# Electric field of a parallel plate capacitor in different geometries

We know from electrostatics that the field of an infinite sheet of surface charge density $\sigma$ is

$$\vec E = \frac{\sigma}{2\epsilon}\hat{x}$$

But the field inside the capacitor is

$$\vec E = \frac{\sigma}{\epsilon}\hat{x}$$

That, in my understanding, is due to the addition of positive and negative plates fields

However in any other geometry such as a cylindrical geometry, we use the field due to one plate only. For instance, the following: The E field is identical to that of only one plate, not the addition of the positive and negative plates fields. Why is this the case?

• Hi and welcome to the Physics SE! The equations become much easier to read, search and edit when mathjax is used. I've proposed an edit to your post this time, but you should use it yourself in your next posts.
– Styg
Dec 23, 2017 at 10:38
• I searched for it in the editor, but without avail. Dec 23, 2017 at 10:39
• Yeah, I see the editor only has Markdown help. You might find this help page useful too.
– Styg
Dec 23, 2017 at 10:43
• More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. Dec 23, 2017 at 12:19