(Ill) A thin glass rod is a semicircle of radius R, Fig.

Chapter , Problem 50

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(Ill) A thin glass rod is a semicircle of radius R, Fig. 21-66.A charge is nonuniformly distributed along the rod with alinear charge density given by A = A0 sin 6, where A0 is apositive constant. Point P is at the center of the semicircle.(a) Find the electric field E(magnitude and direction) atpoint P. [Hint: Remembersin(6) = sin 6, so the twohalves of the rod are oppositelycharged.] (b) Determine theacceleration (magnitude anddirection) of an electron placed atpoint P, assuming R = 1.0 cmand A0 = 1.0/xC/m.

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