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Melawrghk

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## Homework Statement

A very long insulating cylindrical shell of radius 6.90 cm carries charge of linear density 8.40 microC/m spread uniformly over its outer surface. Find the potential of a point located 4.4cm above the surface of the shell

## Homework Equations

## The Attempt at a Solution

I used Gaussian surface to find the electric field created by the shell outside (Q = linear charge density):

QL/[tex]\epsilon_{0}[/tex] = E*2*pi*d*L,

where L is length, d is distance to point from the axis of the cylinder.

The Ls cancel out, rearranging, I got:

E=[tex]\frac{Q}{2pi*d*\epsilon_{0}}[/tex]

I then integrated that to get the potential:

V=[tex]\frac{Q}{2pi*\epsilon_{0}} \int {(1/d)dd} [/tex]

The integral's upper limit is d-r, and lower limit is r.

I got the equation below as a result:

V=[tex]\frac{Q}{2pi*d*\epsilon_{0}} ln(\frac{d-r}{r})[/tex]

Substituting in the numbers I got V=-6.8*10^4, but that's wrong. Is it my method? Should I slice the cylinder into rings and do V from each one, then integrate the whole thing?

Thanks in advance

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