Electric Potential From Rod

  • #1
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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)).
 

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
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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:
  • #3
129
0
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 :tongue:
 

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