714_Physics ProblemsTechnical Physics

714_Physics ProblemsTechnical Physics - 54 Electric...

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54 Electric Potential P25.7 Um v v fi =− × × × L N M O Q P 1 2 1 2 9 11 10 1 40 10 3 70 10 6 23 10 22 3 1 5 2 6 2 18 ej e j e j e j .. . . kg ms J ∆∆ UqV = : = −× −− 6 23 10 1 60 10 18 19 V V 38 9 . V. The origin is at highest potential. P25.8 (a) VE d == × = 5 90 10 0 010 0 59 0 3 . Vm m V bg (b) 1 2 2 mv q V f =∆ : 1 2 9 11 10 1 60 10 59 0 31 2 19 . ×= × af v f v f 455 10 6 . m s P25.9 VV d d d E d yE d x BA A B A C C B −= = −⋅ ° ° = + zz z Es Es Es cos cos . . . . . . 180 90 0 325 0 800 260 0300 0500 0200 0400 a f a f V FIG. P25.9 *P25.10 Assume the opposite. Then at some point A on some equipotential surface the electric field has a nonzero component E p in the plane of the surface. Let a test charge start from point A and move some distance on the surface in the direction of the field component. Then Vd A B z is 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 .

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