midterm 01-solutions

# midterm 01-solutions - Version 020/AABBA – midterm 01 –...

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Unformatted text preview: Version 020/AABBA – midterm 01 – turner – (56725) 1 This print-out should have 19 questions. Multiple-choice 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 x-axis 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 x-axis. 2. aligned with the negative x-axis. 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 y-axis. 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 y-axis. 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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midterm 01-solutions - Version 020/AABBA – midterm 01 –...

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