7-6-00
Sections 16.7 - 16.8
To help visualize how a charge, or a collection of charges, influences the region around it, the concept of an electric field is used. The electric field E is analogous to g, which we called the acceleration due to gravity but which is really the gravitational field. Everything we learned about gravity, and how masses respond to gravitational forces, can help us understand how electric charges respond to electric forces.
The electric field a distance r away from a point charge Q is given by:
Electric field from a point charge : E = k Q / r2
The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by :
F = qE
If q is positive, the force is in the same direction as the field; if q is negative, the force is in the opposite direction as the field.
Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/s2 and points straight down. If you threw a mass through the air, you know it would follow a parabolic path because of gravity. You could determine when and where the object would land by doing a projectile motion analysis, separating everything into x and y components. The horizontal acceleration is zero, and the vertical acceleration is g. We know this because a free-body diagram shows only mg, acting vertically, and applying Newton's second law tells us that mg = ma, so a = g.
You can do the same thing with charges in a uniform electric field. If you throw a charge into a uniform electric field (same magnitude and direction everywhere), it would also follow a parabolic path. We're going to neglect gravity; the parabola comes from the constant force experienced by the charge in the electric field. Again, you could determine when and where the charge would land by doing a projectile motion analysis. The acceleration is again zero in one direction and constant in the other. The value of the acceleration can be found by drawing a free-body diagram (one force, F = qE) and applying Newton's second law. This says:
qE = ma, so the acceleration is a = qE / m.
Is it valid to neglect gravity? What matters is the size of qE / m relative to g. As long as qE / m is much larger than g, gravity can be ignored. Gravity is very easy to account for, of course : simply add mg to the free-body diagram and go from there.
The one big difference between gravity and electricity is that m, the mass, is always positive, while q, the charge, can be positive, zero, or negative.
An electric field can be visualized on paper by drawing lines of force, which give an indication of both the size and the strength of the field. Lines of force are also called field lines. Field lines start on positive charges and end on negative charges, and the direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed at that point. If the charge is positive, it will experience a force in the same direction as the field; if it is negative the force will be opposite to the field.
The fields from isolated, individual charges look like this:
When there is more than one charge in a region, the electric field lines will not be straight lines; they will curve in response to the different charges. In every case, though, the field is highest where the field lines are close together, and decreases as the lines get further apart.
Two charges are placed on the x axis. The first, with a charge of +Q, is at the origin. The second, with a charge of -2Q, is at x = 1.00 m. Where on the x axis is the electric field equal to zero?
This question involves an important concept that we haven't discussed yet: the field from a collection of charges is simply the vector sum of the fields from the individual charges. To find the places where the field is zero, simply add the field from the first charge to that of the second charge and see where they cancel each other out.
In any problem like this it's helpful to come up with a rough estimate of where the point, or points, where the field is zero is/are. There is no such point between the two charges, because between them the field from the +Q charge points to the right and so does the field from the -2Q charge. To the right of the -2Q charge, the field from the +Q charge points right and the one from the -2Q charge points left. The field from the -2Q charge is always larger, though, because the charge is bigger and closer, so the fields can't cancel. To the left of the +Q charge, though, the fields can cancel. Let's say the point where they cancel is a distance x to the left of the +Q charge.
Cross-multiplying and expanding the bracket gives:
Solving for x using the quadratic equation gives: x = 2.41 m or x = -0.414 m
The answer to go with is x = 2.41 m. This corresponds to 2.41 m to the left of the +Q charge. The other point is between the charges. It corresponds to the point where the fields from the two charges have the same magnitude, but they both point in the same direction there so they don't cancel out.