Download PWE 16-5: Determining Charge-to

Example 16-5 Determining Charge-to-Mass Ratio
When released from rest in a uniform electric field of magnitude 1.00 * 104 N>C, a certain charged particle travels 2.00 cm
in 2.88 * 1027 s in the direction of the field. You can ignore any nonelectric forces acting on the particle. (a) What is the
charge-to-mass ratio of this particle (that is, the ratio of its charge q to its mass m)? (b) Other experiments show that the
mass of this particle is 6.64 * 10227 kg. What is the charge of this particle?
Set Up
The only force that acts on the particle is the
electric force given by Equation 16-2. Since
the particle accelerates in the direction of the
s, the force on the particle must
electric field E
s. So the charge on
also be in the direction of E
s is uniform
the particle must be positive. Since E
(it has the same value at all points), the force
on the particle will be constant. Its acceleration
ax will be constant as well, so we can use one
of the constant-acceleration equations from
Chapter 2 to determine ax. We’ll use this with
Newton’s second law and Equation 16-2 to
learn what we can about this particle.
Solve
(a) We are given that the particle travels
2.00 cm = 2.00 * 1022 m in 2.88 * 1027 s.
Use this to determine the particle’s constant
acceleration.
Relate the acceleration to the net external
(electric) force on the particle and solve for the
charge-to-mass ratio.
Electric field and electric force:
s
(16-2)
Fs = qE
Straight-line motion with
constant acceleration:
1
x = x0 + v0x t + axt 2
2
Newton’s second law:
s
s
a Fext = ma
q>0
a
E
x
(2-9)
(4-2)
Take the positive x direction to be the direction in which the particle
moves. The particle begins at rest, so v0x = 0. If we take the initial
position of the particle to be x0 = 0, then Equation 2-9 becomes
1
x = ax t 2
2
Solve for the acceleration:
212.00 * 10-2 m2
2x
= 4.82 * 1011 m>s 2
ax = 2 =
t
12.88 * 10-7 s2 2
The net force on the particle of charge q is the electric force in the
x direction, which from Newton’s second law is equal to the mass m
of the particle multiplied by the acceleration ax:
qEx = max
This says that the acceleration of a particle in an electric field depends
on the particle’s charge-to-mass ratio:
q
E
ax =
m x
In this example we know both ax and Ex, so the charge-to-mass ratio is
4.82 * 1011 m>s 2
q
ax
=
=
m
Ex
1.00 * 104 N>C
Since 1 N = 1 kg # m>s 2, this is
q
= 4.82 * 107 C>kg
m
(b) Given the charge-to-mass ratio and the
mass of the particle, determine the charge q.
Reflect
The charge of the particle is
q
q = ma b = 16.64 * 10-27 kg2 14.82 * 107 C>kg2
m
= 3.20 * 10219 C
The charge on a proton is e = 1.60 * 10219 C; the charge on this
particle is 2e.
The particle in this example has about four times the mass of a proton but only double the charge of a proton. For
historical reasons it’s known as an alpha particle; in fact, it’s the nucleus of a helium atom, which contains two protons
(each with charge e) and two neutrons (each with nearly the same mass as a proton but with zero charge).