Question

A point charge $q_1 = 9.1\mu C$ is held fixed at origin. A second point charge $q_2 = -0.42\mu C$ and a mass $3.2\times10^{-4}$ kg is placed on the x-axis, 0.96 m from the origin. The second point charge is released at rest. What is its speed when it is 0.24 m from the origin?

Answer 1

25.92 m/s

Answer 2

The problem involves the motion of a charged particle in an electrostatic field, which is a conservative field. Therefore, the principle of conservation of mechanical energy can be applied. The total mechanical energy (kinetic energy + potential energy) of the system remains constant.

1. Identify Initial and Final States:

• Initial State (at $r_i = 0.96$ m):

  • The second charge is released from rest, so its initial speed $v_i = 0$.
  • Initial Kinetic Energy: $K_i = \frac{1}{2} m v_i^2 = 0$
  • Initial Potential Energy: $U_i = \frac{k q_1 q_2}{r_i}$

• Final State (at $r_f = 0.24$ m):

  • Let the final speed be $v_f$.
  • Final Kinetic Energy: $K_f = \frac{1}{2} m v_f^2$
  • Final Potential Energy: $U_f = \frac{k q_1 q_2}{r_f}$

2. Apply Conservation of Mechanical Energy:

The principle of conservation of mechanical energy states that $K_i + U_i = K_f + U_f$. Substituting the expressions for kinetic and potential energies: $0 + \frac{k q_1 q_2}{r_i} = \frac{1}{2} m v_f^2 + \frac{k q_1 q_2}{r_f}$

3. Rearrange and Solve for $v_f$:

$\frac{1}{2} m v_f^2 = \frac{k q_1 q_2}{r_i} - \frac{k q_1 q_2}{r_f}$ $\frac{1}{2} m v_f^2 = k q_1 q_2 \left( \frac{1}{r_i} - \frac{1}{r_f} \right)$ $v_f^2 = \frac{2 k q_1 q_2}{m} \left( \frac{1}{r_i} - \frac{1}{r_f} \right)$

4. Substitute Given Values:

• $q_1 = 9.1\mu C = 9.1 \times 10^{-6}$ C
• $q_2 = -0.42\mu C = -0.42 \times 10^{-6}$ C
• $m = 3.2 \times 10^{-4}$ kg
• $r_i = 0.96$ m
• $r_f = 0.24$ m
• Coulomb's constant $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$

First, calculate the term $\left( \frac{1}{r_i} - \frac{1}{r_f} \right)$: $\left( \frac{1}{0.96} - \frac{1}{0.24} \right) = \left( \frac{1}{0.96} - \frac{4}{0.96} \right) = \frac{1 - 4}{0.96} = \frac{-3}{0.96}$ $= \frac{-300}{96} = -\frac{25}{8}$

Now, substitute all values into the equation for $v_f^2$: $v_f^2 = \frac{2 \times (9 \times 10^9) \times (9.1 \times 10^{-6}) \times (-0.42 \times 10^{-6})}{3.2 \times 10^{-4}} \times \left(-\frac{25}{8}\right)$

Notice that $q_1 q_2$ is negative, and the term $\left(\frac{1}{r_i} - \frac{1}{r_f}\right)$ is also negative. The product of two negatives will be positive, as expected for $v_f^2$.

$v_f^2 = \frac{2 \times 9 \times 9.1 \times 0.42 \times 10^{9-6-6}}{3.2 \times 10^{-4}} \times \frac{25}{8}$ $v_f^2 = \frac{18 \times 9.1 \times 0.42 \times 10^{-3}}{3.2 \times 10^{-4}} \times \frac{25}{8}$ $v_f^2 = \frac{68.796 \times 10^{-3}}{3.2 \times 10^{-4}} \times \frac{25}{8}$ $v_f^2 = \left( \frac{68.796}{3.2} \right) \times \left( \frac{10^{-3}}{10^{-4}} \right) \times \left( \frac{25}{8} \right)$ $v_f^2 = 21.49875 \times 10^1 \times 3.125$ $v_f^2 = 214.9875 \times 3.125$ $v_f^2 = 671.8359375$

Finally, calculate $v_f$: $v_f = \sqrt{671.8359375}$ $v_f \approx 25.920$ m/s

Rounding to two significant figures (based on the input values), the speed is approximately 26 m/s.