# Equilateral triangle charged rods

## 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 V

$$dq = Q \frac{dx}{a}$$
$$r = \sqrt{y^2 + (\frac{a}{2} - x)^2}$$

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

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

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

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

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.