Electric Potential for AP Physics B & C (page 2)
Practice problems for these concepts can be found at:
When you hold an object up over your head, that object has gravitational potential energy. If you were to let it go, it would fall to the ground.
Similarly, a charged particle in an electric field can have electrical potential energy. For example, if you held a proton in your right hand and an electron in your left hand, those two particles would want to get to each other. Keeping them apart is like holding that object over your head; once you let the particles go, they'll travel toward each other just like the object would fall to the ground.
In addition to talking about electrical potential energy, we also talk about a concept called electric potential.
Electric potential is a scalar quantity. The units of electric potential are volts. 1 volt = 1 J/C.
Just as we use the term "zero of potential" in talking about gravitational potential, we can also use that term to talk about voltage. We cannot solve a problem that involves voltage unless we know where the zero of potential is. Often, the zero of electric potential is called "ground."
Unless it is otherwise specified, the zero of electric potential is assumed to be far, far away. This means that if you have two charged particles and you move them farther and farther from each another, ultimately, once they're infinitely far away from each other, they won't be able to feel each other's presence.
The electrical potential energy of a charged particle is given by this equation:
Here, q is the charge on the particle, and V is the voltage.
It is extremely important to note that electric potential and electric field are not the same thing. This example should clear things up:
Electric field lines point in the direction that a positive charge will be forced, which means that our positron, when placed in this field, will be pushed from left to right. So, just as an object in Earth's gravitational field has greater potential energy when it is higher off the ground (think "mgh"), our positron will have the greatest electrical potential energy when it is farthest from where it wants to get to. The answer is A.
We hope you noticed that, even though the electric field was the same at all three points, the electric potential was different at each point.
How about another example?
This is a rather simple conservation of energy problem, but it's dressed up to look like a really complicated electricity problem.
As with all conservation of energy problems, we'll start by writing our statement of conservation of energy.
- KEi + PEi = KEf + PEf
Next, we'll fill in each term with the appropriate equations. Here the potential energy is not due to gravity (mgh), nor due to a spring (1/2 kx2). The potential energy is electric, so should be written as qV.
Finally, we'll plug in the corresponding values. The mass of a positron is exactly the same as the mass of an electron, and the charge of a positron has the same magnitude as the charge of an electron, except a positron's charge is positive. Both the mass and the charge of an electron are given to you on the "constants sheet." Also, the problem told us that the positron's initial potential Vi was zero.
- ½ (9.1 × 10–31 kg)(6 × 106 m/s)2 + (1.6 × 10–19 C)(0) =
- ½ (9.1 × 10–31 kg)(1 × 106 m/s)2 + (1.6 × 10–19 C)(Vf)
Solving for Vf , we find that Vf is about 100 V.
For forces, a negative sign simply indicates direction. For potentials, though, a negative sign is important. –300 V is less than –200 V, so a proton will seek out a –300 V position in preference to a –200 V position. So, be careful to use proper + and – signs when dealing with potential.
Just as you can draw electric field lines, you can also draw equipotential lines.
Figure 20.2 shows a few examples of equipotential lines (shown with solid lines) and their relationship to electric field lines (shown with dotted lines):
On the left in Figure 20.2, the electric field points away from the positive charge. At any particular distance away from the positive charge, you would find an equipotential line that circles the charge—we've drawn two, but there are an infinite number of equipotential lines around the charge. If the potential of the outermost equipotential line that we drew was, say, 10 V, then a charged particle placed anywhere on that equipotential line would experience a potential of 10 V.
On the right in Figure 20.2, we have a uniform electric field. Notice how the equipotential lines are drawn perpendicular to the electric field lines. In fact, equipotential lines are always drawn perpendicular to electric field lines, but when the field lines aren't parallel (as in the drawing on the left), this fact is harder to see.
Moving a charge from one equipotential line to another takes energy. Just imagine that you had an electron and you placed it on the innermost equipotential line in the drawing on the left. If you then wanted to move it to the outer equipotential line, you'd have to push pretty hard, because your electron would be trying to move toward, and not away from, the positive charge in the middle.
The potential at point B is higher than at point A; so moving the positively charged proton from A to B requires work to change the proton's potential energy. The question here really is asking how much more potential energy the proton has at point B.
Well, potential energy is equal to qV; here, q is 1.6 × 10–19 C, the charge of a proton. The potential energy at point A is (1.6 × 10–19 C)(50 V) = 8.0 × 10–18 J; the potential energy at point B is (1.6 × 10–19 C)(60 V) = 9.6 × 10–18 J. Thus, the proton's potential is 1.6 × 10–18 J higher at point B, so it takes 1.6 × 10–18 J of work to move the proton there.
Um, didn't the problem say that points A and B were 30 cm apart? Yes, but that's irrelevant. Since we can see the equipotential lines, we know the potential energy of the proton at each point; the distance separating the lines is irrelevant.
Practice problems for these concepts can be found at:
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