Solving for Electric Field at Wire Surface: Homework Exercise

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Homework Help Overview

The problem involves calculating the electric field at the surface of a uniformly charged wire located along the axis of a cylinder. The wire has a radius of 0.02 m, while the cylinder has an inner radius of 0.06 m, and there is a potential difference of 60 volts between them.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using Gauss's law and electric flux equations to relate charge, electric field, and potential difference. There are attempts to derive expressions for charge and electric field, with some questioning the next steps after initial calculations.

Discussion Status

Some participants have provided guidance on solving for charge and suggested substituting it back into the electric field equation. There is an ongoing exploration of the reasoning behind the calculations, but no consensus has been reached regarding the final approach.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the reasoning behind the steps taken rather than simply checking calculations.

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.
 
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Solve for Q.
Then think Gauss.
 
Last edited:
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))
 
Rider4 said:
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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