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Unformatted text preview: 0 b) What is the electric field vector at a point on the positive x axis a distance r from the origin (r<R)? One way of solving this is to remember that the electric field at the surface is the same as that of a point charge Q at the origin and that the electric field is proportional to r for r<R. E = (r/R) x (k e Q/R 2 ) i = k e Qr/R 3 i = 1/(4 ) Qr/R 3 i Another way is to use Gausss law and equate the E field times the surface area to the electric flux from part a. (r 3 /R 3 )Q/ 0 =E4 r 2 which gives the same answer when solved for E...
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This note was uploaded on 08/22/2009 for the course PHYS 7A taught by Professor Guerra during the Spring '08 term at UC Irvine.
 Spring '08
 GUERRA
 Physics, Charge

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