The Motion of Charged particles in Electric & Magnetic Fields

 

We will now take a look at how charged particles move under the influence of electric and magnetic forces.  The basic ideas discussed below have many applications for useful devices & instruments that are used in a number of different ways throughout our society today. 

 

After reading through the basic ideas that follow complete the STSE fields assignment.

 

The Electron Volt

Recall the following:  V = EE/q          1 V = 1 J/C     &    1 C = 6.25 x 1018e 

 

A convenient unit called the “electron volt (eV)”can be used to express energies of particles.  How many Joules is one electron volt?   

 

1 V = 1 J/C = 1 J/ 6.25 x 1018e  =1.6 x 10-19J/e   so,

 

1eV = 1.6 x 10-19J 

 

A) Charges in a non-uniform electric field

 


 


For a small positive test charge in the field of a fixed larger positive charge, it can be shown using the conservation of energy that the electric potential energy that the test charge loses is equal to its gain in kinetic energy.

 

At the point r1, assuming effect of gravity is negligible, the total energy is  ET1= EE1+ Ek1

But if the particle is at rest at r1, then ET1= EE1, since there is no kinetic energy of a particle at rest.

 

At the point r2, the total energy is given by  ET2= EE2+ Ek2

 

Now the conservation of energy requires that DE = 0, which gives


 

 

 


Thus, we see that the gain in kinetic energy of the test charge is equal to its loss of electric potential energy.

 

 

 

 

B) Charges in a uniform electric field (A Charged Particle Gun)

 

 


           


 


In the uniform electric field between two parallel plates a positively charged particle will accelerate constantly toward the negative plate since a constant force is acting on the charged positive particle.  If there is a hole in the negative plate, then the particles will escape with the velocity it has as it reaches the negative plate. The particles act as a beam of particles or a charged particle gun.

 

What speed does a charge q have just before it hits the negative plate, and escapes through the hole?

 

The gain in kinetic energy will be equal to the loss in potential energy, so that


 

Note that the particles velocity depends on the charge of the object, which means that different charges with the same mass will accelerate at different rates in a constant electric field (unlike masses in a constant gravitational field).

 

If starting from rest at the positive end of the plate, a particles change in velocity after time Dt will be

 


 


After time Dt, the charge will have also moved through a displacement (starting from rest) of

 


 

 


C) Deflecting Charged Particles

 


 


If charged particles enter the space between oppositely charged parallel plates with an initial velocity that is parallel to the plates, the charged particles will be deflected toward either plate depending on whether the charge is positive or negative.  If the particles do not collide with the plates, they will emerge from the opposite end of the plates with a final velocity at some angle to the plates.  The particles will then continue to travel along a straight line path unless acted on by some external unbalanced force.

 

Calculate the final deflected velocity v2 of a positively charged particle and the angle q at which it emerges if the particle enters the region with a velocity v1 parallel to the plates midway between them.

 

Horizontal Direction

The particle will travel at a constant velocity of v1, since no forces act in this direction.

The particle will remain between the plates for a total time given by Dt = L/ v1.

 

Vertical Direction

During the time Dt, the particle experiences a force due to the plates perpendicular to its initial velocity so that it accelerates in the perpendicular direction according to

 


Alternatively, by considering the work done to move the charge in the perpendicular direction we can determine the force in terms of the electric potential difference DV (a more easily measurable quantity).

 


 


This suggests that the particle will accelerate in the perpendicular direction with a rate of

 


 


The charge will then have a velocity in the perpendicular direction just before emerging of

 


 


 Resultant final velocity

The magnitude of the final velocity will be

 


and the angle it will emerge from with respect to the horizontal will be given by

 


 

 


D) Forces on Charged Particles in Magnetic Fields

 


 

 


The magnitude of the force exerted on a charged particle is given by

 

FM = q v B sinq ,

 

Where q is the angle between v and B.