HourExamIReviewProblems-02-10-08

HourExamIReviewProblems-02-10-08 - Physics 21 Spring 2008...

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Physics 21 Spring 2008 February 10, 2008 Hour Exam I Review Review I-1. ••11. In Fig. 21-24 , three charged particles lie on an x axis. Particles 1 and 2 are fixed in place. Particle 3 is free to move, but the net electrostatic force on it from particles 1 and 2 happens to be zero. If , what is the ratio ? Solution: 21-11. With rightwards positive, the net force on q 3 is () 13 23 31 32 3 2 2 23 12 23 . qq FF F k k L LL =+= + + We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For example, the first term (the force exerted on q 3 by q 1 ) is negative if they are unlike charges, indicating that q 3 is being pulled toward q 1 , and it is positive if they are like charges (so q 3 would be repelled from q 1 ). Setting the net force equal to zero L 23 = L 12 and canceling k , q 3 and L 12 leads to 11 2 2 0 4.00. 4.00 q q += = Review I-2 ••31. Figure 22-50 a shows a nonconducting rod with a uniformly distributed charge +Q . The rod forms a half-circle with radius R and produces an electric field of magnitude at its center of curvature P . If the arc is collapsed to a point at distance R from P (Fig. 22-50 b ), by what factor is the magnitude of the electric field at P multiplied? Solution: 22-31. First, we need a formula for the field due to the arc. We use the notation λ for the charge density, λ = Q /L . Sample Problem 22-3 illustrates the simplest approach to circular arc field problems. Following the steps leading to Eq. 22-21, we see that the general result (for arcs that subtend angle θ ) is [] arc 00 2s i n (/ 2 ) sin( / 2) sin( / 2) 44 E rr λ λθ θθ πε =− = .
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Now, the arc length is L = r θ if is expressed in radians. Thus, using R instead of r , we obtain arc 2 00 0 2( / )sin( / 2) 2( / )sin( / 2) 2 sin( / 2) 44 4 QL QR Q E rr R θθ π επ ε == = . The problem asks for the ratio E particle / E arc where E particle is given by Eq. 22-3: 2 particle 0 2 arc 0 /4 2 sin( / 2) / 4 2sin( / 2) E EQ R πε θπ . With = π , we have particle arc 1.57. 2 E E =≈ Review I-3 ••51. In Fig. 23-52 , a solid sphere of radius is concentric with a spherical conducting shell of inner radius and outer radius . The sphere has a net uniform charge ; the shell has a net charge . What is the magnitude of the electric field at radial distances (a) , (b) , (c) , (d) , (e) , and (f) ? What is the net charge on the (g) inner and (h) outer surface of the shell?
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HourExamIReviewProblems-02-10-08 - Physics 21 Spring 2008...

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