Equilateral triangle charged rods

AI Thread Summary
The discussion focuses on calculating the voltage at the center of an equilateral triangle formed by three uniformly charged rods, with a charge -q placed at the center. The voltage is derived using integration and results in a formula that accounts for the additive potential from each rod. The potential energy of the charge -q is straightforwardly calculated by multiplying the voltage by -q. For determining the initial speed required for -q to exit the system permanently, conservation of energy principles are applied, emphasizing that the kinetic energy must exceed the potential energy at the center. The methodology suggests expressing variables in terms of known quantities to simplify calculations.
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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

V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r}

U = Vq

K_0 + U_0 = K + U

The Attempt at a Solution



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

I take 1 half of bottom rod and intend to integrate for Vdq = Q \frac{dx}{a}
r = \sqrt{y^2 + (\frac{a}{2} - x)^2}

using the V formula above this integrates(from 0 to a/2) toV = \frac{Q}{4 \pi \epsilon_0}[\ln{\frac{a+\sqrt{\frac{a^2}{4} + y^2}}{y}}]

since this is 1/6th of the potential and they are additive, multiply 6 for total v at centeris this correct solution? U is simple enough to do, simply multiply by -q. What is the methodology for solving part 3?
 
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You might simplify things significantly if you express ##y## via ##a##.
 
y = \sqrt{\frac{3}{4}} a

ln term collapses to
ln{\frac{4}{3}}

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.
 
"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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