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When doing derivation of the electric field between a capacitor, my textbook draws a gaussian surface going through the top plate and going through the space in-between. Then, it calculates the electric field, $E_0=\frac{q}{\epsilon_0 A}$, and calls that the electric field between a capacitor (with no dielectric inserted). diagram of capacitor with gaussian surface going through top plate and space in-between, part of electric field for capacitor derivation using gauss's law

However, in my understanding the electric field counted by the flux only includes the electric field due to the charge enclosed. So $E_0$ only includes the electric field created by the charges in the top plate.

Thus, Shouldn't we have to double our answer to include the electric field created due to the bottom plate?

Hoping someone can resolve this confusion.

Qmechanic
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RS25
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2 Answers2

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If you're studying Gauss's law then you should know that in the equation $\Phi_E = q/\epsilon_0$, $q$ is the enclosed charge. Everything outside does not contribute. That is why you can ignore the bottom plate.

Put another way, the bottom plate does create an electric field, but the net electric flux it creates is zero, because some amount goes in and some amount goes out of the Gaussian surface you draw, and that adds to zero:

enter image description here

This applies regardless of how you draw the box - if you drew something elliptical for example, you'd have to do the surface integral (since flux is defined as electric field normal to the surface), but the in/out fluxes still cancel and you still get zero.

Allure
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(a) Gauss's law relates the net electric flux passing out through a closed surface to the enclosed charge. It does not attribute the field strength at any point on the surface to that charge.

(b) In your case the field between the plates is clearly due to both plates. If the bottom plate weren't there, there'd be equal amounts of flux leaving through the top and the bottom of a Gaussian surface surrounding the top plate, and the field strength would be half what it is with the bottom plate there.

(c) You can confirm these claims by integration of field strengths calculated using the inverse square law, assuming charge to be spread uniformly over plates whose linear dimensions are much greater than their separation.

Philip Wood
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