Two large conducting parallel plates A and B are separated by 2.4 m. A uniform f
ID: 1425225 • Letter: T
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 = 0.00 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 x 10-31 kg)
+3.3 × 10-16 J -2.9 × 10-16 J +2.4 × 10-16 J -3.3 × 10-16 J -2.4 × 10-16 JExplanation / Answer
+2.4*10^-16 J
Use Work-energy theorem,
Workdone on electron = change in kinetic enetgy
q*delta_V = KEf - KEi
q*E*d = KEf - KEi
==> KEf = KEi + q*E*d
= 0.5*m*vi^2 + q*E*d
= 0.5*9.1*10^-31*(1*10^7)^2 + (1.6*10^-19)*1500*1.2
= 2.4*10^-16 J
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