673_Physics ProblemsTechnical Physics

# 673_Physics ProblemsTechnical Physics - Chapter 23 z z Due...

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Chapter 23 13 P23.33 Due to symmetry E dE y y = = z 0 , and E dE k dq r x e = = z z sin sin θ θ 2 where dq ds rd = = λ λ θ , so that, E k r d k r k r x e e e = = = z λ θ θ λ θ λ π π sin cos 0 0 2 a f where λ = q L and r L = π . FIG. P23.33 Thus, E k q L x e = = × × 2 2 8 99 10 7 50 10 0 140 2 9 6 2 π π . . . N m C C m 2 2 e je j a f . Solving, E x = × 2 16 10 7 . N C . Since the rod has a negative charge, E i i = − × = 2 16 10 21 6 7 . ± . ± e j N C MN C . P23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx , has charge Qdx h and produces, at the chosen point, a field d k x x R Qdx h e E i = + 2 2 3 2 e j ± . The total field is E E i i E i i = = + = + = + = + + + L N M M M O Q P P P z z z + = + = + d k Qxdx h x R k Q h x R xdx k Q h x R k Q h d R d h R e d d h e x d d h e x d d h e all charge 2 2 3 2 2 2 3 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 1 2 1 1 e j e j e j b g e j a f e j ± ± ± ± (b) Think of the cylinder as a stack of disks, each with thickness dx , charge Qdx h , and charge- per-area σ π = Qdx R h 2 . One disk produces a field d k Qdx R h
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