Equilateral triangle charged rods

Click For Summary

Homework Help Overview

The problem involves three uniformly charged rods arranged in an equilateral triangle, with a focus on calculating the voltage at the center, potential energy when a charge is placed there, and the initial speed required for that charge to escape the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate to find the voltage at the center of the triangle and questions the correctness of their solution. They also seek methodology for determining the initial speed of the charge to escape.
  • Some participants suggest simplifying variables and express one variable in terms of another, while also discussing the implications of conservation of energy for the escape condition.

Discussion Status

Participants are exploring different approaches to the problem, with some providing insights on simplifying the expressions involved. There is an ongoing discussion about the interpretation of the escape condition and the application of energy conservation principles.

Contextual Notes

There is a lack of explicit values for mass in the discussion, which affects the analysis of the initial speed required for the charge to escape. The original poster's assumptions and setup are also under scrutiny.

oreosama
Messages
51
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
894
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
5K