54 Electric PotentialP25.7∆Umvvfi=−−××−×LNMOQP=×−−12129 11 101 40 103 70 106 23 102231526218ejejejej....kgmsJ∆∆UqV=:+×=−×−−6 23 101 60 101819∆V∆V38 9. V. The origin is at highest potential.P25.8(a)∆VEd==×=5 90 100 010 059 03.VmmVbg(b)122mvq Vf=∆:129 11 101 60 1059 031219.×=×afvfvf455 106. msP25.9VVdddEdyEdxBAABACCB−=−⋅=−⋅−⋅−°−°=+zzzEsEs Escoscos......18090 0325 0 8002600300050002000400afafVFIG. P25.9*P25.10Assume the opposite. Then at some point Aon some equipotential surface the electric field has anonzero component Epin the plane of the surface. Let a test charge start from point Aand movesome distance on the surface in the direction of the field component. Then ∆VdAB⋅zis nonzero.The electric potential charges across the surface and it is not an equipotential surface. The
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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 .