Surface Charge Density of Coaxial Cylinder with Canceling Electric Fields

AI Thread Summary
A long straight wire with a negative charge density of 39 nC/m is enclosed by a non-conducting coaxial cylinder with a radius of 1.7 m, which has a positive charge density that cancels the wire's electric field. The electric field of the wire is calculated using the equation E=λ/2π€r, and the total electric field must equal zero, indicating that the cylinder's field must match the wire's in magnitude but be opposite in sign. The discussion confirms that the approach to finding the charge on the cylinder is correct, but emphasizes the need to convert linear charge density to surface charge density. The conclusion is that the surface charge density of the cylinder should equal the linear charge density of the wire to achieve the desired cancellation of electric fields. Understanding this relationship is crucial for solving the problem accurately.
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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??
 
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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
 
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