• Support PF! Buy your school textbooks, materials and every day products Here!

Gauss' law/electric flux

  • #1
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.


Homework Equations


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

  • #2
173
60
From what I manage to understand, yes your attempt is correct. it should give you the charge on the cylinder. What went wrong then?
 
  • #3
264
26
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.
 
  • #4
thhanks
 

Related Threads on Gauss' law/electric flux

Replies
5
Views
13K
  • Last Post
Replies
5
Views
670
  • Last Post
Replies
13
Views
19K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
4
Views
13K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
9
Views
6K
  • Last Post
Replies
1
Views
5K
Top