Subfield: Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.
Context question:
1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
Context answer:
$\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$


Context question:
2.  The potential due to the line charge could be written as

$$
V(r)=f(r)+K,
$$

where $K$ is a constant. Determine $f(r)$.
Context answer:
\boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$}


Context question:
3.  Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces.
Context answer:
<img_4501>


Extra Supplementary Reading Materials:

For the following questions, ignore any edge effects.

Question: 4.  Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin.

Solution:
From eq.(5) and eq.(6), we see that for any arbitrary potential $V$, the equipotential surfaces of these two equal but opposite lines charge, are cylindrical surfaces. From this observation, we can choose the specific position for each line charge in both cylinders so that the surface of each cylinder is an equipotential surface.

Consider the following figure

<img_4545>

Figure 3: Two line charges with its equipotential surfaces

We would like to find a cylindrical equipotential surface enclose one line charge, let say the $-\lambda$ (if we could find the surface, by symmetry, we surely can find the identical one that enclose the line $\lambda$ ). The potential is given by

$$
\begin{aligned}
V & =-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r_{1}+\frac{\lambda}{2 \pi \epsilon_{0}} \ln r_{2} \\
& =-\frac{\lambda}{4 \pi \epsilon_{0}} \ln \left(l_{1}^{2}+R^{2}-2 l_{1} R \cos \phi\right)+\frac{\lambda}{4 \pi \epsilon_{0}} \ln \left(l_{2}^{2}+R^{2}-2 l_{2} R \cos \phi\right)
\end{aligned}
\tag{7}
$$

Since the surface of the cylinder has to be the equipotential surface, so the potential should not depend on $\phi$, i.e. $\frac{\partial V}{\partial \phi}=0$.

$$
-\frac{\lambda}{4 \pi \epsilon_{0}} \frac{2 l_{1} R \sin \phi}{l_{1}^{2}+R^{2}-2 l_{1} R \cos \phi}  +\frac{\lambda}{4 \pi \epsilon_{0}} \frac{2 l_{2} R \sin \phi}{l_{2}^{2}+R^{2}-2 l_{2} R \cos \phi}=0
\tag{8}
$$

$$
\begin{aligned}
\frac{l_{1}}{l_{1}^{2}+R^{2}-2 l_{1} R \cos \phi} & =\frac{l_{2}}{l_{2}^{2}+R^{2}-2 l_{2} R \cos \phi} \\
l_{1}^{2} l_{2}+R^{2} l_{2}-2 l_{1} l_{2} R \cos \phi & =l_{1} l_{2}^{2}+R^{2} l_{1}-2 l_{1} l_{2} R \cos \phi \\
l_{1} l_{2}\left(l_{1}-l_{2}\right) & =R^{2}\left(l_{1}-l_{2}\right).
\end{aligned}
$$

$$
l_{1} l_{2} =R^{2}
\tag{9}
$$

From the data in the problem, we have

$$
l_{1}+l_{2} =10 a
\tag{10}
$$
$$
l_{1} l_{2} =9 a^{2}
\tag{11}
$$

Solve this quadratic equation to get

$$
l_{1}=5 a \pm 4 a
\tag{12}
$$

However, since $l_{1}>l_{2}$, we have

$$
l_{1}=9 a
\tag{13}
$$
$$
l_{2}=a
\tag{14}
$$

Using this results on eq.(5), we have

$$
V=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}
\tag{15}
$$

This is the potential in all region except inside both cylinders. For cylinders at $x=-5 a$, the potential is constant and equal to

$$
V(x=-2 a, y=0)=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \frac{(4 a+2 a)^{2}+0^{2}}{(4 a-2 a)^{2}+0^{2}}=\frac{\lambda}{2 \pi \epsilon_{0}} \ln 3
\tag{16}
$$

For cylinders at $x=5 a$, the potential is constant and equal to

$$
V(x=2 a, y=0)=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \frac{(4 a-2 a)^{2}+0^{2}}{(4 a+2 a)^{2}+0^{2}}=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln 3
\tag{17}
$$

The potential difference between both cylinders are

$$
\Delta V=\frac{\lambda}{\pi \epsilon_{0}} \ln 3 \equiv V_{0}
\tag{18}
$$



Substituting this results in the potential equation, the potential outside the two cylinders are:

$$
V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}
\tag{19}
$$

And the potential inside the cylinders are:

The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$.

The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$.



Final answer: $\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$, $-V_{0} / 2$, $V_{0} / 2$