# Why is the electric field between two conducting parallel plates not double what it actually is? [duplicate]

This question already has an answer here:

According to here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

We can treat two parallel plates as being a single conductor which has an electric field of $$\frac{\sigma}{\varepsilon_0}$$. They say that equivalently, we can consider the system to be two charge sheets, each one contributing an electric field of $$\frac{\sigma}{2\varepsilon_0}$$.

My question is, why is the electric field between two parallel conducting plates not $$\frac{2\sigma}{\varepsilon_0}$$? Shouldn't each plate be considered to be a conductor (not merely a charge sheet)? In that case, each plate contributes $$\frac{\sigma}{\varepsilon_0}$$, for a net electric field of $$\frac{2\sigma}{\varepsilon_0}$$. Why must we consider each conducting plate to be a charge sheet and not a conductor?

## marked as duplicate by John Rennie, ahemmetter, Kyle Kanos, Jon Custer, MartinMar 22 at 16:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – John Rennie Mar 22 at 6:06

## 1 Answer

First, are you comfortable with the fact that all of the net charge in a conductor will accumulate on its surface (thus, for parallel plates, you have four places to put sheet charges)? Second, have you drawn a diagram of what you imagine to be happening? Third, in that diagram is the electric field inside of both conductors zero? If not, you need to move charge between the sides of each conductor until it is. If you do this right, and you start with each conductor having equal and opposite net charge (density), then you'll find that all of the charge is on the inside surfaces of the plates. Say, to be concrete, the top conductor has $$\sigma = Q/A$$ to divide between its top and bottom surfaces, and the bottom has $$-\sigma$$ to divide. I believe you'll find that the only way to meet all of the constraints is to have $$0$$ on the top conductor's top, $$\sigma$$ on the top conductor's bottom, $$-\sigma$$ on the bottom conductor's top, and $$0$$ on the last surface.

To be concrete, the constraints are:

1. the sum of the charge density on both surfaces of the top plate is $$\sigma$$,
2. the sum of the charge density on both surfaces of the bottom plate is $$-\sigma$$, and
3. the electric field inside the bulk of a conductor (i.e. where the metal fills up) has to be zero.

As a consequence of $$3$$ you can immediately say that all unbalanced charges can only be on the surface(s) of a conductor, so you only need to think about the four surfaces discussed above.

Constraints one and two are set the way they are to ensure that the capacitor has no net charge on it. For a challenge exercise to ensure you know what's going on, consider: what happens if we allow the capacitor to be unbalanced (say, a net of $$\sigma_{t}$$ on the top plate and $$\sigma_b$$ on the bottom)? Draw the diagram, find the charge densities on all four surfaces, and the electric field in the three regions outside of the conductors (above both, between them, and below both).