MODEL Energy is conserved. The glass sphere can be modeled as a charged particle, so the potential energy is that of two point charges. The glass is so much more massive than the proton that it remains at rest as the proton moves. The proton starts “far away,” which we interpret as sufficiently far to make $U_{\text{i}}\approx 0$.

VISUALIZE FIGURE 25.10 shows the before-and-after pictorial representation. To “just reach” the glass sphere means that the proton comes to rest, $v_{\text{f}}=0$, as it reaches $r_{\text{f}}=0.50\text{mm,}$ the radius of the sphere.

FIGURE 25.10 A proton approaching a glass sphere.

SOLVE Conservation of energy $K_{\text{f}}+U_{\text{f}}=K_{\text{i}}+U_{\text{i}}$ is

The proton charge is $q_{\text{p}}=e$. With this, we can solve for the proton’s initial speed:

MODEL Energy is conserved. The particles end “far apart,” which we interpret as sufficiently far to make $U_{\text{f}}\approx 0$.

VISUALIZE FIGURE 25.11 shows the before-and-after pictorial representation. The minimum speed to escape is the speed that allows the particles to reach $r_{\text{f}}=\infty$ with $v_{\text{f}}=0$.

FIGURE 25.11 An electron and a positron flying apart.

SOLVE $U_{\text{elec}}$ is the potential energy of the $\text{electron}+\text{positron}$ system. Similarly, K is the total kinetic energy of the system. The electron and the positron, with equal masses and equal speeds, have equal kinetic energies. Conservation of energy $K_{\text{f}}+U_{\text{f}}=K_{\text{i}}+U_{\text{i}}$ is

Using $r_{\text{i}}=100\text{fm}=1.0\times {10}^{-13}\text{m}$, we can calculate the minimum initial speed to be

REVIEW $v_{\text{i}}$ is a little more than 10% the speed of light, just about the limit of what a “classical” calculation can predict. We would need to use the theory of relativity if $v_{\text{i}}$ were any larger.