extra_credit 01-solutions

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

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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 – fnd 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 fgure. x y + + + + + + - - - - - - In this fgure, at the origin, the net feld v E net is 1. along 45 direction in the frst quadrant. 2. aligned with the positive y -axis. 3. along 60 direction in the frst quadrant. 4. along 225 direction in the frst quad- rant. 5. aligned with the positive x -axis. 6. aligned with the negative y -axis. correct 7. along 135 direction in the frst quad- rant. 8. zero and the direction is undefned. 9. along 30 direction in the frst quadrant. 10. aligned with the negative x -axis. Explanation: At the origin the felds From the bottom slab points towards the negative y axis (towards the negatively charged slab). The feld From the right and leFt slabs have equal magnitudes at the origin and are sym- metric about the y axis. The feld From the right and leFt slabs point away From each pos- itively charged slab. ThereFore the x compo- nent oF the feld cancels at the origin. The net feld From the right and leFt slabs points towards the negative y axis (away From the positively charged slab). Hence, the sum oF the electric felds From all three slabs is aligned with the negative y -axis. 002 10.0 points A line oF charge starts at x = x 0 , where x 0 is positive, and extends along the x -axis to positive infnity. IF the linear charge den- sity is given by λ = λ 0 x 0 /x , where λ 0 is a constant, determine the electric feld at the origin. (Here ˆ ı denotes the unit vector in the positive x direction.) 1. k λ 0 2 x 0 ( - ˆ ı ) correct 2. k λ 2 0 2 x 0 ı ) 3. k λ 0 2 x 0 ı ) 4. k λ 0 x 0 ı ) 5. k λ 0 x 0 ( - ˆ ı ) 6. k λ 0 2 x 2 0 ı ) Explanation: ±irst 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 Coulomb’s law, the electric feld cre- ated by each small element with charge dq is dE = k dq x 2 where dq = λdx = λ 0 x 0 x dx. Now we integrate over the entire distribution (i.e. From x = x 0 to x = + ) and insert our
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marout (ssm979) – extra credit 01 – turner – (56725) 2 dq : | v E | = i k dq x 2 = i x 0 k λ 0 x 0 dx x 3 = - k λ 0 x 0 2 1 x 2 v v v x 0 = k λ 0 2 x 0 .
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extra_credit 01-solutions - marout(ssm979 extra credit 01...

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