Question

An electron is initially at ground level, and the electric potential at that point is assumed to be exactly 0. The electron is immersed in a uniform electric field that points down with a magnitude of 2.7x10^ -11 N/C as well as Earth's graviational field (9.8 N/kg pointing down). Only gravity and the electric force are relevant.

a) If the electron is given an initial upward speed of 30 m/s, what is the maximum height above the ground that it will reach?

b) What is the change in electric potential energy during this motion (final - initial)?

c) How much work is done by the electric force during this motion?

d) What is the voltage at the electron's maximum height?

Answer 1

a)

q = magnitude of charge on electron = 1.6 x 10^-19

E = electric field = 2.7 x 10^-11

m = mass of electron = 9.1 x 10^-31

force equation is given as

qE - mg = ma

(1.6 x 10^-19) (2.7 x 10^-11) - (9.1 x 10^-31) (9.8) = (9.1 x 10^-31) a

a = - 5.1 m/s²

V_i = initial velocity = 30 m/s

V_f = final velocity = 0 m/s

h = maximum height gained

using the equation

V_f² = V_i² + 2 a h

0² = 30² + 2 (- 5.1) h

h = 88.2 m

b)

change in electric potential energy is given as

U = q E h = (- 1.6 x 10^-19) (2.7 x 10^-11) (88.2) = - 3.8 x 10^-28 J

c)

work done by electric force = U = - 3.8 x 10^-28 J

d)

V = E h = (2.7 x 10^-11) (88.2) = 2.4 x 10^-9 Volts