extra_credit 01-solutions

extra_credit 01-solutions - marout (ssm979) extra credit 01...

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Unformatted text preview: marout (ssm979) extra credit 01 turner (56725) 1 This print-out should have 9 questions. Multiple-choice 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 y-axis. 3. along 60 direction in the first quadrant. 4. along 225 direction in the first quad- rant. 5. aligned with the positive x-axis. 6. aligned with the negative y-axis. 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 x-axis. 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 y-axis. 002 10.0 points A line of charge starts at x = x , where x is positive, and extends along the x-axis 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.

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extra_credit 01-solutions - marout (ssm979) extra credit 01...

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