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I find that often times I'll be tripped up by questioning whether or not I can do something mathematically, and be unable to come up with a satisfying answer. This is, unfortunately, one of those times.

I'm told:

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

My first course of action is:

We know... $$|r| = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2} = x = E_0$$ $$\tan^{-1}{\frac{y}{x}} = \theta = 0$$ $$z = z = 0$$

So the electric field only has a component in the $\hat{r}$ direction.

Now, we know that $\vec{E} = -\nabla V(r, \phi, z) = - \frac{\partial V}{r} - \frac{1}{r}\frac{\partial V}{\theta} - \frac{\partial V}{z}$

So, I think "Oh. I just have to integrate." ... but over what? Do I integrate three times, once w.r.t. radius, then phi, then z? I'm pretty sure that won't give me the right answer. If I decide to express $\nabla V$ in terms of cartesian coordinates, I get $- \nabla V(x,y,z) = E_0 \hat{x}$ ... but the question still remains.

I feel like this is definitely the easy part of the problem, and I can often do the more complicated parts—it's just small things like this often throw me off. How would I go about extracting the potential from either of those equations? I know I have to integrate, but... where?

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There are a lot of vector calculus integration identities, but the one you are looking for is – DumpsterDoofus Mar 11 '14 at 0:22
Also, you might find that instead of converting $\mathbf{E}$ from rectilinear to spherical and then computing $V$ from that, it'll be a lot easier to compute $V$ in rectilinear and then convert that to spherical. – DumpsterDoofus Mar 11 '14 at 0:25
up vote 2 down vote accepted

It seems to me that you have more of a conceptual issue than a mathematical one. To hopefully remedy this, let me remind you of a couple of facts.

  1. Given an electric field $\mathbf E$, an electric potential $V$ for $\mathbf E$ is any scalar function $V$ for which \begin{align} \mathbf E = -\nabla V \end{align}
  2. It follows that if $V$ is such a potential, then we can integrate both sides along a curve $C$ to obtain \begin{align} \int_C\mathbf E\cdot d\boldsymbol \ell = -\int_C \nabla V\cdot d\boldsymbol \ell \end{align}
  3. If $C$ is a curve with endpoints $\mathbf a$ and $\mathbf b$, then the gradient theorem tells us that the right hand side can be evaluated in terms of the values of $V$ at these endpoints alone; \begin{align} -\int_C \nabla V\cdot d\boldsymbol \ell = V(\mathbf a) - V(\mathbf b) \end{align}
  4. We now have the freedom to choose a reference point at which we decide what the value of the potential is (this comes from the fact that in step 1, the condition that the field be the gradient of the potential does not uniquely specify the potential) at some chosen reference point $\mathbf b = \mathbf x_0$, and then combining steps 2 and 3 allows us to compute the value of the potential at any other point $\mathbf a = \mathbf x$. In other words, combining these remarks with steps 2 and 3 we obtain \begin{align} V(\mathbf x) = V(\mathbf x_0) + \int_C \mathbf E\cdot d\boldsymbol \ell \end{align} where $C$ is any path from $\mathbf x$ to $\mathbf x_0$.

In short, the electric potential is computed by choosing its value at a certain reference point, and then performing a line integral along any path to another point at which you want to determine its value. In this way, you can obtain the functional form of $V$ at any point $\mathbf x$ you like.

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I agree with joshphysics. This seems to address the issue I sifted out of your question. Step 4 is the big idea to concentrate on. To state it in another way, you shuold integrate from your reference point (which is wherever you want) to an arbitrary point $(x, y, z)$ or $(r,\theta,\phi)$. Most often one uses primed variables $x', y'$, etc as the variable of integration to distinguish from the end point. – BMS Mar 11 '14 at 1:00
@BMS Thanks for that. I think your bold statements might convey the point of #4 better and more succinctly than the way I said it. – joshphysics Mar 11 '14 at 1:07

I've posted an answer describing the derivation of potential energy which you might want to read, as the same argument applies to electrical potential and I think that's what you're missing. Basically, given an electric field, the first step in finding the electrical potential is to pick a point $\vec{x}_0$ to have $V(\vec{x}_0) = 0$. Then, to determine the potential at any point $\vec{x}$, you integrate $\vec{E}\cdot\mathrm{d}\vec{s}$ along any path from $\vec{x}_0$ to $\vec{x}$. The dot product gives you a simple function to integrate, so you don't have to deal with multiple directions. Also note that it doesn't matter what path you pick, the answer will be the same, so you can exploit this freedom to pick a path that is easy to integrate along.

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In Cartesian coordinates, you have $$V(x,y,z)=E_0x,$$ so converting to spherical coordinates (using Mathematica 9.0) yields the potential $V(r,\theta,\varphi)$ of

TransformedField["Cartesian" -> "Spherical", E0 x, {x, y, z} -> {r, \[Theta], \[CurlyPhi]}]

$$E_0 r \sin (\theta ) \cos (\varphi ).$$

I skipped the step where you convert $\mathbf{E}$ to $V$ in Cartesian coordinates because it was pretty obvious what $V$ was in this case. In general, for more complicated fields you determine the potential via the gradient theorem, $$V(\mathbf{r})=\int_{\gamma[\mathbf{r},\mathbf{r}_0]}\mathbf{E}\cdot d\mathbf{r}$$ where $\gamma[\mathbf{r},\mathbf{r}_0]$ is a suitably-chosen path from the reference point $\mathbf{r}_0$ to $\mathbf{r}$.

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