Cylindrical Coordinates: Line Integral Of Electrostatic Field

[QuantumPhoton1]

Homework Statement

An electrostatic field $\mathbf{E}$ in a particular region is expressed in cylindrical coordinates $( r, \theta, z)$ as

$$ \mathbf{E} = \frac{\sin{\theta}}{r^{2}} \mathbf{e}_{r} - \frac{\cos{\theta}}{r^{2}} \mathbf{e}_{\theta} $$

Where $\mathbf{e}_{r}$, $\mathbf{e}_{\theta}$ and $\mathbf{e}_{z}$ are the unit vectors in the directions of $r$ , $\theta$ and $z$ respectively.

Derive an expression for the potential difference $\Delta V = V_{2} - V_{1}$, where $V_{1}$ is the potential at position vector $\mathbf{r}_{1}$ with coordinates $(2R,0,Z)$ and $V_{2}$ is the potential at position vector $\mathbf{r}_{2}$ with coordinates $(4R,\frac{\pi}{2},Z)$.

Homework Equations

$$ \Delta V = V(\mathbf{r}_{2}) - V(\mathbf{r}_{1}) = - \int_{\mathbf{r}_{1}}^{\mathbf{r}_{2}} \mathbf{E} \cdot d \mathbf{l} $$
$$ \mathbf{E} = - \mathrm{grad} \, V $$

The Attempt at a Solution

I have gone through the question in a way that was hinted to me with the following logic. Since

$\mathrm{curl} \, \mathbf{E} = \mathbf{0}$

The electrostatic field is conservative and therefore any line integral from two positions, regardless of the path taken, is path independent. Therefore a suitable path could be suggested to be to traverse a path, from $\mathbf{r}_{1}$ in the radial $\mathbf{e}_{r}$ direction and then, from this new position, traverse the polar $\mathbf{e}_{\theta}$ direction to $\mathbf{r}_{2}$.

This approach I understand but I was wondering if there is another more efficient approach involving a direct linear path for a line integral from $\mathbf{r}_{1}$ to $\mathbf{r}_{2}$. This is where my algebra has taken me.

The aim is to get to an expression, involving the parameter $t$, where

$$
\begin{align}
\Delta V &= - \int_{\mathbf{r}_{1}}^{\mathbf{r}_{2}} \mathbf{E} \cdot d \mathbf{l} \\
&= - \int_{t_{1}}^{t_{2}} \mathbf{E} \cdot \frac{d \mathbf{l}}{dt} ~ dt
\end{align}
$$

Let the path $\mathbf{l}$ be a parametric path connecting $\mathbf{r}_{1}$ to $\mathbf{r}_{2}$ with the following substitutions and conditions.

$$ r = 2R(t+1) ~~~~~~~~~ \theta = \frac{\pi}{2} t ~~~~~~~~~ z= z ~~~~~~~~~ 0 \leq t \leq 1$$

Hence $\mathbf{l}$ can be expressed in these parameters as

$$ \mathbf{l} = 2R(t+1) \mathbf{e}_{r} + \frac{\pi t}{2} \mathbf{e}_{\theta} + Z \mathbf{e}_{z}~~~~~~~~~ 0 \leq t \leq 1$$

This path represents the line integral with $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ at the end points.

Since $\mathbf{l}$ is path in cylindrical coordinates, consider a line element $d \mathbf{l}$ which has the form

$$
\begin{align}
d \mathbf{l} &= dr ~ \mathbf{e}_{r} + r ~ d \theta ~ \mathbf{e}_{\theta} + dz ~ \mathbf{e}_{z} \nonumber \\
\implies \frac{d \mathbf{l}}{dt} &= \frac{dr}{dt} ~ \mathbf{e}_{r} + r ~ \frac{d \theta}{dt} ~ \mathbf{e}_{\theta} + \frac{dz}{dt} ~ \mathbf{e}_{z} \nonumber \\
&= 2R \mathbf{e}_{r} + 2R(t+1) \frac{\pi}{2} \mathbf{e}_{\theta} + 0 \mathbf{e}_{z} \nonumber \\
&= 2R \mathbf{e}_{r} + \pi R(t+1)\mathbf{e}_{\theta}
\end{align}
$$

Also, Expressing the electrostatic field $\mathbf{E}$ with the above parameters becomes

$$
\begin{align}
\mathbf{E} &= \frac{\sin{\theta}}{r^{2}} \mathbf{e}_{r} - \frac{\cos{\theta}}{r^{2}} \mathbf{e}_{\theta} \nonumber \\
&= \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ \left ( 2R(t+1) \right ) ^{2}} \mathbf{e}_{r} - \frac{\cos{\left ( \frac{\pi t }{2} \right )}}{\left ( 2R(t+1) \right ) ^{2}} \mathbf{e}_{\theta} \nonumber \\
&= \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ 4 R^{2} (t+1)^{2}} \mathbf{e}_{r} - \frac{\cos{\left ( \frac{\pi t }{2} \right )}}{4 R^{2} (t+1)^{2} } \mathbf{e}_{\theta} \nonumber \\
\end{align}
$$

Hence, let the line integral be defined as

$$
\begin{align}
\Delta V &= - \int_{\mathbf{r}_{1}}^{\mathbf{r}_{2}} \mathbf{E} \cdot d \mathbf{l} \\
&= - \int_{t_{1}}^{t_{2}} \mathbf{E} \cdot \frac{d \mathbf{l}}{dt} ~ dt \nonumber \\
&= - \int_{0}^{1} \begin{bmatrix}
\frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ 4 R^{2} (t+1)^{2}} \\
- \frac{\cos{\left ( \frac{\pi t }{2} \right )}}{4 R^{2} (t+1)^{2} } \\
0
\end{bmatrix}
\cdot
\begin{bmatrix}
2R \\ \pi R(t+1) \\ 0
\end{bmatrix}
~ dt \nonumber \\
&= - \int_{0}^{1} \frac{2R\sin{ \left ( \frac{\pi t }{2} \right ) }}{ 4 R^{2} (t+1)^{2}} - \frac{\pi R(t+1)\cos{\left ( \frac{\pi t }{2} \right )}}{4 R^{2} (t+1)^{2} } ~ dt \nonumber\\
&= - \int_{0}^{1} \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ 2 R (t+1)^{2}} - \frac{\pi \cos{\left ( \frac{\pi t }{2} \right )}}{4 R (t+1) } ~ dt \nonumber\\
&= - \frac{1}{2R} \int_{0}^{1} \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ (t+1)^{2}} - \frac{\pi \cos{\left ( \frac{\pi t }{2} \right )}}{2 (t+1) } ~ dt \nonumber\\
&= \frac{1}{2R} \int_{0}^{1} \frac{\pi \cos{\left ( \frac{\pi t }{2} \right )}}{2 (t+1) } ~ dt - \frac{1}{2R} \int_{0}^{1} \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ (t+1)^{2}} ~ dt \nonumber\\
&= \frac{\pi}{4R} \int_{0}^{1} \frac{ \cos{\left ( \frac{\pi t }{2} \right )}}{ (t+1) } ~ dt - \frac{1}{2R} \int_{0}^{1} \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ (t+1)^{2}} ~ dt \nonumber\\
\end{align}
$$

This is where I think that I've gone wrong somewhere as I can see no obvious substitution to change either of these integrals into a standard form to be integrated.

Any help would be appreciated. Thank you

 

[Cryo]

I may be missing something, but why are you trying to work with a specific path? Isn't the integral $\int^{\mathbf{r}_2}_{\mathbf{r}_1} \mathbf{E}.d\mathbf{l}$ path-independent if $\mathbf{\nabla}\times\mathbf{E}=\mathbf{0}$

So $\mathbf{E}=-\mathbf{\nabla}V=-\mathbf{\nabla}\left(\frac{\sin\theta}{r}\right)=\frac{\sin\theta}{r^2}\mathbf{e}_r-\frac{1}{r}\cdot\frac{\cos\theta}{r}\mathbf{e}_\theta$

$\int^{\mathbf{r}_2}_{\mathbf{r}_1} \mathbf{E}.d\mathbf{l}=-V\left(\mathbf{r_2}\right)+V\left(\mathbf{r_1}\right)$

etc...

 

[TSny]

$$
\begin{align}
\Delta V &= - \int_{\mathbf{r}_{1}}^{\mathbf{r}_{2}} \mathbf{E} \cdot d \mathbf{l} \nonumber\\
... \nonumber\\
&= \frac{\pi}{4R} \int_{0}^{1} \frac{ \cos{\left ( \frac{\pi t }{2} \right )}}{ (t+1) } ~ dt - \frac{1}{2R} \int_{0}^{1} \frac{\sin{ \left ( \frac{\pi t }{2} \right ) }}{ (t+1)^{2}} ~ dt \nonumber\\
\end{align}
$$

This is where I think that I've gone wrong somewhere as I can see no obvious substitution to change either of these integrals into a standard form to be integrated.

I think your final expression is correct. But the integrals are difficult. Mathematica did not seem to be able to evaluate them in terms of elementary functions. However, numerical integration does give the correct result.

The path you chose is not a straight line. It's an arc of an Archimedean spiral. https://en.wikipedia.org/wiki/Archimedean_spiral

You can find the expression for the straight path in polar coordinates. It's then still somewhat messy to set up the integration in polar coordinates. But the integrals turn out to be elementary.

Of course, @Cryo 's method is the quickest route to the answer. But, maybe you are just practicing doing line integrals in polar coordinates.

By the way, welcome to PF!

 

[vela]

$$-\frac{1}{2R} \int_0^1 \left[ \frac{\sin \frac{\pi t }{2}}{(t+1)^2} - \frac{\frac{\pi}{2} \cos \frac{\pi t}{2}}{(t+1)}\right]\,dt $$

Note that the integrand is the derivative of $\left(-\frac 1{t+1} \sin \frac{\pi t}{2}\right)$.

 

[TSny]

Note that the integrand is the derivative of $\left(-\frac 1{t+1} \sin \frac{\pi t}{2}\right)$.

Nice. I was too dense to consider combining the two separate integrands.

I guess no matter what path you choose, you should always be able to express the overall integrand as the differential of V.