Two particles, each having a mass of 3.0 μg and having equal but opposite charges of magnitude of 6.0 nC, are released simultaneously from rest when they are a very large distance apart. What distance separates the two at the instant when each has a speed of 5.0 m/s?
A) 4.3 mm
B) 8.6 mm
C) 7.3 mm
D) 5.6 mm
E) 2.2 mm

Question

Quizplus · Accepted Answer

The energy conservation principle can be applied here, where the initial potential energy due to the electric force between the charges is converted into kinetic energy as they move towards each other. The initial potential energy (U) when the particles are infinitely apart is considered to be zero, and the final potential energy when they are at a distance $r$ apart is given by $U = k \frac{q_1 q_2}{r}$, where $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2$), and $q_1$ and $q_2$ are the charges of the particles. The kinetic energy (K) of each particle when they have a speed of $v = 5.0 \, \text{m/s}$ is given by $K = \frac{1}{2}mv^2$, where $m$ is the mass of the particle.Since the total energy is conserved, the sum of the kinetic energies of both particles equals the potential energy at the final position:$$2 \times \frac{1}{2}mv^2 = k \frac{q^2}{r}$$Given that $m = 3.0 \, \mu\text{g} = 3.0 \times 10^{-9} \, \text{kg}$, $q = 6.0 \, \text{nC} = 6.0 \times 10^{-9} \, \text{C}$, and $v = 5.0 \, \text{m/s}$, we can solve for $r$:$$mv^2 = k \frac{q^2}{r}$$$$r = k \frac{q^2}{mv^2}$$Plugging in the values:$$r = \frac{8.99 \times 10^9 \times (6.0 \times 10^{-9})^2}{3.0 \times 10^{-9} \times (5.0)^2}$$$$r \approx 4.3 \, \text{mm}$$Therefore, the correct answer is 4.3 mm, which corresponds to choice A.

Two Particles, Each Having a Mass of 3