Download Uniform Electric Fields

Uniform Electric Fields
Uniform Electric Field
 We have already looked at what electric fields look like when there is a dipole. Electric fields point
in the direction that a positive charge would experience a force if it were placed at a position in the
field. From the diagram, any would be repelled by the (+) positive charge and attracted to the (–)
charge. So the lines are drawn from left to right in the diagram.
 Is the electric field uniform? NO. The lines have different density at different points. It is strongest
near the poles. I have chosen this diagram because each pole has the same charge.
 What if, like in a capacitor, you had a lot of particles with positive charge on one side lined up with
negative charge on the other? What would that field look like?
 If we took some positive charges like the diagram above and lined them up across from the same
number of equal negative charges on the other sides, (protons and electrons), superimposed it would
look something like this:
 Cleaned up a lot, you get the image of a uniform electric field BETWEEN two parallel plates.
 The electric field is uniform? Yes. The lines are equally spaced and parallel between the plates. At
the ends of the plates, you will get some bowing, like in the diagram on the first page. But in
between the plates, the field is constant.
 Don’t loose sight that we are talking about the area between the plates of a capacitor! We are not
talking about the capacitor itself.
 That means anywhere in the field, a particle will experience a constant force. Each particle with the
same charge below, would experience the same force, because the field is constant. The equation
relating the force and the field is still, F = qE. Only there are no kQq’s in these problems because we
are looking at plates, not point charges. These are charges between the plates, not in the wires.
 That means anywhere in the field, a particle will experience a constant acceleration, F = ma.
 The electric field between the parallel plates is determined by the potential difference (voltage)
between the plates and the distance they are apart. More voltage means stronger field. Closer plates
means stronger field. The equation for the uniform electric field is
.
 The uniform force experienced by a charged particle depend on the strength of the field and the
amount of charge. The equation is F = qE.
 The potential energy in a uniform electric field is given by the equation UE = qV.
 So we still have the same equations that relate F = qE, and U = qV. q is the charge of the particle
between the plate, little q. Q was on Friday’s lesson was the charge on the plates, BIG Q. Now we
have a way to relate field and force with voltage and energy, V = Ed (I rearrange the equation
above). V = Ed ONLY works with uniform fields, which is why we didn’t see it before.
Particles moving from one plate to the other.
 The image to the right shows the solid electric field lines
between a set of parallel plates. It also shows the
equipotential lines dotted between the plates. These dotted
lines are similar to places of the same height above the
Earth’s surface that would have the same potential energy.
As a particle moves from one plate to the other, they are like
rocks falling to Earth. The further they fall, the faster they
will get.
 A popular scenario is that of an “electron gun” or “particle
accelerator” as shown to the right. This electron can pass from left
to right across the field and through the hole on the other side. The
potential energy of the particle is U = qV. After it passes through
the hole, as shown below, the potential energy is converted to
kinetic energy.
qV = ½ mv2
 To determine the acceleration of the particle, you would have to use Newton’s Second Law, F = ma,
and the uniform force in the field, F = qE.
ma = qE
 If the electric field strength isn’t given, you would have to know the distance between the plates and
calculate the electric field first.
 When you look at a pair of plates, remember that the BIGGEST voltage (more positive) is the
positive plate. In this picture, 0 V > -5000 V, so the 0 V plate is (+) and the -5000 V plate is (-). The
electron is fired from negative to positive.
Particles fired between the plates.
 A particle moving between parallel plates is treated as a horizontally fired projectile which
experiences a uniform acceleration as it passes between the plates. In these problems we pretend that
gravity is turned off.
0 V is less than 230 V. So, the top plate is negative, and the bottom plate is positive.
While the electron is between the plates, it will experience a force to the bottom plate that is constant
because the electric field is constant. Since the field is constant, and the force is constant, the
acceleration of the particle downward will be constant.
This is a horizontally fired tiger jumping off an electric cliff. The acceleration is not g.
E = V/d and F =qE. So, F = qV/d. Set that equal to ma and you get
ma = qV/d
a = qV/md
By pluggin into that you can calculate the acceleration of the particle between the plates.
1. An electron, initially at rest, is accelerated through a potential difference of 600 V as shown.
a) Calculate the electric field strength.
b)
Calculate the force on the electron.
c) Determine the velocity in which the electron will strike the top plate.
2. A pair of square parallel conducting plates, having sides of length 0.05 meter, are 0.01 meter apart
and are connected to a 200-volt power supply, as shown above. An electron is moving horizontally with
a speed of 3 × 107 meters per second when it enters the region between the plates. Neglect gravitation
and the distortion of the electric field around the edges of the plates.
(a) Determine the magnitude of the electric field in the region between the plates and indicate its
direction on the figure above.
(b) Determine the magnitude and direction of the acceleration of the electron in the region between the
plates.
(c) Determine the magnitude of the vertical displacement of the electron for the time interval during
which it moves through the region between the plates.
(d) On the diagram below, sketch the path of the electron as it moves through and after it emerges from
the region between the plates. The dashed lines in the diagram have been added for reference only.