Download Lesson 17 - Motion of a Charged Particle in a Uniform Field

Motion of a Charged
Particle in a Uniform
Field
Physics 12 Adv
Charged Particle placed in an EField


When a charged
particle is placed in
an electric field, it
experiences a force
based on the field
strength and charge
Determine an
equation that solves
for acceleration of a
particle in terms of q,
m and E

 qE
a
m
Charged Particle moving in an EField


When a charged
particle is moving in
an electric field, it will
experience an
acceleration parallel
to the electric field
This requires that we
treat the motion in two
dimensions using
trigonometry
Motion in 2D
Previously we have constrained objects to
move in one dimension
 We are now going to remove that
constraint and investigate the motion of a
charged particle in a uniform electric field,
where the particle is free to move in both
the x and y direction

Charged Particle Motion with an
Initial Velocity
When we consider a charged particle
moving in an electric field, we will consider
either the x or y axis to be in the direction
of the electric field
 We can then consider the velocity using
vector components and write equations of
motion to describe the charged particle

EOMs

 
v (t )  at  v0

2



at
d (t ) 
 v0t  d 0
2



vx (t )  a x t  v0 x
 2



axt
 v0 x t  d 0 x
d x (t ) 
2



v y (t )  a y t  v0 y
 2


a yt

 v0 y t  d 0 y
d y (t ) 
2
Charged Particle moving in an EField


An electron, moving
with an initial velocity
enters an electric field
as shown in the
diagram at the right
and will follow a
parabolic path as a
result of the e-field
We can solve this
problem through the
use the 2D EOM’s
Problem

A cathode ray tube is created using a potential
difference of 5.0kV between A and B. An electron
is emitted from A and accelerated toward B where
A and B are separated by 9.5cm. After passing B,
the electron travels at a constant velocity until it
enters the electric field created by C and D. C and
D are separated by 2.5cm and the plates are
5.0cm long; what is the maximum voltage that can
be applied to the plates so that the electron does
not strike either plate.
Charged Particle placed in a BField


When a charged
particle is placed in a
magnetic field, it
experiences a force
based on the cross
product of its velocity
and the magnetic field
intensity
Therefore, a charged
particle experiences
no force if it is not
moving
Circular Motion
When a charged particle is moving in a
magnetic field, it always experiences a
force that is at right angles to the velocity
 This results in a change in the direction of
the velocity but not its magnitude
 As a result, this force will provide a
centripetal acceleration towards the centre
of the circular path

How can we calculate centripetal
acceleration?
r v

r
v
d  vt
r  vt
vt v

r
v
2
v
v

r
t
v2
ac 
r
Centripetal Force


Like the centripetal
acceleration, the
centripetal force is always
directed towards the
centre of the circle
The centripetal force can
be calculated using
Newton’s Second Law of
Motion


F  ma
2
v
ac 
r
2
mv
Fc 
r
Charged Particle moving in an BField


A charged particle,
moving with an initial
velocity enters a
magnetic field as shown
in the diagram at the
right and will follow a
circular path as a result
of the b-field
We can solve this
problem through the use
the centripetal motion
Radius of Curvature
mv 2
Fc 
r

 
F  qv  B
F  qvB
mv 2
 qvB
r
mv
r
qB
Problem

An electron is accelerated through a
potential difference of 2.5kV before
entering a uniform magnetic field of
strength 0.50T. What is the radius of
curvature of the electron?