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.

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 r_{1}, assuming effect of gravity is
negligible, the total energy is E_{T1}=
E_{E1}+ Ek_{1}

But if the particle is at rest at r_{1}, then E_{T1}=
E_{E1}, since there is no kinetic energy of a particle at rest.

At the point r_{2}, the total energy is given
by E_{T2}= E_{E2}+ Ek_{2}

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.

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

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 v_{2} of a
positively charged particle and the angle q
at which it emerges if the particle enters the region with a velocity v_{1}
parallel to the plates midway between them.

The particle will travel at a constant velocity of v_{1},
since no forces act in this direction.

The particle will remain between the plates for a total time
given by Dt = L/ v_{1}.

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

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

F_{M }= q v B sinq
,

Where q is the angle between v and B.