# Equipotential Lines and Electric Fields Labratory Experiment

1. Sep 14, 2009

### llauren84

1. The problem statement, all variables and given/known data
Our lab is not online, but this is so similar to what we did. http://physics.fullerton.edu/~SAM/PDF/Lab%20Manuals/212/Individual%20Experiments/Equipotential%20Surfaces%20E6.pdf" [Broken] Instead of the ten that they set their voltage to, we set ours to 12.

2. Relevant equations

Eq. 1: E=Fq , where E and F are vectors, E is the electric field, F is the force on the charge, and q is the small positive test charge.
Eq. 2: $$\Delta$$V = Ed, where $$\Delta$$V is the potential difference, E is the electric field strength, and d is the distance between potentials.

3. The attempt at a solution

I am so confused and unfortunately, I can't ask the professor at this time. I have a few questions.
(1) I am not sure what q is exactly. Is that the number that the multimeter reads or is it the number 12 that we set our voltage to?
(2) How can I calculate $$\Delta$$V?
(3) Do you think I should be doing different calculations for points on different equipotential lines if the q is the readout from the multimeter at those points or along the curve?

Basically, I just have no idea what to calculate. I am so used to charts as our data and I'm having a hard time taking the info from the curves that we drew to actual calculations.

Last edited by a moderator: May 4, 2017
2. Sep 14, 2009

### ideasrule

(1) You don't know q, and don't need to worry about it.
(2) You measure it with a voltmeter.
(3) I'm not sure what you mean here. Along any equipotential curves, delta-V should be equal. You can approximate the electric field by measuring the distance between two equipotential curves and using the formula V=Ed; I think this is what the lab wants you to do.

3. Sep 14, 2009

### mikelepore

Not E=Fq, but F=qE.

4. Sep 15, 2009

### llauren84

Thank you =)

So the multimeter readout is the $$\Delta$$V?