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 18.2 , 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.
In the diagram above, point A and point B are separated by a distance of 30 cm. How much work must
be done by an external force to move a proton from point A to point B?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.