Equilateral triangle charged rods

  • Thread starter oreosama
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  • #1
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Homework Statement



Three rods each of uniform charge Q and length a are formed into an equilateral triangle. Given[Q,a,q] determine:
1. The voltage at the center of the triangle.
2. If a charge -q is placed at the center of the triangle determine potential energy.
3. Determine the initial speed of the -q in order that it exits the system and never returns.

Homework Equations




[tex]V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r}[/tex]

[tex]U = Vq[/tex]

[tex]K_0 + U_0 = K + U[/tex]

The Attempt at a Solution



http://i.imgur.com/7IZ3PbP.png

I take 1 half of bottom rod and intend to integrate for V


[tex]dq = Q \frac{dx}{a}[/tex]
[tex]r = \sqrt{y^2 + (\frac{a}{2} - x)^2}[/tex]

using the V formula above this integrates(from 0 to a/2) to


[tex]V = \frac{Q}{4 \pi \epsilon_0}[\ln{\frac{a+\sqrt{\frac{a^2}{4} + y^2}}{y}}][/tex]

since this is 1/6th of the potential and they are additive, multiply 6 for total v at center


is this correct solution? U is simple enough to do, simply multiply by -q. What is the methodology for solving part 3?
 

Answers and Replies

  • #2
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You might simplify things significantly if you express ##y## via ##a##.
 
  • #3
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[tex]y = \sqrt{\frac{3}{4}} a[/tex]

ln term collapses to
[tex]ln{\frac{4}{3}}[/tex]

Is this correct? What about part 3? I only know of 1/2mv^2 which I assume was at least greater than potential it would get out.. but I have no m given.
 
  • #4
6,054
391
"Never returns" means that it goes infinitely far away, and the velocity "at the infinity" is at least zero.

With this in mind, apply conservation of energy.
 

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