# Gauss' law/electric flux

a long straight wire has fixed -ve charge density of 39nC/m. the wire is enclosed by thin wall non conducting coaxial cylinder of radius 1.7m. the shall has positive charge density and its Field is such as that it will cancel the field due to wire. what will be the surface charge density of cylinder.

E=lamda/2pi€r
gausses law

## The Attempt at a Solution

first i have found the the electric field of wire enclosed in cylinderusing equation E=lamda/2pi€r. then as given total electric field must be zero so E for cylinder must be same in magnitude as of wire but opposite sign. but i have no idea i am attempt towards solution is right or wrong any help??

## Answers and Replies

From what I manage to understand, yes your attempt is correct. it should give you the charge on the cylinder. What went wrong then?

Since electric field lines start and end on charges, the shell must have the same charge per unit length
as the wire. So it appears that you need to convert a linear charge density to a surface charge density.

thhanks