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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.