5.2.2: Explorations

Exploration 1: Fields and Test Charges

If interactions between charges can be described by the forces on them, and the electric field is related to the electric force, why talk about electric fields at all? We will consider this issue in the following exploration.

The animation shows a positive charge of \(4\text{ C}\) located at (\(0\text{ m},\: 0\text{ m}\)) and a test charge of \(1\text{ C}\) located at (\(1\text{ m},\: 1\text{ m}\)). The distance between the test charge and the charge at the origin is shown along with a vector representing the force on the test charge (position is given in meters and charge is given in coulombs). The slider allows you to change the charge on the test charge. Restart.

1. Drag the test charge to \(x = 2\text{ m}\) and \(y = 2\text{ m}\). Relative to the force on the charge in its initial position, is the force now greater, less, or the same? How will the force on the test charge compare to its initial force when it is at (\(-1\text{ m},\: 1\text{ m}\))? What about when it is at (\(0.5\text{ m},\: -1\text{ m}\))? What determines the relative magnitude of the force on the test charge?
2. Now consider leaving the test charge at its initial location but increasing the positive charge to \(2\text{ C}\). How does the force on the test charge change? How will the force on the test charge compare to its initial force when the test charge is \(0.5\text{ C}\)? What about when it is \(-1\text{ C}\)? What determines the relative magnitude of the force on the test charge?

Now suppose you wanted to describe the force that would be felt due to the charge at the center. Can you describe the force in general terms, without specifying the value of the charge to be acted upon? Your answer to (a) and (b) should have shown you that you must know both where the charge is located and its value in order to talk about the force it would feel. If you could talk about the force in terms of the force on a test charge, you will run into difficulty. The force is entirely dependent on the test charge! Change the value of the test charge and you will find a different force. We need a description of the field that is independent of our detecting device.

3. Can you think of a way of quantitatively describing the region around the center charge that does not depend on the properties of a test charge? Write down at least one proposal BEFORE going to (d).
4. The physicist's answer to this dilemma is the creation of the concept of the electric field. The electric field is defined to be the force on a test charge divided by the value of the test charge, \(E = F/q\). For a single charge like the one at the center of the animation, \(F = (kQq)/r^{2}\), so the electric field becomes \(E = kQ/r^{2}\).
5. Repeat (a), (b), and (c) considering the electric field instead of the force.
6. Does the force depend on the value of the test charge? Does the electric field depend on the value of the test charge? Test your predictions by changing the value of the test charge and watching the effect on the force vector and the electric field vectors. (Turn the electric field vectors on using the link.)
7. In your own words, why is it useful to define the electric field in the way it is defined? Why not just talk about forces?

Exploration authored by Melissa Dancy.

Exploration 2: Field Lines and Trajectories

The animation shows two fixed charges and a test charge (position is given in meters and time is given in seconds). The electric field lines due to the fixed charges and the force vector on the test charge are shown. The test charge will move under the action of the electric field when the animation is played. Restart.

1. Using Configuration A, drag the test charge to the approximate position of (\(-0.8\text{ m},\: 0\text{ m}\)). Write down a prediction for the path the charge will follow after being released at this point. After you have made your prediction, play the animation. Was your prediction correct? If not, what caused your error?
2. Reset the applet and then drag the test charge to the approximate position of (\(1\text{ m},\: 0.35\text{ m}\)). As before, write down a prediction for the path the charge will follow after being released. If your prediction was incorrect, explain the flaw in your reasoning.
3. Repeat using Configuration B with the charge being released from the point (\(-0.5\text{ m},\: 0.5\text{ m}\)).
4. Repeat using Configuration B with the charge being released from the point (\(0\text{ m},\: 1.3\text{ m}\)).

Exploration authored by Melissa Dancy.

Exploration 3: Adding Fields

A charged bead is placed on a circular wire frame as shown. The center of the circle is at the point (\(0\text{ m},\: 1\text{ m}\)). In addition to gravity, you can add a uniform electric field in the \(x\) direction (position is given in meters, time is given in seconds, and the electric field strength is given in newtons/coulomb). Restart. The force field is shown using arrows as in Illustration 23.1.

Enter a value for the electric field and click the "set value and play" button to begin the animation. The bead will move unless it is at an equilibrium position. You can set the instantaneous velocity to zero, but the bead will again begin to move unless you happen to damp it at an equilibrium point. Pause the animation, zero the velocity, drag the bead, and play the animation as many times as you like. If the electric field is small enough so it is similar in size to the gravitational field, you can see the field vector at an angle with the horizontal because it is the vector sum of the gravitational force (\(m\mathbf{g}\)) and the electric force (\(q\mathbf{E}\)). Determine the charge on the bead if the mass is \(10\) grams.

1. Notice that the bead oscillates about an equilibrium position. Find a value of the electric field that gives you an equilibrium position somewhere on the wire. Zero the velocity to get the bead to stop at that equilibrium position.
2. Draw a force diagram and show that the \(y\) component of the normal force of the wire on the bead must be equal to the weight of the bead, while the \(x\) component of the normal force must be equal to the force due to the electric field (\(q\mathbf{E}\)).
3. Since the normal force is perpendicular to the wire (and therefore points to the center of the wire circle), find the angle that the normal makes with either the horizontal or vertical and then show that the ratio of the gravitational force to the electric force is simply the tangent of the angle that the normal force makes with the horizontal. Therefore, you can find the electrical force (\(q\mathbf{E}\)) required to keep the bead at equilibrium.

Exploration authored by Anne J. Cox, Mario Belloni, and Wolfgang Christian.
Script authored by Mario Belloni and Wolfgang Christian.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.