Electricity

(3)

where k is a constant of proportionality. In MKS units, k is given by

This is frequently written in terms of another constant, the permittivity constant, e₀ (e₀= 8.8542 x 10^-12 C²/N­m²)
 
 

.

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    Notice that this equation has the same form as that for gravity, once we replace  by , supporting our previous statement that m can be viewed as a charge for gravity. Since it has an inverse square distance relationship, we see that this is a long range force, i.e. it falls off to zero only at infinity. Also note that this is a vector force law. For multiple charges, the total force on one charge is the vector sum of the force due to each one individually.

Example:
    Three positive charges lie on the x-axis; q₁ = 25 mC is at the origin, q₂ = 10 mC is at x = 2 m, and q₃ = 20 mC is a x = 3 m. Find the resultant force on q₃.

The force on q₃ due to q₂, which is 1 m away, is in the positive x direction and has a magnitude of

The force on q₃ due to q₁, which is 3 m away, is also in the positive x direction. Its magnitude is

Since both forces are in the positive direction, the resultant is also in that direction and just the sum of the magnitudes:

Example:
    Two electrons are located 1 m apart. What is the ratio of the magnitudes of their electrical and gravitational forces?

    The ratio of the electrical and gravitational force can be written as

Notice that this result is independent of distance.

Example:
    As an example of the power in the Maxwell equations and the Lorentz force law, we derive the Coulomb force law. The Coulomb force law is valid in the absence of any magnetic phenomenon, so we can set B = 0 in (2). Similarly, we will see that we can define the electric field of a single charge as . Combining these results in (2), we get

But, this is just Coloumb's force law!

    For now we will be considering electrostatics; that is, when all of the charges in the region are not moving and the electrical field is constant throughout time. In this case, there is no magnetic field (B = 0).

Electric Fields

This defines the force on charge 1 by charge 2. We can rewrite this as

Thus we see that the field due to charge 2 is given by
 
 
 

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Since the force involved is electrical, this is called the electric field.
    How does this approach differ from the traditional concept of action at a distance that we have been using up till now? The most basic difference is that the concept of interactions via action at a distance implies that, as one particle moves, the change in the force acting on another particle is felt immediately (to see this, notice the absence of a velocity factor or time dependence in all of the above equations). But, we know from experiment that nothing can travel than the speed of light, c. Therefore, we are forced to accept the notion of fields. One could argue that the absence of a velocity or time dependence in the definition of the fields yields the same problem, except that now we have particle 2 acting instantaneously on the field. The change in the field can occur at the speed of light (since so far, we have not investigated the properties of the field itself), and then the change in the force at particle 1 also occurs instantaneously. Note that the two instantaneous changes are now allowed since the field is "in contact" with the particle.
    What we have defined above is the electric field for a point particle. In order to generalize this to any situation, we define the field at a point occupied by a test charge q₀ to be
 
 
 

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where F is the force that acts upon the test charge by all the other charges in the region of interest. Note that this definition is a completely general one. We can apply it to gravitation as easily as electromagnetism by taking the particle's mass, m, as the test charge. In the case of a static electric field (which is all that we are considering currently), the field is the electric field and is denoted by ƒ = E. In MKS units, the electric field has units of Newtons/Coulomb (N/C).

Example:
    a) When a 5 nC test charge is placed at a point, it experiences a force of 2x10^-4 N in the x-direction. What is the electric field E at that point?

    From the definition, the electric field is

where i is the unit vector in the x direction.
    b) What is the force on an electron placed at the point where the electric field is 4x10⁴ i N/C?

    Since the charge on an electron is -e = -1.6 x 10^-19 C, the force is

In most situations, we do not know the force involved. Instead, we are given a system of charges and distances between them. In order to calculate the electrical field in these situations, we must go back to the field of a point charge

Since the electrical field is a vector quantity, whenever we have more than one charge we can find the resultant field by adding the individual fields vectorially
 
 

 
Example:
    What is the total electrical field of a dipole at a distance r along the perpendicular bisector?

A dipole consists of two charges separated by a distance, d = 2a. The total electrical field is given by

where

and similarly for E₂. Writing this in component form and adding, we get

where
 
 
 

So we have that

For large distances (r >> a) this reduces to

where j is the unit normal in the y direction. The product 2qa is defined to be the electric dipole moment and is denoted by the symbol, p.

Lines of Force

1. The number of lines leaving a positive point charge or entering a negative charge is proportional to the charge.
2. The lines are drawn symmetrically leaving or entering a point charge.
3. Lines begin or end only on charges.
4. The density of lines (number per unit are perpendicular to the lines) is proportional to the field.
5. No two field lines can cross.

Example:
    Sketch the lines of force for a point charge.

Example:
    Sketch the lines of force for two equal positive point charges, q, separated by a distance a.

Example:
    Sketch the lines of force for two charges of equal magnitude and opposite signs separated by a distance a.

Motion in a Field

or
 
 
 

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Example:
    An electron is projected into a uniform electric field who strength is E = 2000 N/C with an initial velocity of v₀ = 10⁶ m/sec perpendicular to the field. By how much is the electron deflected after it has traveled 1 cm?

    It takes the electron a time

to travel a distance of 1 cm perpendicular to the field. In this time it is deflected a distance antiparallel to the field given by

    It is possible to have rotational motion as well as linear motion in an external field. However, the rotational motion requires the existence of at least a dipole rather than a single charge (this is easy to see since there can be no torque on a single point). Consider a dipole of moment p (recall that p = 2qa) placed at an angle q₀ to a field, ƒ. We know the force on each charge in the dipole is given by F = qƒ. Therefore, by definition, the torque of each charge about the center of the dipole is given by
 
 

where a is the vector pointing from one charge to the other. This torque acts to align the dipole along the direction of the field.