Equilateral triangle charged rods

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SUMMARY

The discussion focuses on calculating the voltage at the center of an equilateral triangle formed by three uniformly charged rods, each with charge Q and length a. The voltage is derived using the formula V = (1/4πε₀) ∫(dq/r), leading to V = (Q/4πε₀) * [ln(a + √(a²/4 + y²))/y] after integration. The potential energy U of a charge -q placed at the center is calculated as U = Vq. For determining the initial speed of -q to escape the system, conservation of energy principles are applied, emphasizing that the kinetic energy must exceed the potential energy for the charge to move infinitely far away.

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  • Familiarity with calculus, particularly integration techniques.
  • Knowledge of conservation of energy principles in physics.
  • Basic concepts of electric fields and forces from charged objects.
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  • Study the derivation of electric potential from continuous charge distributions.
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Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in solving problems related to electric potentials and charged particle dynamics.

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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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