Introductory experience:
Take two pieces of Scotch tape, each about 3 inches long, and stick them together, with the sticky side of one piece in contact with the non-sticky side of the other Rip the two pieces apart. Hold one piece in each hand and bring them close to one another. You will see clearly that the two pieces attract.
Stick these two pieces at two different places, not too close to one another, on the side of a non-metallic tabletop so that the lower ends of both pieces hang freely.
Stick two more pieces of tape together in the same way and rip them apart. Take one piece of tape and bring it near each of the hanging pieces, in turn. You will see that the new piece attracts one of the hanging pieces and repels the other. The other new piece will also attract one of the hanging pieces and repel the other, but its effects on each of the hanging pieces will be the opposite of the first new piece.
The attractive and repulsive forces are due to positive and negative electric charges distributed on each piece of tape. Each of these charges is nearly approximated by a point charge, but each piece of tape has a large number of point charges distributed over it, and the interaction between the many charges on one piece and the many charges on the other is complicated. However the interaction between a single point charge and another is fairly simple.
We begin the study of electricity with the interaction of two point charges. This interaction is governed by Coulomb's Law.
In general if we have point charges q1 and q2 separated by distance r, the charges exert equal and opposite forces on one another, and the forces have magnitude
Note that | F | stands for the magnitude of the force.
The above statement, in addition to the statements below which define the directions of the forces, are known as Coulomb's Law.
This is analogous to the gravitational force F = G m1 m2 / r^2 between two masses, with which you should be familiar from first-semester physics. However, in the case of masses the force is always attractive, whereas with charges the force is attractive if the charges are like (i.e., of the same sign) and repulsive if unlike (i.e., of opposite signs).
The two charges lie on a straight line. The forces they exert on one another are equal and opposite and both forces are directed along this line. Being equal and opposite the two forces are at 180 degrees with one another.
Vector statement of Coulomb's Law: Physics 202 should note, and Physics 232 students should understand and be able to use the following. Physics 122 students will not be required to use this notation, but might find it of interest:
Using vector notation, if the charges q1 and q2 are located at respective points (x1, y1) and (x2, y2), the magnitude of the force will be
- | F | = k | q1 q2 | / r^2 = k | q1 q2 | / | r12 |^2
where r12 is the vector displacement | r12 | = < (x2 - x1), (y2 - y1) > from charge 1 to charge 2.
The direction of the force exerted on charge 2 by charge 1 will be parallel to the vector r12, in the direction of this vector when the charges are like and opposite when unlike.
The vector r12 has direction defined by the unit vector u12 = r12 / | r12 | (this is a vector of magnitude 1, directed from q1 toward q2). The force exerted on charge 2 by charge 1 is thus the product of the magnitude of the force, the unit vector u12, and if the charges are unlike the factor -1 (this factor will reverse the direction of the vector u12, so that the vector is directed toward q1 from q2, corresponding to an attractive force):
- F12 = ( k q1 q2 / | r12 | ^ 2 ) * u12 = ( k q1 q2 / | r12 | ^ 2 ) * r12 / | r12 |.
The x component of this force is
- F12x = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ] * [ (x2 - x1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ],
and the y component is
- F12x = k q1 q2 / [ (x2 - x1)^2 + (y2 - y1)^2 ] * [ (x2 - x1) / `sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) ].
If the charges reside on masses, then these masses can be accelerated by the resulting forces according to Newton's Second Law (a = F_net / m or, equivalently, F_net = m a).
If charges move according to the electrostatic forces they exert on one another, then work is done according to the definition of work (work = force * displacement in the direction of the force, sometimes expressed in over-simplified form as work = force * distance) and the work-energy theorem applies (work by nonconservative forces = change in PE + change in KE, or `dW_NC_on = `dPE + `dKE).
Before you need to use the ideas of acceleration, work and energy you will be directed to some review materials to refresh your memory.
Introductory Problem Set 1 problems 1-5 explore the application of Coulomb's Law. Subsequent text assignments will amplify these ideas.
Electric Field
The electric field at a point is the force per unit test charge, with the test charge located at the point:
To find the electric field at point (x2, y2) due to a charge Q at point (x1, y1), we first find the force on a 'test charge' q at (x2, y2):
- F12x = F12 (x2 - x1) / r and
- F12y = `F12 (y2 - y1) / r,
- where r = `sqrt [ (x2 - x1)^2 + (y2 - y1)^2 ].
The electric field at (x2, y2) is the force per unit of test charge at that point: E = F12 / q. The magnitude of the electric field is therefore
Physics 202 and particularly Physics 232 students should understand that having found the components of the force F12 we find the components E_x and E_y of the electric field:
- E_x
= k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (x2 - x1) / `sqrt( (x2-x1)^2 + (y2 - y1)^2)
= k Q q * (x2 - x1) / ( (x2-x1)^2 + (y2 - y1)^2) ^ (3/2)
and
- E_y
= k Q q / ( (x2-x1)^2 + (y2 - y1)^2) * (y2 - y1) / `sqrt( (x2-x1)^2 + (y2 - y1)^2)
= k Q q * (y2 - y1) / ( (x2-x1)^2 + (y2 - y1)^2) ^ (3/2).
Work and Energy in electrostatics
When a charge Q, in Coulombs, is subject to an electric field E, in N/C, it experiences a force of magnitude | F | = | Q * E | Newtons. The direction of the force is in or opposite to the direction of the field, depending on whether the product Q * E is positive or negative. If the charge moves through a distance `ds parallel to the force, then the work done by this field force on the charge has magnitude
This work is positive if, as in the present case, the force is parallel to the displacement and negative if the force is in the opposite direction.
The work required to move the charge is equal in magnitude but of opposite sign to the work done on the charge by the field. So the work required to move the charge has magnitude
This work is negative if, as in the present case, the force exerted by the field is parallel to the displacement and positive if the force is in the opposite direction.
When a charge Q moves from a point at potential V1 to a point at potential V2, it passes through a potential difference `dV = V2 - V1. The potential difference is measured in volts = Joules / Coulomb; the charge is measured in Coulombs.
If a charge Q (measured in Coulombs) moves from one point to another, requiring work `dW (in volts, or Joules / Coulomb), then the potential difference between the points is `dV = `dW / Q (the number of Joules divided by the number of Coulombs).
The average strength of an electric field is measured as the average force per unit of charge.
If a charge Q moves between two points and `dW Joules of work are done as the charge moves through a distance `ds, then we easily determine the average force F = `dW / `ds on the charge. The electric field is the force per unit charge (measured in Newtons/Coulomb), or in this case
Note that since `dW / Q = `dV, the potential difference, we have
This shows that the electric field can also be expressed in units of volts per unit distance (i.e., volts / meter).
As found in the preceding problem, the force on a charge of q Coulombs in a potential gradient dV/dx is (q Coulombs) * dV/dx.
The force per Coulomb is found by dividing by the number q of Coulombs, which gives force per Coulomb dV / dx.
Thus dV / dx, which measures volts per meter, is also a measure of Newtons per Coulomb.
Since a volt is a Joule per Coulomb, a volt per meter turns out to be identical to a Newton per Coulomb.