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If we have the electric field on the surface of a conductor being $E=\sigma / \epsilon_0$ where $\sigma$ is the charge per unit area then why is the combined $E$ field from two parallel plates not 2x this result?

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  • $\begingroup$ This question does not show any research effort, e.g., Electric Field: parallel plates $\endgroup$ – Alfred Centauri May 12 '18 at 1:16
  • $\begingroup$ More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. $\endgroup$ – Qmechanic May 12 '18 at 5:01
  • $\begingroup$ @AlfredCentauri Had seen it. Didnt understand it, therefore asked myself. Sorry, will include a reference list of all the posts on the topic I didnt understand next time. $\endgroup$ – Jake Rose May 13 '18 at 16:57
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The field for one infinite plane is half of what you typed (there are two surfaces to include in gauss's law). Your thought is correct. Note that the planes must be oppositely charged in order for the charge to be $\sigma/\epsilon_0$.

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