Two large, metallic, planar, parallel, charged capacitor plates have an electric

Question

Two large, metallic, planar, parallel, charged capacitor plates have an electric potential difference of V_1 - V_2 = +2500V, where V_1 and V_2 are the electric potentials on the top and bottom plate, respectively, as shown here: An electron is shot through a small hole in the top plate, into the space between the two plates. The electron, while traveling from the top to the bottom plate, ... will gain 4.0 times 10^-16J in kinetic energy between top and bottom plate. will lose 2.0 times 10^-16J in kinetic energy between top and bottom plate. will gain 2.0 times 10^-16J in kinetic energy between top and bottom plate. will lose 8.0 times 10^-16J in kinetic energy between top and bottom plate. must have a kinetic energy of at least 4.0 times 10^-16J, as it passes through the top plate, in order to reach the bottom plate.

Explanation / Answer

Since the upper plate is at a higher potential, the electric field between the plates will be pointing vertically downwards. Which means a positive charge when placed inside this field would accelerate in the downward direction. But since we are placing an electron which has a negative charge to it, it will experience retardation. And hence will slow down. So, there needs to be a minimum Kinetic energy initially in order to even reach the bottom plate.

The change in energy when a charge is moved through a potential difference of V, is given by U = qV

So, if at all the electron hits the bottom plate, the loss in its energy will be

U = e( V1 - V2) = 1.6 x 10-19 x 2500 = 4 x 10-16 J

So, this is the minimum energy that the electron should initially have, in order to reach the bottom.

Hence option E.

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