Unformatted text preview: ε 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 πε ) Qr/R 3 i Another way is to use Gauss’s law and equate the E field times the surface area to the electric flux from part a. ⇒ (r 3 /R 3 )Q/ ε 0 =E4 π r 2 which gives the same answer when solved for E...
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 Spring '08
 GUERRA
 Physics, Charge, Magnetic Field

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