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Screwdriver
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Homework Statement
Find an expression for the electric potential a distance z away from the center of a thin uniformly charged rod of length L on a line that bisects the rod.
Homework Equations
[tex]V=k\frac{Q}{r}[/tex]
The Attempt at a Solution
Determine the electric potential due to one point charge i.
[tex]V_i=k\frac{\Delta Q}{r}[/tex]
[tex]r=\sqrt{y^2+z^2}[/tex]
Where y is some distance along the rod to i and z is the distance along the axis to the point away.
[tex]\Delta Q=\lambda \Delta y[/tex]
Then,
[tex]V_i=k\lambda \frac{\Delta y}{\sqrt{y^2+z^2}}[/tex]
Then,
[tex]V_{tot}=k\lambda \sum_{i=1}^{n}\frac{\Delta y}{\sqrt{y^2+z^2}}[/tex]
Now take the limit as n goes to infinity and delta y goes to zero.
[tex]V_{tot}=k\lambda \int_{\frac{-L}{2}}^{\frac{L}{2}}\frac{d y}{\sqrt{y^2+z^2}}[/tex]
[tex]V_{tot}=k\lambda \; ln({\frac{2\sqrt{{\frac{L^2}{4}}+z^2}+L}{2\sqrt{{\frac{L^2}{4}}+z^2}-L}})[/tex]
Is this correct? I feel like the integration was too hard for this simple problem (I had to substitute y = ztan(x)).