Potential energy of point-charges: The figure shows an arrangement of two

Question

Potential energy of point-charges: The figure shows an arrangement of two - 4.5 nC charges, each separated by 5.0 mm from a proton. If the two negative charges are held fixed at their locations and the proton is given an initial velocity v as shown in the figure, what is the minimum initial speed v that the proton needs to totally escape from the negative charges? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C², e = 1.60 × 10^-19 C, m_proton= 1.67 × 10^-27 kg)

Quizplus · Accepted Answer

The potential energy of the system can be calculated as:
$$U = \frac{kq_1q_2}{r} + \frac{kq_1q_3}{r} + \frac{kq_2q_3}{r}$$
where $q_1 = q_2 = -4.5$ nC, $q_3 = 1.6$ $\mu$C, and $r = 5$ mm. Substituting the values, we get:
$$U = -1.61 	imes 10^{-17} 	ext{ J}$$
The initial kinetic energy of the proton is given by:
$$K = \frac{1}{2}mv^2$$
where $m$ is the mass of the proton and $v$ is its initial velocity. The minimum initial speed required for the proton to completely escape from the negative charges is when its kinetic energy is equal to the potential energy of the system.
Therefore, equating $K$ and $U$, we get:
$$\frac{1}{2}mv^2 = 1.61 	imes 10^{-17}$$
Substituting the values, we get:
$$v = \sqrt{\frac{2 	imes 1.61 	imes 10^{-17}}{1.67 	imes 10^{-27}}} = 1.8 	imes 10^6 	ext{ m/s}$$
Therefore, the minimum initial speed required for the proton to completely escape from the negative charges is 1.8 $	imes$ 10$^6$ m/s, which is option A.