Electric potential between two parallel plates

[hype_chicky]

5. Two parallel plates labeled W and X are separated by 5.2 cm. The electric potential between the plates is 150V. An electron starts from rest at time tW and reaches plate X at time tX. The electron continues through the opening and reaches point P at time tp (remember e = - 1.6 X 10^-19 C and the mass of an electron is 9.1 X 10^-31 kg)

a.) Sketch the speed-time graph on the axes below
b.) Determine the kinetic energy of the electron as it arrives at plate X

 

[hype_chicky]

FE = Ek
Q delta V/r = 1/3 mv^2

1.6 x 10^-19 C (150V) / 0.05a = ½ (9.1X10^-31 kg) (v^2)
V= 2.9 X 10^-23 m/s

 

[wais]

a.) The speed-time graph would show an initial point at tW with a speed of 0 m/s, followed by a straight line with a positive slope representing the acceleration of the electron due to the electric field between the plates. At tX, the line would reach a maximum speed and then continue at a constant speed until reaching point P at tp.

b.) Using the equation for kinetic energy, KE = 1/2mv^2, we can calculate the kinetic energy of the electron as it arrives at plate X. The mass of an electron is 9.1 X 10^-31 kg, and the maximum speed it reaches is found by using the equation for acceleration, a = qE/m, where q is the charge of an electron (-1.6 X 10^-19 C) and E is the electric field between the plates (150V/0.052m = 2884.62 N/C). Plugging these values into the equation, we get a maximum speed of 2.03 X 10^6 m/s. Plugging this value into the equation for kinetic energy, we get KE = 1/2(9.1 X 10^-31 kg)(2.03 X 10^6 m/s)^2 = 1.86 X 10^-16 J. This is the kinetic energy of the electron as it arrives at plate X.