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724_Physics ProblemsTechnical Physics

# 724_Physics ProblemsTechnical Physics - 64 Electric...

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64 Electric Potential P25.43 (a) α λ = L N M O Q P =⋅ F H G I K J = x C mm C m 2 1 (b) Vk dq r k dx r k xdx dx kL d L d ee e L e == = + =− + F H G I K J L N M O Q P zz z αα 0 1 ln FIG. P25.43 P25.44 V kdq r k xdx bLx e e +− 2 2 2 bg Let z L x 2 . Then x L z 2 , and dx dz Lz d z bz dz k zdz zzb k zb V L x L xb k L V LLL b LLb e e e e e e L e L e = −− + + + + + + F H I K ++ F H G I K J F H G I K J + L N M M O Q P P F H G I K J + −+ + L N M M M O Q P P P z 2 22 2 2 2 2 2 0 2 2 0 2 2 2 2 af ln ln ln F H G I K J F H G I K J + L N M M O Q P P L N M M M O Q P P P k L Lb L b V bL L L e e 2 42 2 2 2 2 ln ej P25.45 Vd V dq r 1 4 0 π All bits of charge are at the same distance from O . So V Q R = F H G I K J −× F H G I K J 1 4 899 10 750 10 151 0 9 6 ππ . . . Nm C C 0.140 m MV . P25.46 dV rx e = + where dq dA rdr σσ 2 rdr kx bx a e a b e = + =+ + L N M O Q P z πσ σ FIG. P25.46
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