Electric Potential voltage and conservation of energy

1. Jul 23, 2006

therealkellys

Hi,

I'm a little stuck on a problem involving voltage and conservation of energy. The answer that I came up with involved a negative value for kinetic energy, which doesn't make sense to me because mass and v^2 are never negative.

Problem: A proton is accelerated from rest by a uniform electric field. The proton experiences a potential decrease of 100 V. Find its final kinetic energy.

What I've done so far:
KE_i = 0 because accelerated from rest
V_b - V_a = -100 V
Change in voltage = Work/charge = [-q(int)E dot ds] / q = Change in potential energy

-100 = -Change in PE --> 100 J = change in potential energy

Therefore, under law of conservation of energy... change in kinetic energy must equal -100 J. Since KE_i = 0 J, KE_f= -100 J.

Thanks in advance for the help.

2. Jul 23, 2006

therealkellys

Bump... I'm still really unsure about my answer. If anybody could point out anything that I might have overlooked, that would help a lot. Thanks!

3. Jul 24, 2006

Hootenanny

Staff Emeritus
Your suspicions are correct. Change is usually expressed as final value minus the initial value, hence leading to a positive value for the change in potential energy. Just for reference a perhaps easier method would be to consider the definition of potential difference or voltage; which is work done per unit charge, thus;

$$V = \frac{W}{q}$$

From which we can obtain the expression $W = Vq$, substituting your values in we obtain $W = 100 \times 1 = 100\; J$.

Last edited: Jul 24, 2006
4. Jul 24, 2006

sdekivit

$$q \Delta V = -\Delta E_{kin}$$

5. Jul 24, 2006

SGT

Since your proton is at a lower potential, it has a negative potential energy (-100J), so its kinectic energy must be +100J.
Think of it in terms of mechanics. You have a mass of 1kg at a height of 10m. You have a potential energy of 100J (actually 98) in reference to ground. If you drop the mass, when it hits the ground its potential energy will be zero and the kinectic energy 98J.