# Equipotential surgaces

1. Jan 31, 2009

### gunitsoldier9

1. The problem statement, all variables and given/known data
Consider a region in space where a uniform electric field 6700 N/C points in the negative x direction.
What is the orientation of the equipotential surfaces?
a.Parallel to the xz-plane.
b.Parallel to the yz-plane.
c.Parallel to the xy-plane.

2. Relevant equations

E=V/r

3. The attempt at a solution

2. Jan 31, 2009

### v0id19

i'll try to rephrase the question:
a continuous, 2 dimensional shape in which of the following planes (a, b, or c) would have an equal potential throughout its entire surface

3. Jan 31, 2009

### gunitsoldier9

would that make it xy because a z would mean 3 dimensions are involved.

4. Jan 31, 2009

### Hannisch

There are three dimensions involved.

Remember, an equipotential surface is perpendicular to the electric field--if you draw (or imagine, whichever you find easier) the electric field and define the x-axis, the y-axis and the z-axis, you can find the plane that would always be perpendicular to the field.

5. Jan 31, 2009

### gunitsoldier9

I know that its perpendicular to the eclectic field but it just doesnt make sense if the electric field is on the x axis i know the equipotential surface would be vertical. but i just dont get what plane it would be parallel to.

6. Jan 31, 2009

### Hannisch

Yes, it'd be vertical, but remember it has a thickness as well.

A new wording might help: The electric field is along the x-axis. If you have the xy plane, as an example, it would go along the x-axis as well, but with a height or something. If two things go along the same axis, can they ever be perpendicular?

7. Jan 31, 2009

### gunitsoldier9

no they cant. so does that rule out the xy plane

8. Jan 31, 2009

### Hannisch

Yes, because something going along the same axis as another thing will never be perpendicular to each other. Hopefully you see why precisely it's ruled out. Try to use the same logic with the other alternatives and you might come up with the correct answer. But try to really understand what you are doing!

9. Jan 31, 2009

### gunitsoldier9

i think its yz. its really hard to visualize the plane in 3 dimensions and my book doesnt even talk about 3 dimensions for that equipential surfaces

10. Jan 31, 2009

### Hannisch

Three dimensions isn't the easiest thing to imagine and it's not the easiest thing to write either. You are correct in that it's the yz plane.

Try to think of it like this or something:

Use a piece of cardboard or something similar--if you put it on the table on the edge and it's facing you, you have a xy plane. If you lie it down on the table you have a xz plane. If you put it on the edge again but with the thin side facing you, that's a yz plane.

Or sort of just think of the z-axis to come out of the paper or something.

And electric fields always exist in three dimensions. If you have a field around a point charge you only draw the field lines parallel to the paper, but you sort of have to imagine that there are field lines going up from the paper and down through the table as well, because the field isn't only limited to going in the two dimensions you can draw (easily).

Therefore, equipotential surfaces also have to exist in three dimensions, right? :)

11. Jan 31, 2009

### gunitsoldier9

Yeah thanks. I wish that they would actually draw a picture in the book and point it out.