- #1

Physgeek64

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## Homework Statement

Consider a cylindrical hole of radius a and infinite length cut into a dielectric medium with relative electric permittivity ε (the interior can be treated as a vacuum). Inside the hole there are two line charges of infinite length with line charge densities λ and −λ, respectively. These line charges are arranged parallel to the z−axis, but displaced from it by an amount ±d/2 along the direction φ = 0 (where d ≪ a).

Show that the electrostatic potential due to the line charges in the range d ≪ r ≪ a (i.e. ignoring the effects of the dielectric medium outside) is given to lowest order by

V(r,φ)≃ ##\frac{λ dcosφ}{2πε_0 r}##

Determine the electrostatic potential everywhere for r ≫ d.

## Homework Equations

## The Attempt at a Solution

for the first part

##\int{E\dot dS}=\frac{Q}{\epsilon_0}##

##E=\frac{\lambda}{2\pi \epsilon_0 r'}##

##V=\int{E\dot dr'}##

##V=\frac{\lambda}{2\pi \epsilon_0}\ln{r'}##

##V_{total}=\frac{\lambda}{2\pi\epsilon_0}[\ln{|\vec{r}-\frac{d}{2}\vec{x}}-\ln{|\vec{r}+\frac{d}{2}\vec{x}}]##

##|\vec{r}+\frac{d}{2}\vec{x}| ~r+\frac{d}{2}cos\phi##

##V_{total}=\frac{\lambda}{2\pi \epsilon_0}[\ln{r-\frac{d}{2} cos\phi}-\ln{r+\frac{d}{2} cos\phi}] ##

##V_{total}=-\frac{\lambda}{2\pi \epsilon_0}\frac{cos\phi}{r} ##

I don't know why I get a minus here- i assume I've done something wrong.

Im really struggling with the next part of the question. I know the boundary conditions are that the parallel component of E must be continuous and also the perpendicular component of D must be continuous since we have no free charges at the boundary.

Many thanks

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