# How to measure circular displacement?

1. Sep 24, 2009

### TwoTruths

1. The problem statement, all variables and given/known data
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.

2. Relevant 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
3. 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)?

2. Sep 24, 2009

### kuruman

Treat this as if it were a pendulum with the electric force qE replacing the gravitational force mg.

3. Sep 24, 2009

### TwoTruths

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?

4. Sep 24, 2009

### kuruman

You say correctly and there is nothing fundamentally wrong with your approach. That's the way I would do it if I had to.