Voltmeter cylindrical shell help

[kimm]

Homework Statement

A very long insulating cylindrical shell of radius 6.00 cm carries charge of linear density 8.90*10^-6 C/m spread uniformly over its outer surface.
*What would a voltmeter read if it were connected between the surface of the cylinder and a point 4.70 above the surface. and What would a voltmeter read if it were connected between the surface and a point 1.00 from the central axis of the cylinder?

Homework Equations

delta V= ( lemda/ 2pi epslion) (ln(rb/ra))

The Attempt at a Solution

delta V= ( lemda/ 2pi epslion) (ln(rb/ra))

 

[thebigstar25]

if you don't show your attempt to solve the question ,no one would be able to help , please show your work even if it is wrong..

 

[kimm]

Homework Statement

A very long insulating cylindrical shell of radius 6.00 cm carries charge of linear density 8.90*10^-6 C/m spread uniformly over its outer surface.
*What would a voltmeter read if it were connected between the surface of the cylinder and a point 4.70 above the surface. and What would a voltmeter read if it were connected between the surface and a point 1.00 from the central axis of the cylinder?

Homework Equations

delta V= ( lemda/ 2pi epslion) (ln(rb/ra))

The Attempt at a Solution

delta V= ( lemda/ 2pi epslion) (ln(rb/ra))
what i did was
(8.90* 10^-6/ 2 pi 8.85*10^-12)ln (4.5/6)
i got -39084.73
but the answer was wrong

 

[thebigstar25]

I think that the equation you used is applied for charged conducting cylinder, and your question involves cylindrical shell.. so are they the same?

I suggest you try to think if you can somehow relate the question you have to a ring of charge in order to find the voltage..

 

[kimm]

Guys I reallly thought about it but i have not found an answer can some one solve it.

 

[Redbelly98]

Guys I reallly thought about it but i have not found an answer can some one solve it.

No, this forum does not work that way.

The Attempt at a Solution

delta V= ( lemda/ 2pi epslion) (ln(rb/ra))
what i did was
(8.90* 10^-6/ 2 pi 8.85*10^-12)ln (4.5/6)
i got -39084.73
but the answer was wrong

Where does the 4.5 come from? The point is "4.70 above the surface", but you should use distance to the cylinder's axis in that formula.