Calculating Voltage Difference in a Cylindrical Shell

[kimm]

.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?

I started with this equation
delta V= ( lemda/ 2pi epslion) (ln(rb/ra))

 

[uart]

distance units? 4.7 ?, 1.0 ?

I started with this equation
delta V= ( lemda/ 2pi epslion) (ln(rb/ra))

Ok so presumably you've substituted your numbers into this formula and got an answer. What exactly is the problem?

 

[uart]

BTW. Assuming the above are both in units of cm, then your equation is valid for the first measurement but not for the second one. Do you know why?

 

[kimm]

i did like this
(8.90* 10^-6/ 2 pi 8.85*10^-12)ln (4.5/6)
i got -39084.73
but the answer was wrong
what i am not sure about is ln ra and ln rb
i substituted ra cm and rb 4.5 cm

 

[kimm]

ra is 6 cm

 

[uart]

i did like this
(8.90* 10^-6/ 2 pi 8.85*10^-12)ln (4.5/6)
i got -39084.73
but the answer was wrong
what i am not sure about is ln ra and ln rb
i substituted ra cm and rb 4.5 cm

Well for one I don't even get that answer when I use your numbers. Double check your calculator usage.

In any case the numbers you're using are not correct. I think you should be using Ra = 6cm and Rb = 10.7 cm.

Are the distances of 4.7 and 1.0 methioned in you question in cm ?

 

[uart]

ra is 6 cm

Oh thanks for clarifing that. It was the ONLY distance in your original question that you did include units for. So what a great idea to repeat that one instead of clarifying the units of the other distances that you didn't include units for. I'm going to give up here.

 

[kimm]

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