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 Ed E yy == z 0 , and E k dq r xe zz sin sin θ 2 where dq ds rd λλ , so that, E k r d k r k r x ee e = z λ θθ π sin cos 0 0 2 af where = q L and r L = . FIG. P23.33 Thus, E kq L x e ×⋅ × 2 2 8 99 10 7 50 10 0140 2 96 2 .. . Nm C C m 22 ej e j . Solving, E x 216 10 7 . N C . Since the rod has a negative charge, Ei i =− × = − 216 7 . ± . ± NC MNC . 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 kx xR Qdx h e = + 32 ± . The total field is EE i i E ii + =+ = + = + ++ L N M M M O Q P P P z + = + = + d kQxdx hx R kQ h x d x h h dR dh R e d dh e xd e e all charge 12 2 2 2 2 2 11 bg ± ± ±± (b) Think of the cylinder as a stack of disks, each with thickness dx , charge Qdx h , and charge- per-area σ = Qdx Rh 2 . One disk produces a field d kQdx x e + F H G G I K J J 2 1 2 ± . So, i + F H G G I K J J = + d x e all charge
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