Motion of a Charged Particle in an Electric Field
Download
Report
Transcript Motion of a Charged Particle in an Electric Field
Motion of a Charged Particle
in an Electric Field
Physics 12
Joke of the day:
Review:
kq
E
d
Review:
1.
2.
3.
If you have a charge of 1.0C, how many electrons
have been removed from the object?
What voltage is created if a charged particle is
placed 2.5cm from a -1.0mC charge?
What electric field is created between two plates,
separated by 5.0cm with a voltage of 250V?
6.25x1018
-360 000 000V
5 000N/C
Answers:
1.
2.
3.
6.25x1018
-360 000 000V
5 000N/C
Electron in a uniform electric field:
1.
2.
Draw a uniform electric field (contains a positive and
negative plate) where the field is directed into the ground
Draw a free body diagram for an electron placed in a
uniform electric field where the field is directed into the
ground
Fe
Fg
Consider:
What happens when a charged particle is
placed (with zero initial velocity) inside a
uniform electric field?
Draw a picture to show an electron placed at
the negative plate that is creating a uniform
field
Fe
What if we wanted to find the velocity of the electron?
A derivation…..using formulas you know!
V=Ed , E =Fonq , F=ma , Vf2 = Vo2 +2ad
q
Reminder: me= 9.1x10-31kg , qe= 1.6x10-19C
Example #1:
If an field is created using 250V with plates separated
by 1.0cm, determine how fast the electron is moving
when it gets to the positive plate (ignore gravity)?
Problem #1:
You’ve been hired by Stanford to
determine the voltage required to
accelerate an electron or positron to
0.95c
a.
b.
What voltage is required if the accelerator is
3.2km long?
Find the electric field required in order to
accomplish this final velocity for one of the
particles.
Answers: 2.3x105V , 72 N/C
Crooke’s Tube:
Invented by Sir William Crookes (1875) consisting essentially of a
sealed glass tube from which nearly all the air has been removed
and through the walls of which are passed two electrodes.
When a high voltage is applied between the two electrodes,
electrons are emitted from the cathode and are accelerated
toward the anode.
Many of these electrons (aka cathode rays), miss the anode and
strike instead the glass wall of the tube, causing it to exhibit
fluorescence.
Review:
Loss of potential energy is equal to a
gain in kinetic energy
-ΔPe = Δke
Electric potential (V) is a type of potential
energy!
We can use this to calculate the velocity
of particles….
Ke = ½ mv2
Problem #2:
a.
How fast would an electron be moving
in Crooke’s tube if:
b.
V = 5000V
d = 0.025m
How much kinetic energy would the
electrons have?
Answers: 4.2x107m/s , 8.0 x10-16J
LHC
Interested??
Large Hadron Collider
http://www.youtube.com/watch?v=TIeY7
Zj27IM
Problem #3:
The first stage of the LHC accelerates
protons to .75c determine:
a.
b.
The voltage required if the electric field
exists over 1.5km
The electric field required
Answers: 1.4x105V , 96 N/C
To do:
In class Assignment