- #1

rshalloo

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## Homework Statement

Find the potential difference between two oppositely charged, infinite cylinders of radii R whose axes lie at $$y =+\frac{d}{2}$$ and $$y = -\frac{d}{2}$$ They have surface charge densities of magnitude $$\sigma$$

## Homework Equations

The family favourite - Gauss' Law:

$$\int \textbf{E.da} = \frac{q_{encl}}{\epsilon_{0}}$$

and of course how could we forget

$$V = -\int \textbf{E.dl}$$

## The Attempt at a Solution

Using gauss' law and a cylindrical gaussian surface I obtained an electric field for a single infinite cylinder located at the origin to be:

$$\textbf{E} =\frac{\sigma R}{\epsilon_{0} r}\hat{\textbf{r}}$$

I now use this to find the E-field on the line connecting the centers of the cylinders only. Obviously I am also confining my region further to exclude the radii of the cylinders. So my range of applicability is $$[-\frac{d}{2}+R,+\frac{d}{2}-R]$$

So placing the positive cylinder in the lower half of the plane I get:

$$\textbf{E_{+}} = \frac{\sigma R}{\epsilon_{0} (y+d/2)}\hat{\textbf{y}}$$

and placing the negative sphere in the upper half of the plane I get:

$$\textbf{E_{-}} = \frac{\sigma R}{\epsilon_{0} (d/2-y)}\hat{\textbf{y}}$$

Using superposition the total field is simply the sum of these.

I then use the formula for the potential listed above and use the integration limits:

$$[-\frac{d}{2}+R,s]$$

that is finding the potential difference assuming zero potential at the lower sphere. s is my variable height

I get the following answer but it doesn't match the solution.

$$V = -\frac{\sigma R}{\epsilon}ln\left (\frac{(s+d/2)(d-R)}{(d/2-s)(R)}\right )$$

Is my reasoning correct or have I made a fundamental error somewhere?

Thanks in advance for your help!