# Symmetries of non-parallel infinite conducting planes

Suppose I have semi-infinite conducting planes that intersect at some angle $\theta_0$ and have a potential difference of $V$ (the axis of intersection is somehow insulated so they are not actually in contact). If we consider the space between the plates that is subtended by the angle $\theta_0$, my textbook says that we can say that "because of the symmetry of the problem, any plane that passes through the axis where semi-infinite planes intersect is an equipotential surface", and thus the "potential between the plates is only a function of angular position". If we adopt a cylindrical coordinate system where the z-axis is aligned with the axis where the semi-infinite planes intersect, then this means that

\begin{equation} \varphi(r,z,\theta) = \varphi(\theta) \end{equation}

since $r$ and $z$ do not effect the potential because of the alluded to symmetry.

When I look at this problem, the independence of $z$ position is a lot more obvious than the $r$ independence. The geometry of this problem doesn't change for different surfaces of constant $z$, so then the potential shouldn't depend on $z$ -- that makes sense to me. However, can someone please explain how there is a such a symmetry in $r$? Since the planes are only semi-infinite, it seems to me that for different values of $r$, the geometry of the problem should be different, so then how is there a symmetry?

• If you have two semi-infinite conducting planes that intersect, wouldn't they be at the same potential?
– jim
Jul 3, 2016 at 15:28
• Yeah we just suppose those that they are held at a potential difference of $V$. The inclusion of the origin as a point where the potential is valid is exactly why I am suspicious of the argument @user47033 provided in his answer, I'm trying to sort out the image charges without relying on the origin right now since I'm pretty sure it should work out. Jul 3, 2016 at 15:35

The symmetry arises because of the boundary conditions, which are independent of $z$ and $r$. Let us place one conducting plate at $\theta=0$ with potential $\varphi=0$, and another at $\theta_0$ at potential $\varphi=V$. We now want to find $\varphi(r, \theta, z)$ in the gap.

You are right that while $\varphi(z) = \varphi(-z)$ at arbitrary $(r,\theta,z)$, the same is not true for $\varphi(r) \neq \varphi(-r)$ . However, in cylindrical coordinates ($r$, $\theta$, $z$) we have $r\geq0$. Therefore, to establish a radial symmetry in cylindrical coordinates used here, we need $\varphi(r_1, \theta,z) = \varphi(r_2, \theta,z)$ for all unequal $r_1, r_2 > 0$ , which is true along each conducting boundary plate.

For the curious, the Laplacian now gives $\nabla ^2 \varphi = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial \varphi}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 \varphi}{\partial \theta^2} + \frac{\partial^2 \varphi}{\partial z^2}=\frac{1}{r^2}\frac{\partial^2 \varphi}{\partial \theta^2}=0$. This differential equation has a general solution that is linear in $\theta$. Matching the boundary conditions gives $\phi(\theta)=V \frac{\theta}{\theta_0}$ in the gap.

• How does having $\varphi(r_1,\theta,z) = \varphi(r_2,\theta,z)$ true only along the two plates means that it must be true between the plates as well? I agree that the boundary conditions set the symmetry, but I think you can get away with just saying that because the electric field tangential to the surface is 0 at the plates, and the tangential components of the electric field must be continuous, this means that everywhere between the plates the radial components vanish. Since $E_r = -\frac{\partial V}{\partial r} =0$, then the potential cant depend on $r$. Jul 5, 2016 at 22:22

Consider a planar crossection of the configuration. Let O be the origin at the intersection of the 2 planes. Suppose that there are electric field lines in the radial direction. This would imply that non zero work must be done to bring a charge q from infinity along the radial direction to O or in other words the potential at O (just above the plates ,not on it) must be non zero. However, we can see that the image charges of q lie on a circle of radius r when q is at a distance r from O. Also sum of the magnitudes of the image charges and q is zero. Now, since all these charges are at the same r from O, the potential at O is always zero. Contradiction. Hence there is no radial field, only tangential. The field lines are circular arcs centred at O. Thus , equipotential surfaces being perpendicular to field lines, will include all radial surfaces.

• Hi, can you please comment on how you know where the image charges will be induced? Thanks. Jul 3, 2016 at 6:55
• The image charges are pretty much like that due to optical reflection of an object from 2 mirrors inclined with each other. These images lie on a circle. Same here Jul 3, 2016 at 7:04
• My only concern with this approach right now is the inclusion of the origin in the calculation, if one plate is at V = 0, and the other is at V = V, then how can we say that the point of intersection must be at V = 0 ? Jul 3, 2016 at 15:47
• First of all, i mentioned O is just above the plate. Not at the exact intersection. Moreover the 2 plates must be at the same potential . Because metals in contact have the same potential. Jul 3, 2016 at 18:29
• For this specific problem they are held at a potential difference, I suppose they are insulated at the intersection then. Jul 3, 2016 at 18:34