# Calculating electric potential from constant E-Field

1. Mar 11, 2014

### AmagicalFishy

Hey, folks.

I've typed out the question here. I hope you don't mind my just linking this (if that's against the rules or something, I'll type it out).

I'm still pretty confused about what to do. I assume, as usual, it's probably something... well... stupid—but I can't figure out what.

Thanks for any help you can offer.

2. Mar 11, 2014

### BvU

Hello Fishy,

I'm unaware of the rules about linking and such and want to keep it that way.
I am more than astonished by your coordinate transformations: How do you know that y and z are zero when you say "we know ... r = x" ? And ... $\theta = 0$ z=0 as well.

If E is constant all over the place, it should have the same value at z = 100000 as at z= 123 or z=0. Who made you believe that z = 0 ?

3. Mar 11, 2014

### BvU

There are more things you claim "we know" that are definitely not true. $E(x_0-x)=V(x)-V(x_0)$ ???? Can't be! E are Volts/m and V are Volts. No way they can appear with an = in between. It just doesn 't make sense!

4. Mar 11, 2014

### AmagicalFishy

My coordinate transformations come from the fact that the E-field only moves in the x-direction; thus, when we integrate it (to get a voltage), we only get something dependent on the x-coordinate. This means that the potential function is only a function of x. It has no y or z dependence. So, when we transform the voltage into cylindrical coordinates, we know y = 0 and z = 0. I'm... not sure what's wrong with that (but apparently something is).

In regards to your second comment, note that $E_0(x_0-x)$ is the magnitude of the electric field multiplied by $(x_0 - x)$ (which give the correct units). It's not a function, but two quantities multiplied together.

5. Mar 12, 2014

### BvU

1. Coordinate transformations have nothing to do with anything except coordinate systems. Not with whatever you want to describe in those coordinate systems.
2. The E-field is constant. It does not move.
3. Correct, so V(x,y,z) = V(x), (4) Correct, V is a function of x only. (5) It has no y, z dependence.

6. No! You do not transform the voltage. You really don't. The only thing you transform is the way you describe the coordinates in space. V is NOT affected at all. If the temperature at the top of the Eiffel tower is 23$^\circ$ it is 23$^\circ$ also at 48.8582° N, 2.2945° E, 324 m height.

( voltage transformers are known in the AC world, though. But they have little to do with coordinate transforms...)

7. There is something very wrong with that! If you are at y = 123 mm, there is a V, whether or not you describe it in cartesian or cylinder or polar or other coordinates. V is a scalar. It does not have any y-ish-ness. What is zero is the y component of $\vec E$. And its z component.

Oops, my misinterpretation. Sorry. Physics wise, dimensions are correct.

$\vec E$ is a vector, V is a scalar. So this is a scalar equality. Means that on the right there is a scalar (with the physical dimension of energy/charge) , e.g. 10 Volt. So on the left there is necessarily also a scalar. As you state, $E_0$ (clearly a scalar) times a dot product $\int \hat x d\vec l$ integrated from vector position $\vec x_0$ to $\vec x$.

The integral gives you ${\rm x}(\vec x) - {\rm x}(\vec x_0)$ As you have seen, the usual reference point at $\infty$ is not usable, so let's choose the voltage at the y-z plane as a reference and choose V = 0 there. (So: V(0, y, z) = 0 Volt, and V(x, y, z) = x Volt). Choice does not influence the electric field.

Pick a point. Anywhere in space. What is V ? Answer from the user (named Cas) of a cartesian coordinate system: V = x Volt. Let's say you picked the point that C describes as (48.8582, 2.2945, 324). So cas immediately knows that V there is 48.8582 Volt.

Next to Cas sits a smart but stubborn guy named Cyril who uses a cylindrical coordinate system. C&C have agreed to use the same origin for their respective coordinate systems, and not to move that origin any more either.

The coordinate transformation is what Cyril needs to do to convert Cas's designation of that point in space you picked to something that designates, describes and and uniquely determines that point for him in his cylindrical coordinate system.

The exercise you were given is to express the potential there (48.8582 Volt) in an expression in terms of Cyril's $s, \phi$ and $z$.

Being smart and stubborn, Cyril says: Well, whatever, I don 't need z. Cas's z and my z are identically the same. If he don't need it, I don't either. I don't need his y either, but I do need his x because that is his voltage expression.

And Cyril continues: I know (he leaves Fish's "we know" for what it is...see below under "Sophie") that my $\rho$ (which in the exercise is called $s$ for some untraceable reason -- but in an unexplicable mood of demurity he uses $s$ from here on) is $\sqrt(x^2+y^2)$ And that the cosine of my $\phi$ is his $x$ divided by my $s$. Meaning that his x is my $s \ \sin \phi$. Meaning that $V(s,\phi, z) = s\ \sin \phi$ Volt. Done.

[Note to moderators: I am aware that dishing out a solution is a crime in PF. My plea is that the solution is not what this OP needs most. He (she?) needs to learn to read thoroughly what the helpers at StackExchange and PF try to tell him. Wading through this long story is also a good exercise for this particular OP. And anyway, the Stackers have already spoiled completely]

C&C compete for the favours of next-door beauty Sophie, who is living in other spheres altogether and consistently uses spherical coordinates. Fed up with C&C's nerdiness she offers Fish her friendship if he helps her with a similar exercise:

A uniform electric field, $\vec E= E_0 \hat x$. What is the potential, expressed using spherical coordinates, $V(r, \theta, \phi)$ ?

If you got this far, you also have the stamina to re-read what you harvested on stackexchange. Everything you need is really there, so posting on PF was a sign of what, exactly ?

joshphysics reply is excellent. Doofus spoils -- funny enough not your exercise, but Sophie's.

Last edited: Mar 12, 2014
6. Mar 13, 2014

### AmagicalFishy

It took me a while, man, but thank you very much. I was getting my... well, my everything pretty badly mixed up.

I was mixing up the $\hat{x}$ with the $x$ component of the E-field and the variable $x$ when converting from Cartesian Coordinates to Cylindrical. I don't know what in God's name went wrong in my head—but thanks very much. :)