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Unformatted text preview: marout (ssm979) extra credit 01 turner (56725) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Given symmetrically placed rectangular insu lators with uniformly charged distributions of equal magnitude as shown in the figure. x y + + + + + + + + + + + +      In this figure, at the origin, the net field vector E net is 1. along 45 direction in the first quadrant. 2. aligned with the positive yaxis. 3. along 60 direction in the first quadrant. 4. along 225 direction in the first quad rant. 5. aligned with the positive xaxis. 6. aligned with the negative yaxis. correct 7. along 135 direction in the first quad rant. 8. zero and the direction is undefined. 9. along 30 direction in the first quadrant. 10. aligned with the negative xaxis. Explanation: At the origin the fields from the bottom slab points towards the negative y axis (towards the negatively charged slab). The field from the right and left slabs have equal magnitudes at the origin and are sym metric about the y axis. The field from the right and left slabs point away from each pos itively charged slab. Therefore the x compo nent of the field cancels at the origin. The net field from the right and left slabs points towards the negative y axis (away from the positively charged slab). Hence, the sum of the electric fields from all three slabs is aligned with the negative yaxis. 002 10.0 points A line of charge starts at x = x , where x is positive, and extends along the xaxis to positive infinity. If the linear charge den sity is given by = x /x , where is a constant, determine the electric field at the origin. (Here denotes the unit vector in the positive x direction.) 1. k 2 x ( ) correct 2. k 2 2 x ( ) 3. k 2 x ( ) 4. k x ( ) 5. k x ( ) 6. k 2 x 2 ( ) Explanation: First we realize that we are dealing with a continuous distribution of charge (as opposed to point charges). We must divide the dis tribution into small elements and integrate. Using Coulombs law, the electric field cre ated by each small element with charge dq is dE = k dq x 2 where dq = dx = x x dx. Now we integrate over the entire distribution (i.e. from x = x to x = + ) and insert our marout (ssm979) extra credit 01 turner (56725) 2 dq :  vector E  = integraldisplay k dq x 2 = integraldisplay x k x dx x 3 = k x 2 1 x 2 vextendsingle vextendsingle vextendsingle x = k 2 x ....
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This note was uploaded on 10/24/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

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