7-5-00

Sections 16.5 - 16.6

The force exerted by one charge q on another charge Q is given by Coulomb's law:

r is the distance between the charges.

Remember that force is a vector, so when more than one charge exerts a force on another charge, the net force on that charge is the vector sum of the individual forces. Remember, too, that charges of the same sign exert repulsive forces on one another, while charges of opposite sign attract.

Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 x 10^{-6} C. The charges in the other two corners are -3.0 x 10^{-6} C. What is the net force exerted on the charge in the top right corner by the other three charges?

To solve any problem like this, the simplest thing to do is to draw a good diagram showing the forces acting on the charge. You should also let your diagram handle your signs for you. Force is a vector, and any time you have a minus sign associated with a vector all it does is tell you about the direction of the vector. If you have the arrows giving you the direction on your diagram, you can just drop any signs that come out of the equation for Coulomb's law.

Consider the forces exerted on the charge in the top right by the other three:

You have to be very careful to add these forces as vectors to get the net force. In this problem we can take advantage of the symmetry, and combine the forces from charges 2 and 4 into a force along the diagonal (opposite to the force from charge 3) of magnitude 183.1 N. When this is combined with the 64.7 N force in the opposite direction, the result is a net force of 118 N pointing along the diagonal of the square.

The symmetry here makes things a little easier. If it wasn't so symmetric, all you'd have to do is split the vectors up in to x and y components, add them to find the x and y components of the net force, and then calculate the magnitude and direction of the net force from the components. Example 16-4 in the textbook shows this process.

An electric field describes how an electric charge affects the region around it. It's a powerful concept, because it allows you to determine ahead of time how a charge will be affected if it is brought into the region. Many people have trouble with the concept of a field, though, because it's something that's hard to get a real feel for. The fact is, though, that you're already familiar with a field. We've talked about gravity, and we've even used a gravitational field; we just didn't call it a field.

When talking about gravity, we got into the (probably bad) habit of calling g "the acceleration due to gravity". It's more accurate to call g the gravitational field produced by the Earth at the surface of the Earth. If you understand gravity you can understand electric forces and fields because the equations that govern both have the same form.

The gravitational force between two masses (m and M) separated by a distance r is given by Newton's law of universal gravitation:

A similar equation applies to the force between two charges (q and Q) separated by a distance r:

The force equations are similar, so the behavior of interacting masses is similar to that of interacting charges, and similar analysis methods can be used. The main difference is that gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive. The charge (q or Q) plays the same role in the electrostatic case that the mass (m or M) plays in the case of the gravity.

A good example of a question involving two interacting masses is a projectile motion problem, where there is one mass m, the projectile, interacting with a much larger mass M, the Earth. If we throw the projectile (at some random launch angle) off a 40-meter-high cliff, the force on the projectile is given by:

F = mg

This is the same equation as the more complicated equation above, with G, M, and the radius of the Earth, squared, incorporated into g, the gravitational field.

So, you've seen a field before, in the form of g. Electric fields operate in a similar way. An equivalent electrostatics problem is to launch a charge q (again, at some random angle) into a uniform electric field E, as we did for m in the Earth's gravitational field g. The force on the charge is given by **F** = q**E**, the same way the force on the mass m is given by **F** = m**g**.

We can extend the parallel between gravity and electrostatics to energy, but we'll deal with that later. The bottom line is that if you can do projectile motion questions using gravity, you should be able to do them using electrostatics. In some cases, youšll need to apply both; in other cases one force will be so much larger than the other that you can ignore one (generally if you can ignore one, it'll be the gravitational force).