so considering two limitless parallel plans of opposite charge thickness let"s to speak +σ because that the left plan and -σ because that the appropriate planWhy is the electrical field calculation this way :$$ E = σ/2εo + σ/2εo = σ/εo $$I understand that between the plan the vector(E+) will allude to the ideal toward the negatively charged plan. The very same goes for vector(E-) the goes toward the negatively fee plan.

What i don"t know is why perform we not take into consideration the "-σ" worth in the equation?


*

*

This is nice trouble that plenty of a first year stayinfiji.com undergraduate will get wrong. We usually recommend utilizing Gauss"s theorem come relate the ar to the surface charge density. If you do that, then you discover that the field is $sigma/epsilon_0$ external a conductor having actually surface charge density $sigma$. That"s it. That"s the ideal answer for the capacitor. But then the college student notices the other plate and they start adding another donation .... Problems. The method to think around the combination of both plates is to argue that you have actually two airplane of charge and also each airplane of charge sends out ar lines in both directions, $sigma/2epsilon_0$ each way. In between the plates, the direction agree and add up come the full field. Anywhere else the contribute from the 2 planes that opposite fee cancel out. (I am treating big plates and also ignoring edge effects as usual).

You are watching: Electric field between two plates with different charge densities


re-publishing
mention
improve this prize
follow
reply Sep 27 "18 at 20:22
*

Andrew SteaneAndrew Steane
39.9k33 yellow badges6060 silver badges158158 bronze badges
$endgroup$
4
add a comment |
0
$egingroup$
Refer come the number below. We have the right to look in ~ the field between (and outside) the plates indigenous the perspective of Gauss’s regulation (left drawing), or from the perspective of the contributions of two charged sheets (right drawing). Recall the law states the the complete of the electrical flux out of a closed surface is equal to the fee enclosed separated by the permittivity.

Looking in ~ the top cylinder the the left figure, we check out that the just charge fastened is the optimistic charge top top the left plate. This continues to be the situation as we extend the cylinder toward the best surface. When the cylinder includes the fee on the right plate the net fee enclosed is zero thus the field is zero external the plates. You have the right to do the same thing starting from the best plate.

See more: 2700 1St Ave,Seattle, Wa 98121, 2700 1St Ave

The figure at the right supplies the method of the ar contributions of each plate which can be synthetic up to obtain the same ar as from the Gaussian surface. Note that the direction that the field vectors is constantly that of the force that a confident charge would certainly experience.