Ok so the answer to this question is D, if someone could please just help me und

Question

Ok so the answer to this question is D, if someone could please just help me understand why using electric potential energy and kinetic enery that would be great... the equation we use in class is (delta)U=q(delta)V

6. Two large, metallic, planar, parallel, charged capacitor plates have an electric potential difference of Vi 2500V, where Vi and V2 are the electric potentials on the top and bottom plate, respectively, as shown here 1 Plate 1 Fig. 2.37 Plate 2 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, A. will lose 2.0 x 10 16 in kinetic energy between top and bottom plate. B. will lose 8.0 x 16 in kinetic energy between top and bottom plate. C. will gain 2.0 x 10 16 in kinetic energy between top and bottom plate. D. must have a kinetic energy of at least 4.0 x 10 16 as it passes through the top plate, in order to reach the bottom plate. E. will gain 4.0 x 10 16 kinetic between top and bottom plate. in energy

Explanation / Answer

as we know U = qdelta V= -1.6×10^-19 × (V2-V1) = 4×10^-16J

so this much electric potential energy change has to be there. as electric potential energy is increasing for electron as sign of U is positive so some other energy must decrease so that total energy remains constant. hence that why kinetic energy must decrase amount equal to 4×10^-16J. thats why option d is correct

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