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# Electric Potential for AP Physics B & C (page 2)

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By McGraw-Hill Professional
Updated on Feb 12, 2011

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:

Electrostatics Practice Problems for AP Physics B & C

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