How Is Electric Potential Calculated for a Charged Rod?

AI Thread Summary
The discussion focuses on calculating the electric potential at a distance z from the center of a uniformly charged rod of length L. The potential is derived by integrating the contributions from differential charge elements along the rod, leading to the expression V_tot = kλ ln((2√(L²/4 + z²) + L) / (2√(L²/4 + z²) - L)). The integration process involves substituting y = ztan(x) to simplify the calculations. Participants confirm the correctness of the final expression and discuss minor adjustments to the formula. The overall approach and solution are validated by the contributors.
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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



V=k\frac{Q}{r}

The Attempt at a Solution



Determine the electric potential due to one point charge i.

V_i=k\frac{\Delta Q}{r}

r=\sqrt{y^2+z^2}

Where y is some distance along the rod to i and z is the distance along the axis to the point away.

\Delta Q=\lambda \Delta y

Then,

V_i=k\lambda \frac{\Delta y}{\sqrt{y^2+z^2}}

Then,

V_{tot}=k\lambda \sum_{i=1}^{n}\frac{\Delta y}{\sqrt{y^2+z^2}}

Now take the limit as n goes to infinity and delta y goes to zero.

V_{tot}=k\lambda \int_{\frac{-L}{2}}^{\frac{L}{2}}\frac{d y}{\sqrt{y^2+z^2}}

V_{tot}=k\lambda \; ln({\frac{2\sqrt{{\frac{L^2}{4}}+z^2}+L}{2\sqrt{{\frac{L^2}{4}}+z^2}-L}})

Is this correct? I feel like the integration was too hard for this simple problem (I had to substitute y = ztan(x)).
 
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It looks correct to me. :approve: [strike](Of course you could simplify it slightly by at least getting rid of the '2's)[/strike]

[Edit: Never-mind about the 2's'. They're fine the way they are. Sorry about that.]
 
Last edited:
It looks correct to me.

Really? Well thank you!

Never-mind about the 2's'. They're fine the way they are. Sorry about that.

Okay :-p
 
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