Symmetries of non-parallel infinite conducting planes

Question

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?

Answer 1

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.

Answer 2

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.