- #1
JWHooper
Homework Statement
An electron has a mass of 9.11 x 10^(-31) kg. Its charge is -1.6 x 10^(-19) C. Suppose it is given a initial horizontal velocity of 8.0 x 10^7 m/s. It then travels through a vertical (upward pointing) electric field with a constant strength of 2.0 x 10^4 N/C. In a time T it has a horizontal displacement of 10 cm.
(a) How much force acts on the electron? In which direction is the force?
(b) What is the acceleration of the electron?
(c) What will ber the horizontal velocity of the electron at the end of time T?
(d) What will be the vertical velocity of the electron at the end of time T?
(e) At the end of time t, what angle will the electron's velocity make with its initial velocity?
(f) What will be the vertical displacement of the electron during time T?
Homework Equations
E = F_q / q = kQ / d^2
F = kq_1q_2 / d^2
x(t) = v_0t + 1/2at^2
The Attempt at a Solution
(a) Since E = F_q / q, this can arrange to F_q = Eq = (2.0 x 10^4 N/C, upward)(-1.6 x 10^(-19)) = 3.2 x 10^(-15) N, downward.
(b) Since v(t) = 8.0 x 10^7 m/s, v'(t) = a(t) = 0 m/s^2.
(c) Since x'(t) = v(t) = v_0 + at, v(T) = 8.0 x 10^7 m/s, right + (0 m/s^2)T = 8.0 x 10^7 m/s, right.
(d) Since no velocity is given in the problem, 0 m/s.
The last two questions are that I don't get. Please help!
Thanks,
J.