Question 16.5: ELECTRIC POTENTIAL ENERGY AND DYNAMICS GOAL Apply conservati...
ELECTRIC POTENTIAL ENERGY AND DYNAMICS
GOAL Apply conservation of energy and electrical potential energy to a configuration of charges.
PROBLEM Suppose three protons lie on the x-axis, at rest relative to one another at a given instant of time, as in Figure 16.9. If proton q_{3} on the right is released while the others are held fixed in place, find a symbolic expression for the proton’s speed at infinity and evaluate this speed when r_{0}=2.00 \mathrm{fm}. (Note: 1 \mathrm{fm}=10^{-15} \mathrm{~m}.)
STRATEGY First calculate the initial electric potential energy associated with the system of three particles. There will be three terms, one for each interacting pair. Then calculate the final electric potential energy associated with the system when the proton on the right is arbitrarily far away. Because the electric potential energy falls off as 1 / r, two of the terms will vanish. Using conser vation of energy then yields the speed of the particle in question.
Calculate the electric potential energy associated with the initial configuration of charges:
P E_{i}=\frac{k_{e} q_{1} q_{2}}{r_{12}}+\frac{k_{e} q_{1} q_{3}}{r_{13}}+\frac{k_{e} q_{2} q_{3}}{r_{23}}=\frac{k_{e} e^{2}}{r_{0}}+\frac{k_{e} e^{2}}{2 r_{0}}+\frac{k_{e} e^{2}}{r_{0}}
Calculate the electric potential energy associated with the final configuration of charges:
P E_{f}=\frac{k_{e} q_{1} q_{2}}{r_{12}}=\frac{k_{e} e^{2}}{r_{0}}
Write the conservation of energy equation:
\Delta K E+\Delta P E=K E_{f}-K E_{i}+P E_{f}-P E_{i}=0
Substitute appropriate terms:
\begin{aligned}&\frac{1}{2} m_{3} v_{3}{ }^{2}-0+\frac{k_{e} e^{2}}{r_{0}}-\left(\frac{k_{e} e^{2}}{r_{0}}+\frac{k_{e} e^{2}}{2 r_{0}}+\frac{k_{e} e^{2}}{r_{0}}\right)=0 \\&\frac{1}{2} m_{3} v_{3}^{2}-\left(\frac{k_{e} e^{2}}{2 r_{0}}+\frac{k_{e} e^{2}}{r_{0}}\right)=0\end{aligned}
Solve for v_{3} after combining the two remaining potential energy terms:
v_{3} =\sqrt{\frac{3 k_{e} e^{2}}{m_{3} r_{0}}}
Evaluate taking r_{0}=2.00 \mathrm{fm} :
\begin{aligned}v_{3} &=\sqrt{\frac{3\left(8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(1.60 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(2.00 \times 10^{-15} \mathrm{~m}\right)}}=1.44 \times 10^{7} \mathrm{~m} / \mathrm{s}\end{aligned}
REMARKS The difference in the initial and final potential energies yields the energy available for motion. This calculation is somewhat contrived because it would be difficult, although not impossible, to arrange such a configuration of protons; it could conceivably occur by chance inside a star.