At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
It looks like you worded your question wrong but if taken literally and assuming a conducting plate (ie some sort of metal) at steady state, the entire plate will be at the same potential. The charges will not be evenly distributed on its surface. The conducting plate cannot have any electric field within it at steady state, the reason being an electric field implies a force on charges and since charges are free to move in a conductor, a field would mean the charges are still moving and are not at steady state. Because the conducting plate has no E field in it at steady state, the path integral of E dot dL between any two points inside the plate is zero and therefore the voltage of every point in the plate is the same. The charges will *not* distribute evenly on the rectangular plate's surface. This is because the curvature on the plate's surface is not constant. A rectangular plate has corners, where charges will collect in higher density. This seems odd but if you draw a corner on paper and look at the forces between two charges on the two sides near the corner, the forces they apply on each other are almost normal to the rectangular surface rather than repulsive in a direction parallel to the metal surface which can cause movement. Compare that to the forces on a flat surface between neighbour surface charges where most of the coulomb force is directed parallel to the surface.
I think maybe your question is about the existence of an E field inside a metal box because this is the kind of question most people ask after seeing that the E field inside a charged hollow metal sphere is zero, which is found using an application of Gauss' Law. Later in the lecture series you will hear Prof Lewin go on about how textbooks incorrectly use Kirchoff's voltage law when applied to circuits containing inductors. He's technically right -- the more general Faraday's Law is correctly applied -- but electrical engineers always model inductance emf as voltage drops in a circuit so that KVL can be applied. This thing where Gauss' Law is used to argue the steady state field inside a hollow metal sphere is zero I am convinced is also an error that all textbooks make and even Prof Lewin too. The real reason the field inside a hollow conducting sphere is zero is that the sphere is conducting. Suppose the field inside the hollow sphere is not zero. This means there are E field lines inside the hollow sphere. These E field lines must terminate on surface charges that can only be located on the interior of the metal sphere. But if there is an E field line that starts at one point on the metal interior and terminates on another point in the metal interior, there is a voltage difference between those two points in the metal conductor. This is not possible at steady state so at steady state there must be no charges on the interior of the hollow metal sphere and no field in the hollow either. This argument can also be applied to any shape metal conducting shell, including a rectangular box. This is known as a Faraday Cage. In the case of the charged hollow metal sphere, the argument using Gauss' Law that there is no charge on the interior surface of the sphere is flawed. The Gauss surface is drawn as a sphere that passes through the interior of the metal shell. Gauss' Law states the net charge enclosed is zero because there is no field on the Gaussian surface. Then the textbook conclusion is there is no charge on the interior surface. This is wrong, the correct conclusion is there is no net charge on the interior surface. This does not rule out an equal amount of + and - charge on the interior surface of the shell.