There is a similar answer posted online but does not give all the equations. I h
ID: 1398875 • Letter: T
Question
There is a similar answer posted online but does not give all the equations. I havent taken Physics 1 for a long time so have forgotten kinetic/potential equations. Also please explain reason for signs (-/+). Please give a detailed explanation. I have a test soon.
Two large conducting parallel plates and are separated by 2.4 m. A uniform field of 1500 V/m, in the positive -direction, is produced by charges on the plates. The center plane at = 0.00 m is an equipotential surface on which = 0. An electron is projected from = 0.00 m, with an initial velocity of 1.0 × 107 m/s perpendicular to the plates in the positive -direction, as shown in the figure. What is the kinetic energy of the electron as it reaches plate ? ( = 1.60 × 10-19 C, el = 9.11 x 10-31 kg)
-2.9 × 10-16 J +2.4 × 10-16 J -3.3 × 10-16 J -2.4 × 10-16 J +3.3 × 10-16 JExplanation / Answer
Here ,
initial speed , u = 1.0 *10^7 m/s
distance , d = 1.2 m
Now , for the electron to reach the plate B ,
final kientic energy = 0.5 * m*u^2 - e*E * d
final kientic energy = 0.5 * 9.11 *10^-31 * (1 *10^7)^2 - 1.602 *10^-19 * 1.2 * 1500
final kientic energy = -2.428 *10^-16 J
as kinetic energy is coming as negative , electron will not reach plate B ,
it will collide with plate A ,
final kinetic energy = 0.5 * m*u^2 +e*E * d
final kientic energy = 0.5 * 9.11 *10^-31 * (1 *10^7)^2 + 1.602 *10^-19 * 1.2 * 1500
final kinetic energy = 3.3 *10^-16 J
the final kinetic energy of the electron when it reaches the plate is 3.3 *10^-16 J
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