Charge and mass of the electron

fiftysixpowersElectronics - Devices

Oct 18, 2013 (3 years and 7 months ago)

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Reyes


Charge and mass of the electron

Background:

Electric Force

(AKA Coulomb’s Force,
F
e
) is defined as:




(1)

Electric Field
E

(defined as electric force per unit of charge) is:


(2)

1
C

is 1 C
oulomb

Replacing the expression for the
electric force (1)

in to the

electric field (2)

we get:


(3)

Using the electric field (3), we can calculate the electric force (1) exerted upon any charge as:

(4)

Wh
en electric charges move, they generate a magnetic field which exerts an additional force, therefore, the total force
over a charged particle is:


(5)

The additional magnetic force depends on the charge itself (
q
), its speed (
v
)
and the magnitude and direction of the
magnetic field (
B
). It has been found that
F
m

is perpendicular to
v

and to
B
,

and it can be calculated as:




(5)


(
θ

is the angle formed by the vectors

and
)

If

and

are perpendicular (
θ

= 90°
), then

the magnetic force
:


(6)


(
charge
x

speed
x

magnetic field
)


Replacing the equation (4
)

for

electric force,

and (6)

for magnetic force

on equation (
5
)

for the total force
, we see that
the electromagnetic force acting upon a charge
q

will be given by:



(7) (AKA Lorentz’s Force.)


Joseph John Thomson’s experi
ment:

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. If an electric field is applied, then the electrons
would be de
viated towards the opposite charge of the electric field.


A magnetic field would also exert a force on the electrons. If the magnetic field is homogeneous, the electrons are
forced into a circular trajectory with radius R. In the circular motion, the part
icle feels the centripetal acceleration:


(8)

If we apply Newton’s 2
nd

law

(
F

=
ma
)
, the magnetic force (6) has to be equal to the mass, multiplied by the
centripetal
acceleration (8):



(9)

Rearranging this
expression (9) we get the relation
e/m

for the particles found in the cathode rays:


(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:


or

(10)

From which:




(11);
t
herefore
, eq 9a may be rewritten as
:


(12).

Since the radius
R
, the electric and magnetic fields (
E

and
B)

can be measured, measuring
e

or
m

the other one could
be obtained. This led to
Robert Andrews Millikan’s Experiment
: Mullikan observed oil drops electrically charged
and suspende
d on air. An electric field would tend to make the
m

go up, and the gravitational field would pull them
down; when both forces balance each other:



(13)

Since
F
g

= m
g

(weight equals the mass times the acceleration

of gravity
), then
:


(14)
;
where
m

is the
mass of a drop and
q

its charge. Millikan found that the charge of the droplets

q

was always a multiple of:


e

= 1.591 x 10
-
19

C


or


q = ne

(
15
)
;


(where
n

= 1, 2, 3…
)


The value currently accepted is:
e

= 1.6022 x 10
-
19

C, and then
, from (12)

m =

9.1095 x
10
-
28

g.