In the figure below, a uniform, upward electric field of magnitude 2.00 x 103 N/
ID: 2047004 • Letter: I
Question
In the figure below, a uniform, upward electric field of magnitude 2.00 x 103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 10.9 cm and separation d = 2.72 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity 0 of the electron makes an angle theta = 59 degree with the lower plate and has a magnitude of 5.86 106 m/s. How far horizontally from the left edge will the electron strike? cmExplanation / Answer
The magnitude of the upward electric field is E = 1.70 * 103 N/C The length of the plates is L = 12.0 cm = 12.0 * 10-2 m The separation between the plates is d = 1.80 cm = 1.80 * 10-2 m The electron is shot at an angle ? = 45.0o from the left edge of the lower plate with a velocity u = 7.00 * 106 m/s (a)The force acting on the electron is F = E * q or m * a = E * q or a = (E * q/m) where q = 1.6 * 10-19 C and m = 9.1 * 10-31 kg The maximum height reached by the electron is H = (u2 * sin2?/2a) When the maximum height reached by the electron is greater than the separation between the plates then the electron strikes the upper plate. (c)The horizontal distance covered by the electron from the left edge is S = ut + (1/2)at2 where t = (u * sin?/a)