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Unformatted text preview: Version 020/AABBA – midterm 01 – turner – (56725) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 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 λ = λ x /x , where λ is a constant, determine the electric field at the origin. (Here ˆ ı denotes the unit vector in the positive x direction.) 1. k λ x (ˆ ı ) 2. k λ 2 2 x (ˆ ı ) 3. k λ 2 x ( ˆ ı ) correct 4. k λ x ( ˆ ı ) 5. k λ 2 x 2 (ˆ ı ) 6. k λ 2 x (ˆ ı ) 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 Coulomb’s law, the electric field cre ated by each small element with charge dq is dE = k dq x 2 where dq = λdx = λ x x dx. Now we integrate over the entire distribution (i.e. from x = x to x = + ∞ ) and insert our dq :  vector E  = integraldisplay k dq x 2 = integraldisplay ∞ x k λ x dx x 3 = k λ x 2 1 x 2 vextendsingle vextendsingle vextendsingle ∞ x = k λ 2 x . Since the distribution is to the right of the point of interest, the electric field is directed along the x axis if λ is positive. That is, a positive charge at the origin would experience a force in the direction of ˆ ı from this charge distribution. In fact, the direction of an elec tric field at a point P in space is defined as the direction in which the electric force acting on a positive particle at that point P would point. So vector E = k λ 2 x ( ˆ ı ) . 002 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. aligned with the positive xaxis. 2. aligned with the negative xaxis. 3. along 135 ◦ direction in the first quad rant. 4. along 30 ◦ direction in the first quadrant. 5. along 45 ◦ direction in the first quadrant. 6. aligned with the positive yaxis. 7. zero and the direction is undefined. 8. along 60 ◦ direction in the first quadrant. Version 020/AABBA – midterm 01 – turner – (56725) 2 9. along 225 ◦ direction in the first quad rant. 10. aligned with the negative yaxis. correct 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)....
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 Fall '10
 TURNER,J
 Charge, Electrostatics, Electric charge

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