Is the electric field between two positively charged parallel infinite plates one with a charge density twice the other effect the electric field on the outside of the plates? I am thinking no, because the field lines cannot cross in the field between the plates since they have positive charges. What does the electric field in the middle look like? How would I draw electric field lines to geometrically interpret this situation? If I have a Gaussian cylinder between the plates will the field go out the walls of the cylinder like in the case of two positive point charges within a Gaussian cylinder?
[Physics] Electric Field Between Two Parallel Infinite Plates of Positive Charge and a Gaussian Cylinder
electric-fieldselectrostaticsgauss-law
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Best Answer
Yes the additional positive charge to one plate increases electric field behind the other one.
To see this let's look at general Gauss's law:
$$\oint_{S}\vec{E} \,d\vec{A}=Q/\epsilon_0$$
The $S$ is the Gaussian surface which we are free to choose therefore it could include both plates even if the surface goes to infinity with both plates.
The charge Q however could be written as charge density $\sigma$ times surface area or in general:
$$Q = \int_A \sigma \, dA $$
where $\sigma$ could be $\sigma_1$, $\sigma_2$ or $\sigma_1 + \sigma_2$.
Now we are ready to use symmetry of configuration. Lets assume that field is homogeneous at the middle of of both infinities and rewrite the left side of Gauss law as:
$$\oint_{S} \vec{E}\, d\vec{A} = \oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} $$
The right side with the same division:
$$\int_A = \int_{in} +\int_{out}$$
Now we are ready for the main part:
$$\oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} =\int_{in}\sigma \, dA + \int_{out}\sigma \, dA$$
And the Gauss law says that charges which is outside from closed surface does not affect the flux or $\vec{E} = \vec{E}_{in} + \vec{E}_{out}$ and the $\oint_{out} \vec{E}_{in} \, dA = 0 $ which makes to satisfy:
$$\oint_{S_{out}} \vec{E}\, d\vec{A} = \int_{out}\sigma \, dA$$
And the only thing which is left is always satisfied:
$$\oint_{S_{in}} \vec{E}\, d\vec{A} = \int_{in}\sigma \, dA$$
which for homogeneous field means:
$E_{top} + E_{bottom} = \sigma / \epsilon_0$
where the signs for field is positive if it is directed out of surface.