Calculating the electric potential in cylindrical coordinates from constant E-field

Question

I am having so much trouble with this problem. I feel like I shouldn't be, but I am.

A uniform electric field, $\vec{E} = E_0\hat{x}$. What is the potential, expressed using cylindrical coordinates, $V(s,\phi,z)$?

I asked a question related to this yesterday, and got a solid answer, but when I carry things out, it still doesn't make sense physically (though, after a bit of studying, I now understand the Gradient Theorem).

We know that: $$\vec{E} = E_0\hat{x} = -\nabla V \implies -\int_{C[x_0,x]} E_0\hat{x}\cdot dl = V(x) - V(x_0)$$ Since the electric field only has quantity in the $\hat{x}$ direction, the dot product inside of the integral comes out to $E_0~dx$, giving us: $$E_0(x_0 - x) = V(x) - V(x_0)$$ Now, this makes sense. Since the electric field is a conservative one, the path we take shouldn't depend on any other variable but $x$. If we move in some $y$ direction and some $z$ direction, there will be zero change in potential.

Let $V(x_0) = 0$ be our point of reference, so: $$V(x) = E_0(x_0 - x)$$ So far (I think) this all seems fine and dandy. The reason I don't set $E_0x_0$ to zero is because there is no $1/x$ term; otherwise I'd be able to make the reference point at $x=\infty$, and things would cancel out nicely.

Now, I try to convert to cylindrical coordinates. We know: $$r = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2} = x \\ \theta = \tan^{-1}{\frac{y}{x}} = 0 \\ z = 0 $$

So $V(r) = E_0(r_0 - r)$

But... this doesn't make any sense. With this potential function, moving $r$-distance in the $-x$ direction will give you the same potential as moving $r$-distance in the positive $x$ direction (which will give you the same potential as moving $r$-distance in the $y$ direction, even, since $r$ is just a radial distance from, say, the $z$-axis.

I am doing something terribly wrong and I have no idea what it is, haha.

Answer 1

Let's assume you already know the potential in Cartesian coordinates:

$V(x,y,z)$ with $\vec E(x,y,z)=-\nabla V(x,y,z)$.

Now you only have to substitute x,y,z with your cylindrical coordinates:

$x=r cos(\alpha)$

$y=r sin(\alpha)$

$z=z$

Which leads to $V'(r,\alpha,z)=V(r cos(\alpha), rsin(\alpha), z)$ with $V'$ being the potential in cylindrical coordinates.

If you want to get the $\vec E$-Field in cylinder coordinates you just have to use the cylindrical version of the gradient:

I hope this answer could help you.

All the best!

Answer 2

You're so close to the answer I'm not sure how to nudge you along without basically just giving you the answer. I'll try anyways. There are some troubling conceptual mistakes you've made in an otherwise straightforward derivation.

For starters :

The reason I don't set $E_0 x_0$ to zero is because there is no $1/x$ term; otherwise I'd be able to make the reference point at $x=\infty$, and things would cancel out nicely.

This argument makes no sense. The only thing you can set is the point $\vec{r}$ at which $V(\vec{r})$ is $0$, which in your case as you rightfully point out boils down to a choice of $x_0$. You can't set $E_0 x_0$ to anything, it's just a product of numbers, so you might as well choose $x_0 = 0$ to simplify things. Your potential, in Cartesian coordinates, then becomes :

$$ V(x,y,z) = V(x) = - E_0 x $$

Now you want to convert this to cylindrical coordinates for some reason. At this point, you went about things backwards. You looked at the formulas for converting cylindrical coordinates to Cartesian coordinates, rather than the opposite. Even though this step is useless, it's worth pointing out you did this completely wrong on 2 levels :

First, the radial coordinate in cylindrical coordinates is $s = \sqrt{x^2 + y^2}$. There's no $z$ term involved.

Second, just because your physical system has some symmetry doesn't mean your coordinate system magically changes. $y$ and $z$ are not $0$ just because you don't care about them. You can decide to work on the line $y = z = 0$, but that's different from what you did. You seem to be operating under the impression that at all coordinates, even for non zero $y$ and $z$, we have $y = z = 0$, which is just nonsense.

Finally, let's take a look at the transformations you do need, those that transform Cartesian coordinates to cylindrical ones. In fact, because your potential only depends on $x$, we only need one formula :

$$ x = s \cos \phi $$

I won't write the down the answer for you, but you have everything you need to know now. You have an expression for $V$ in Cartesian coordinates, and the formula that transforms Cartesian coordinates into cylindrical ones. I hope I don't have to spell it out more than that.

Answer 3

You assumed that $y=0$ and $z=0$ when you did the following steps, $$r = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2} = x$$ As a result, you are now looking at all the set of points which are having coordinates of form $(x,0,0)$ that is, the x-axis. So you cannot find the potential at any other point except the ones on the x-axis using your potential formula in cylindrical coordinates.

If you want to find the general potential in cylindrical coordinates, the you have to use the following transformation: $$x=r\cos(\phi)$$ $$\Longrightarrow V(r,\phi,z) = E_0(x_0 - r\cos(\phi))$$ Which is the correct form of the potential in cylindrical coordinates