Solving for Electric Field at Wire Surface: Homework Exercise

[Rider4]

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/\epsilon0
Electric flux=integral of E dA
V=integral of E dr

The Attempt at a Solution

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

 

[rude man]

Solve for Q.
Then think Gauss.

 

[Rider4]

I solved for Q and got Q=(3.29\ast10^8)/L
then should I plug Q back in for the Q in the equation E=(Q/ϵ0)(1/((2pi)(r)(L))

 

[rude man]

I solved for Q and got Q=(3.29\ast10^8)/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.