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).

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reply Sep 27 "18 at 20:22

Andrew SteaneAndrew Steane