 2004, 4:00 am – Inst: Sonia Paban 2 002 (part 1 of 1) 10 points A rod 10 . 1 cm long is uniformly charged and has a total charge of- 17 μ C. Determine the magnitude of the electric field along the axis of the rod at a point 32 . 6243 cm from the center of the rod. Correct answer: 1 . 47076 × 10 6 N / C. Explanation: Given : ‘ = 10 . 1 cm , Q =- 17 μ C , and r = 32 . 6243 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 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

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hw2sol - Flores Michael – Homework 2 – Due Feb 3 2004...

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