Download Charge and mass of the electron

Charge and mass of the electron
Background:
Electric Force (AKA Coulomb’s Force, Fe) is defined as:
Fe  k
Electric Field E (defined as electric force per unit of charge) is:
E
qq'
r2
Fe
1C
(Eq. 1)
(Eq. 2)
On equation (2) the denominator is “1 C”, which is a charge of 1 Coulomb. Replacing the expression for the electric
Ek
force (Eq. 1) into the electric field (Eq. 2) we get:
q'
r2
(Eq. 3)
Notice how 1C and q cancel out.
Using the electric field (Eq. 3) as our new definition for electric field, we can calculate the electric force (Eq. 1)
exerted upon any charge simply as: Fe  qE (Eq. 4)
Notice how q is the only difference between Eq. 1 and Eq. 3.
When electric charges move, they generate a magnetic field which exerts an additional force, therefore, the total force
over a charged particle is given by the expression:
FT  Fe  Fm
(Eq. 5), where Fm is the magnetic force.
The additional magnetic force (Fm) depends on the charge itself (q), its speed (v) and the magnitude and direction of
the magnetic field (B). It has been found that if Fm is perpendicular to v and to B, it can be calculated as:
Fm  q sin  (Eq. 5a)
If the vectors
and
(θ is the angle formed by vectors
and )
are also perpendicular, then θ = 90° (and sin 90° = 1). Replacing these values into Eq. 5a, we
get a simplified expression of the magnetic force given by: Fm  qB
(Eq. 6)
(charge x speed x magnetic field)
Inserting equations (4 and 5), for the electric force and magnetic forces, into (Eq. 5) for the total force acting upon an
electrical charge that moves, we see that the total electromagnetic force acting upon a charge q will be given by:
FT  qE   B  (Eq. 7) (AKA Lorentz’s Force.)
Joseph John Thomson’s experiment:
From the cathode, electrons with mass m and charge e are emitted. Some pass through the hole in the anode with a
speed v. If there is no other interaction, they would show a linear path; however, if an electric field is applied, then the
electrons would be deviated towards the opposite charge of the electric field.
A magnetic field would also exert a force upon the electrons. If the magnetic field is homogeneous, the electrons are
forced into a circular trajectory with radius R. In the circular motion, the particle feels the centripetal acceleration:
a
Reyes
2
R
(Eq. 8)
If we apply Newton’s 2nd law (which states that F = ma), the magnetic force (Eq. 6) has to be equal to the mass
multiplied by the centripetal acceleration (8): Be  ma
m 2
Be 
R
or
(Eq. 9)
Rearranging this expression (Eq. 9) we get the relation e/m for the particles found in the cathode rays as:
e


m BR
(Eq. 9a)
Unfortunately, the initial velocity v is not known, but if we apply the electric and magnetic field at the same time and
balance them, then the particles return to a straight pattern, and that is the case: Fe  Fm or Ee  Be
Using (Eq. 10), we can solve for the initial – unknown – velocity as:
above may be rewritten as:
e
E
 2
m B R
(Eq. 12).

E
B
(Eq. 10)
(Eq. 11); therefore, Eq. 9a
[This is what J.J. Thomson found!]
Since the radius R, and the electric and the magnetic fields (E and B) can be measured, if one of the two (either e or m
– the charge or mass of the electron) is measured, the other variable could be obtained. This result led to Robert
Andrews Millikan’s Experiment: Mullikan observed oil drops electrically charged and suspended in air. An electric
field would tend to make them go up, and the gravitational field would pull them down; when both forces balance
each other there is no net movement and the situation is described by:
Fe  Fg
(Eq. 13)
Since Fg = mg (the weight or force of gravity equals the mass times the acceleration of gravity), then:
Eq  mg
(Eq. 14). On this equation, m is the mass of a drop and q its charge.
Millikan found that the charge (q) of the droplets was always a multiple of:
e = 1.591 x 10-19 C
or
q = ne (Eq. 15);
(where n = 1, 2, 3…)
The value currently accepted is: e = 1.6022 x 10-19 C, and then, from (Eq. 12) m = 9.1095 x 10-28 g.
Reyes