PHY C17: Solving Problems involving Uniform Electric Fields (AS)

In my previous post, I explained uniform electric field strength & derived its formulae. Now let’s look at a few applications:

  • Problems involving basic forces
  • Problems involving motion of charges
    • Utilising the conservation of energy

Problems involving forces

A. Finding the electric force acting on a charge within an electric field
This is pretty straightforward – utilise the formulae we derived:

F = Eq
or
F = Vq/d

Just find the values you need, sub them in, & you can get an answer!

Remember to check for the direction of the force – field lines show the force acting on a POSITIVE charge. If your charge is negative, be sure to flip the direction of the force.

B. Multiple forces act on a charged object
This requires a bit more cleverness.

  1. Draw a free-body diagram & draw all the forces acting on your object
  2. You can now find individual forces & find resultant forces using vector addition
  3. You could also use resolution of forces to work backwards & solve for specific values

Always be sure to see whether the forces are in equilibrium or not.
If they are, you know that the net force is 0.

An example of using this skill in real science is measuring the charge on a particle.
What if you wanted to find the charge of an object with a known mass?

  1. If you know the mass, you know the weight.
  2. If the object is charged, an electric field should exert a force on it.
  3. If you place the object in an electric field, the net force acting on it would be the sum of its weight & the electric force.
  4. Thus, you can levitate charged objects by applying an electric field in the vertical direction.

When the object remains stationary:

Felectric = Weight
qE = mg
or
Vq/d = mg

If you place a charged object of known mass between 2 plates, & vary the potential difference until it remains in place, you can now calculate the charge!

This is exactly what Robert Millikan did in 1912 to measure the electron charge.


The Electron Charge
The set-up:

Oil droplets are sprayed through a hole at the top plate, & become negatively charged through friction with the nozzle.

At the right potential difference, a negatively-charged oil drop is balanced between the plates due to a downward gravitational force & an upwards electric force.

Just as before,

Felectric = W

qE = mg
or
Vq/d = mg

Repeating the experiment, Millikan found that the values of q (the charge) were all WHOLE MULTIPLES of a specific number: 1.6 x 10-19 C.

Thus, we learnt that charge is QUANTISED: it cannot take values less than 1.6 x 10-19 & must be a whole multiple of that value.

The charge of the electron is usually labelled as ‘e’, the ELEMENTARY CHARGE.


Problems involving the motion of charges

Charged particles in uniform fields experience a CONSTANT force. Thus, they experience a CONSTANT acceleration in the parallel to the field (the direction depends on the sign of the charge).

This is similar to masses travelling in a uniform gravitational field (which we have seen through projectile motion). For simplicity, let’s start by considering objects which only move in the direction of the field lines:

Objects moving parallel to the electric field
If a charged object is at rest OR is moving parallel to the electric field, it will experience a constant acceleration in the direction of the field. It will not travel perpendicular to the field.

There are a few calculations you can do in this situation:

  • Using the equations of linearly accelerated motion (AKA the “suvat” equations)
  • Using conservation of energy

Using the suvat equations is identical to what we’ve learnt before, so let’s look at the conservation of energy.

You have used the conservation of energy when solving problems about falling objects due to gravity.

If the objects begins at rest, gravitational potential energy is converted into kinetic energy in the fall:

Change in GPE = Change in KE

mgΔh = ½ mv2

In this topic, we have a different form of energy: ELECTRICAL POTENTIAL ENERGY.

As a charge travels parallel to the electric field between 2 points, it experiences a CHANGE in potential energy. It travels from a high potential to a low potential, gaining kinetic energy: It accelerates!

Change in Electric PE = Change in KE

This change in electric potential energy is the same value as the work done to travel that distance in an electric field, remember:

W = Vq

Therefore,

Vq = ½ mv2

To summarise:
V = potential difference between 2 points
q = charge of object
Vq = difference in potential energy as object travels between 2 points in an electric field

This is equal to the gain in kinetic energy it experiences, so
Vq = ½ mv2

If the object travels from one plate to another, then the value of V is the potential difference between the 2 plates, AKA the VOLTAGE between the plates.


Charged objects entering a uniform electric field perpendicular to the field at a constant velocity
The object maintains its velocity perpendicular to the field, but accelerates parallel to the field.

Consider this example, we will say the electric field is vertical:

The positive charge maintains its horizontal velocity but accelerates in the vertical direction. Its trajectory follows a parabola, & is a form of projectile motion.

A negative charge would experience an acceleration in the opposite direction:

Things to remember when problem-solving:

  1. Check if you need to find the RESULTANT velocity or just the velocity in a certain direction. Although the charges above maintain their horizontal component of velocity, the resultant velocity still increases. Use vector addition (Pythagoras’ Theorem or trig) to calculate the resultant velocity!
  2. Check the sign of the charge, which affects the direction of force. Negative charges oppose the field arrows!

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