The electric field between two parallel square metal plates is 130 N/C. The plates are 0.85 m on a side and are separated by 3.0 cm. What is the charge on each plate (assume equal and opposite)? Neglect edge effects.
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This is Giancoli Answers with Mr. Dychko. Here's a drawing of the two plates in black and they have this constant uniform electric field between them and we can pick any Gaussian surface we like to answer this question and let's choose this green dotted rectangle here. The charge within this Gaussian surface is going to be the charge on one plate and the charge that it encloses divided by permittivity of free space is gonna equal the total flux through the surface and so the flux will be the area of this side multiplied by the electric field strength. So electric fields during times area is Q enclosed over Epsilon naught and then we can multiply both sides by Epsilon naught and we get the charge enclosed as permittivity of free space times electric fields strength times area. Turning this sum into something that looks simple like this is only possible when we have convenient geometry like we have here where the electric field is perpendicular to the area and all the little bits of area can be easily calculated what we have here is we're looking at the edge of a rectangle here with field lines coming out of the page towards us here and each little bit of area that can be added up into the total area of this rectangle. And we know that the plates are 0.85 meters squared. So the area is going to be that squared. So we have permittivity of free space 8.85 times ten to the minus 12 coulombs per Newton meter squared times 130 Newtons per Coulombs electric fields strength times 0.85 meters squared which is 8.3 times ten to the minus ten coulombs. It's gonna be the charge on one of the plates and the other plate will have the same charge of opposite sign.
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