Potential difference of a conducting pipe

[Suziii]

Homework Statement

A conducting cylindrical pipe has an inner radius Ra and an outer radius Rb. a charge q is released inside it. Assume the length of a cylinder L>>Rb. Find the potential difference between the axis of the cylinder and outer surface.

Homework Equations

Vb-Va=-∫E*dl (vector multiplication and the integral is bounded between a and b)

The Attempt at a Solution

I found the electric field in the outer surface to be E=q/ε2πRL
Vb-Va= q/ε2πL ∫1/rdr= q/ε2πL (lnRb-lnRa) (note: the integral is definite between Ra and Rb)
Since the question asks for the potential difference between the axis and the outer surface I need to run R between 0 and R, so i get a ln(0) which does not make sense...

Pleaseeeeeeeeee helppppp

 

[anathema]

Thank you for your question. Your solution is on the right track, but there are a few things that need to be clarified.

First, you are correct in finding the electric field on the outer surface to be E=q/ε2πRL. However, this is only true for points on the outer surface, not points on the axis. For points on the axis, the electric field will be zero since the charge is located at the center of the cylinder and there is no electric field inside a conductor.

Next, your integral should not be bounded between Ra and Rb, but rather between 0 and L. This is because the potential difference between the axis and outer surface is the same as the potential difference between any two points on the axis and outer surface, regardless of the distance between them. So, we can choose any two points on the axis and outer surface to find the potential difference, and it will be the same as the potential difference between the axis and outer surface.

Finally, when integrating, you should use the natural logarithm function (ln), not log base 10. Also, when integrating 1/r, you should use the limit as r approaches 0, not the limit as r approaches R.

So, the correct solution would be:

Vb-Va= q/ε2πL ∫1/rdr= q/ε2πL (lnL-ln0) = q/ε2πL (lnL-(-∞)) = q/ε2πL (lnL+∞) = ∞

This means that the potential difference between the axis and outer surface is infinite. This makes sense because the electric field is strongest at the surface of a conductor, and since the potential difference is directly proportional to the electric field, the potential difference will also be infinite.

I hope this helps clarify your solution. Let me know if you have any further questions.