- #1

Suziii

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## 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