726_Physics ProblemsTechnical Physics

# 726_Physics - 66 Electric Potential*P25.50(a Both spheres must be at the same potential according to where also q1 q 2 = 1.20 10 6 C Then q1 = k e

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66 Electric Potential *P25.50 (a) Both spheres must be at the same potential according to kq r r ee 1 1 2 2 = where also qq 12 6 120 10 += × . C . Then q qr r 1 21 2 = r q q q V r e 2 2 6 2 6 6 1 666 1 1 96 2 5 1 0300 10 1 2 01 0 0 3 0 0 0 9 0 0 8 9 91 0 0 9 0 0 61 0 135 10 = × + × = × == ×⋅ × × −−− . . . .. . . C C 6 cm 2 cm C on the smaller sphere C C Nm C C m V 22 ej e j (b) Outside the larger sphere, Er r r 1 1 1 2 1 1 5 6 225 10 = × r V r e ±± . ± . V 0.06 m V m away . Outside the smaller sphere, 2 5 6 674 10 = × . ± . V 0.02 m V m away . The smaller sphere carries less charge but creates a much stronger electric field than the larger sphere. Section 25.7 The Milliken Oil Drop Experiment Section 25.8 Application of Electrostatistics P25.51 (a) E kQ r rr V r max max .
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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