# Homework Help: Equipotential surfaces electric field problem

1. Jan 27, 2010

1. The problem statement, all variables and given/known data
A given system has the equipotential surfaces shown in the figure
What is the magnitude of the electric field?
What is the direction of the electric field? (degrees from + x axis
What is the shortest distance one can move to undergo a change in potential of 5.00

2. Relevant equations
E= V/d

3. The attempt at a solution
I seem to be getting the same answer and its wrong I used the pythagoreon therom to find distances of each and divided the volts by the respecitive distances. I do not know what I am doing wrong I keep getting 2.24*10^2 V/m. Moreover, when I do the angle part I just do some trig and use that. Either way I try to do it and get 26.6 or 63.2. What am I doing wrong? Is the first part just the slope

2. Jan 27, 2010

### rl.bhat

Where is the figure?

3. Jan 27, 2010

### merryjman

I can't see the figure yet either, but unless you're dealing with a uniform electric field then you can't use the equation you listed. As you said, electric field is the slope of a potential vs. position graph. If you're using calculus, it's the derivative with respect to position.

4. Jan 27, 2010

I think I uploaded it now. Sorry!
It is an attachment

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5. Jan 27, 2010

### rl.bhat

Slope of the equipotential lines is tan(theta) = 2/4 = 1.2.
Find theta.
Distance between the equipotential lines d is x*sin(theta). Here x = 4 cm.

6. Jan 28, 2010