Two large conducting parallel plates A and B are separated by 2.4 m. A uniform f

Question

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on the plates. The center plane at x = 0.0 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m, with an initial kinetic energy K = 300 eV, in the positive x-direction as shown.


In Fig. 23.4, the x-coordinate, at which the electric potential is V = +600 V, is closest to:

-.4m

+.6m

-.6m

+.4m

+.2m

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on the plates. The center plane at x = 0.0 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m, with an initial kinetic energy K = 300 eV, in the positive x-direction as shown. In Fig. 23.4, the x-coordinate, at which the electric potential is V = +600 V, is closest to: -.4m +.6m -.6m +.4m +.2m

Explanation / Answer

if the electric field is pointing in the positive x direction, it means that the side at -1.2m has the higher potential. now we can find that voltage by multiply E by the distance. this is 1500V/m * 1.2m (only dealing with the left half, since x=0 has 0 V) which equals 1800 V since the field is uniform, we can set up a ratio 600V/1800V = x/1.2m x = .333m away from ground. since ground is located at x = 0, 600V is at x=-.333m, which is closest to -.4m

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