Electric field

  • Thread starter Rider4
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Homework Statement


A uniformly charged wire or radius R1=0.02m runs down the axis of a cylinder of inner radius R2=0.06m. The potential difference between the wire and cylinder is 60 volts. Find the electric field at the surface of the wire.

Homework Equations


Electric flux=Q/[itex]\epsilon0[/itex]
Electric flux=integral of E dA
V=integral of E dr

The Attempt at a Solution


Q/[itex]\epsilon0[/itex]=E(2pi)(r)(L)
Applying the above equations, I got E=(Q/[itex]\epsilon0[/itex])(1/((2pi)(r)(L))
I then integrated as a function of r, and got V=(Q/[itex]\epsilon0[/itex])(1/((2pi)(L))(ln(r2/r1))
I'm not too sure where to go from here, so any help would be greatly appreciated.
 

Answers and Replies

  • #2
rude man
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Solve for Q.
Then think Gauss.
 
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I solved for Q and got Q=(3.29[itex]\ast[/itex]10^8)/L
then should I plug Q back in for the Q in the equation E=(Q/ϵ0)(1/((2pi)(r)(L))
 
  • #4
rude man
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I solved for Q and got Q=(3.29[itex]\ast[/itex]10^8)/L
then should I plug Q back in for the Q in the equation E=(Q/ϵ0)(1/((2pi)(r)(L))

Yes, with of course the appropriate vaue for r.

BTW I'm not checking your math, just your reasoning. Hope you understand the steps you took well.
 

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