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.
Electric flux=integral of E dA
V=integral of E dr
The Attempt at a Solution
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.