Chapter 23 - Electric Potential - Problems - Exercises - Page 777: 23.5

(a) $v_b = 12.3 \,\text{m/s}$ (b) $r_{min} = 0.32 \,\text{m}$

(a) The difference in potential energy $\Delta U$ between the two spheres is given by \begin{align*} \Delta U &= U_a - U_b \\ & = \dfrac{1}{4\pi \epsilon_o} (q_1q_2) \left[\dfrac{1}{r_a} - \dfrac{1}{r_b} \right]\\ &= (9\times 10^{9} \mathrm{~N\cdot m^2/C^2})(-2.8 \times 10^{-6}) (-7.8 \times 10^{-6}) \left[\dfrac{1}{0.8 \,\text{m}} - \dfrac{1}{0.4 \,\text{m}} \right]\\ &= -0.25 \,\text{J} \end{align*} The change in potential energy $\Delta U$ is equal to the change in kinetic energies of the two protons and we calculate the speed $v$ of the proton from this concept by \begin{gather*} \Delta U = \Delta K \\ \Delta U =\dfrac{1}{2} mv_b^2 - \dfrac{1}{2} mv_a^2 \\ \dfrac{1}{2} mv_b^2 = \dfrac{1}{2} mv_a^2 + \Delta U \\ v_b = \sqrt{v_a^2 + 2 \dfrac{\Delta U}{m}} \\ v_b = \sqrt{(22 \,\text{m/s})^2 + 2 \dfrac{ -0.25 \,\text{J}}{0.0015 \,\text{kg}}}\\ \boxed{v_b = 12.3 \,\text{m/s}} \end{gather*} (b) At the closest distance, the kinetic energy of $q_2$ is $K_b $ = zero So, the potential energy is maximum at this distance and could be calculated by \begin{align*} U_{max} &= K_a + U_a \\ &= \dfrac{1}{2} mv_a^2 + \dfrac{1}{4\pi \epsilon_o} \left[\dfrac{q_1q_2}{r_a} \right]\\ &= \dfrac{1}{2} (0.0015 \,\text{kg})(22 \,\text{m/s})^2 +(9\times 10^{9} \mathrm{~N\cdot m^2/C^2}) \left[\dfrac{(-2.8 \times 10^{-6}) (-7.8 \times 10^{-6})}{0.8 \,\text{m}} \right]\\ &= 0.61 \,\text{J} \end{align*} So, the minimum distance is at the maximum potential energy and we calculate it by \begin{gather*} U_{max} = \dfrac{1}{4\pi \epsilon_o} \left[\dfrac{q_1q_2}{r_{min}} \right]\\ r_{min} = \dfrac{1}{4\pi \epsilon_o} \left[\dfrac{q_1q_2}{U_{max}} \right]\\ r_{min} = (9\times 10^{9} \mathrm{~N\cdot m^2/C^2}) \left[\dfrac{(-2.8 \times 10^{-6}) (-7.8 \times 10^{-6})}{ 0.61 \,\text{J}} \right]\\ \boxed{r_{min} = 0.32 \,\text{m}} \end{gather*}