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.
Ki + Ui = Kf + UfNext, 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 kx 2 ). The potential energy is electric, so it should be written as qV .
½ mvi 2 + qVi = ½ mvf 2 + qVfFinally, 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 V (^) i was zero.
½ (9.1 × 10−31 kg)(6 × 10^6 m/s)^2 + (1.6 × 10−19 C)(0) =
½ (9.1 × 10−31 kg)(1 × 10^6 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.
Equipotential Lines: Lines that illustrate every point at which a charged particle would experience a
given potential
Figure 18.2 shows a few examples of equipotential lines (shown with solid lines) and their relationship
to electric field lines (shown with dotted lines):
Figure 18.2 Two examples of equipotential lines (in bold) and electric field lines (dotted).
On the left in Figure 18.2 , 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—