How to measure circular displacement?

Click For Summary

Homework Help Overview

The problem involves a charged particle connected to a pivot by a string in a uniform electric field. The particle's initial position is at an angle T degrees from the field axis, and the task is to determine its speed when it reaches that axis, starting from rest. The context includes concepts from energy conservation and electric potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles, equating initial and final energy states. They question how to measure displacement in the context of the electric field and consider using angular ratios to relate displacement to the circular path.

Discussion Status

Some participants have provided guidance on treating the system similarly to a pendulum, suggesting that the electric force can be considered analogous to gravitational force. The discussion is exploring the application of energy methods without reaching a consensus on the best approach to measure displacement.

Contextual Notes

Participants are navigating the assumptions about the system's setup, including the nature of the forces acting on the particle and the implications of using energy conservation in this context. There is a focus on the relationship between electric potential energy and kinetic energy.

TwoTruths
Messages
37
Reaction score
0

Homework Statement


A particle with charge q and mass m is tied by a (I assumed stiff) string of length L to a pivot point P, all of which lie on a horizontal plane. A uniform electric field E is placed over this system. If the initial position of the particle is at a point where the string is displaced T degrees from the axis parallel to the electric field, what will be the speed of the particle when it reaches this axis? The particle is initially at rest. Assume no outside work is done on the particle.

We are given E in V/m, q (the charge of the particle) in C, T in degrees, L in m, and mass in kg.

Homework Equations


Energy(final) = Energy(initial)
In a uniform E, potential difference = -E*displacement
K (kinetic energy) = 1/2 mv^2 for v <<<< c (v way way less than c, so neglect the change in mass)
U = qV

The Attempt at a Solution


It seems a pretty simple problem up to the point I'm stuck at. I set up the general energy equation:

K(initial) + U(initial) = K(final) + U(final)
K(initial) = 0 (it starts at rest)
K(final) = (mv^2)/2 and contains the final v I want
Moving U(final) to the left side yields:
U(initial)-U(final), or qV(initial) - qV(final), or q[V(initial)-V(final)], or q(deltaV)

From this, we can see that by solving for v, we can easily find it. But wait! What's the potential difference (deltaV)? Easy - the work done to move the particle from its initial position to the final position, or just -E * displacement for a uniform electric field.

Where I'm stuck: how should I go about measuring the displacement? I know I could solve a line integral, but I honestly think there's an easy answer to this, and I feel like I've done it before, but I just can't remember it. Help?

Edit: Looking at this a little more, I thought of trying the ratio of the angles and the arcs. So, can I use:

(T/360) = (displacement/circumference)?
 
Physics news on Phys.org
Treat this as if it were a pendulum with the electric force qE replacing the gravitational force mg.
 
Unfortunately, the only way I've solved pendulum problems is by using energy, much like I'm doing for this problem. I would say the kinetic energy at the bottom (from which I can get the velocity) is equal to the change in potential energy. In this problem, I say the kinetic energy is equal to the change in electric potential energy. Is there something fundamentally wrong with this approach?
 
You say correctly and there is nothing fundamentally wrong with your approach. That's the way I would do it if I had to.
 

Similar threads

Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
938
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
1K
Replies
5
Views
2K
Replies
12
Views
2K