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.
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