How to use symmetry to determine the directions in which the electric field is zero?

Question

I learned that if the charge distribution has a plane of symmetry, then the electric field lies on the plane of symmetry. If that is correct, then if two positively charged plates are kept opposite to each other in the x-y plane, then won't the electric field be zero perpendicular to the xy plane?

Answer 1

You are being not clear in your formulation.

Clarifications.

There are two things to consider when talking about vector fields:

• the point in which you evaluate (where you look) the field, noted $\mathbf{r}$
• the vector value (magnitude and direction) of field - or just, "the field" - on that point, noted $\mathbf{E}$.

In vector field notation (and cartesian coordinates), we write the field on a point as \begin{equation} \mathbf{E}(\mathbf{r}) = E_{x}(x,y,z)\mathbf{\hat{x}} + E_{y}(x,y,z)\mathbf{\hat{y}} + E_{z}(x,y,z)\mathbf{\hat{z}} \end{equation}

Charge distribution and field symmetry.

The correct general word-formulated statement would be something like

If the charge configuration has a reflection symmetry wrt to a plane of symmetry $\pi$, the field will have a reflection symmetry wrt to that plane $\pi$.

A reflection symmetry consists on flipping the sign of the component of the vector perpendicular to the plane of symmetry. Hence, for two points $1$ and $2$ that are relections wrt to $\pi$, we have \begin{align} \mathbf{r{ _{2\parallel}} } &= \mathbf{r{_{1\parallel}}} \\ r_{ {2\perp} } &= - r_{ {1\perp} } \end{align} where the first equation refers to the components parallel to the plane and the second one to the perpendicular component. We can write both points as \begin{align} \mathbf{r_{1}} &= (\mathbf{r_{\parallel}}, r_{\perp}) \\ \mathbf{r_{2}} &= (\mathbf{r_{\parallel}}, - r_{\perp}) \end{align}

The reflection of the field is \begin{align} \mathbf{E}_{\parallel}(\mathbf{r}_{\parallel}, r_{\perp}) &= \mathbf{E}_{\parallel}(\mathbf{r}_{\parallel}, -r_{\perp}) \\ E_{\perp}(\mathbf{r}_{\parallel}, r_{\perp}) &= - E_{\perp}(\mathbf{r}_{\parallel}, -r_{\perp}) \end{align}

Particular case.

It will become clearer with your example. The plane of symmetry is $xy$ and, therefore, the component perpendicular to it is $z$. The above statement reads \begin{align} E_{x}(x,y,z) &= E_{x}(x,y,-z) \\ E_{y}(x,y,z) &= E_{y}(x,y,-z) \\ E_{z}(x,y,z) &= - E_{z}(x,y,-z) \end{align}

But it seemed you were more preciselly concerned about what happened in the plane of symmetry $xy$ ($z=0$). There, we have \begin{align} E_{x}(x,y,z) &= E_{x}(x,y,0) \\ E_{y}(x,y,z) &= E_{y}(x,y,0) \\ E_{z}(x,y,0) &= - E_{z}(x,y,0) \\ \end{align} You can see that the first two equations for the components of the field on the plane ($E_{x}$ and $E_{y}$) are trivially satisfied and contain no information. However, the last equation for $E_{z}$ does tells us something: \begin{equation} E_{z}(x,y,0) = 0 \end{equation}

Correctly word-formulated would be

In the $xy$ plane, the component of the electric field perpendicular to the $xy$ plane is $0$.

or

In the $xy$ plane (evaluation), the field (it's direction) is contained on the $xy$ plane.

Answer 2

I am not entirely sure what you mean by "lies on the plane of symmetry"... Maybe you could reformulate and I'll expand my answer?

Concerning your question: Yes, the z-component electric field vanishes between two equally charged parallel plates (ignoring edge effects). You can easily calculate this by regarding the plates as infinite flat sheets of charge.