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 = 0.00 m is an equipotential surface on which V = 0.An electron is projected from x = 0.00 m, with an initial velocity of 1.0 × 10⁷ m/s perpendicular to the plates in the positive x-direction,as shown in the figure.What is the kinetic energy of the electron as it reaches plate A? (e = 1.60 × 10^-19 C,m_el= 9.11 × 10^-31 kg) A) +2.4 × 10^-16 J B) +3.3 × 10^-16 J C) -2.4 × 10^-16 J D) -2.9 × 10^-16 J E) -3.3 × 10^-16 J

Question

Quizplus · Accepted Answer

The potential difference between the two plates can be found using the electric field and plate separation:
ΔV = EΔx = (1500 V/m)(2.4 m) = 3600 V
The electron moves from a potential of 0 V to a potential of 3600 V, so its change in potential energy is:
ΔU = qΔV = (1.60×10^-19 C)(3600 V) = 5.76×10^-16 J
Since no work is done on the electron by external forces, the change in potential energy equals the kinetic energy gained by the electron:
K = ΔU = 5.76×10^-16 J
Note that the answer is positive because the electron is gaining kinetic energy. Choice B is the closest to this value.

Two Large Conducting Parallel Plates a and B Are Separated