Question

A particle with a charge of \(+5.0 \mu C\) is released from rest at a point on the \(x\) -axis, where \(x=0.10 \mathrm{~m}\). It begins to move as a result of the presence of a \(+9.0-\mu C\) charge that remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point \(x=0.20 \mathrm{~m} ?\)

Short answer

Answer: The kinetic energy of the particle at the instant it passes the point x = 0.20 m is 2.022 x 10^{-3} J.

Step-by-step solution

Determine the initial and final positions of the particle

The initial position of the particle is x=0.10 m, and the final position is x=0.20 m.

Calculate the initial potential energy

The potential energy between two charged particles is given by the formula: \(U = \frac{kq_1q_2}{r}\), where \(k=8.9875\times 10^9 Nm^2/C^2\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges of the particles, and \(r\) is the distance between them. At the initial position (x=0.10 m), the potential energy is: \(U_i = \frac{(8.9875\times 10^9 Nm^2/C^2)(5.0\times 10^{-6} C)(9.0\times 10^{-6} C)}{0.10 m}\). Calculate the initial potential energy: \(U_i = 4.044\times 10^{-3} J\).

Calculate the final potential energy

At the final position (x=0.20 m), the potential energy is: \(U_f = \frac{(8.9875\times 10^9 Nm^2/C^2)(5.0\times 10^{-6} C)(9.0\times 10^{-6} C)}{0.20 m}\). Calculate the final potential energy: \(U_f = 2.022\times 10^{-3} J\).

Apply the conservation of mechanical energy

Since the initial kinetic energy of the particle is zero, the initial mechanical energy is equal to the initial potential energy: \(E_{mech_i} = U_i\). The final mechanical energy is the sum of the final potential energy and the final kinetic energy: \(E_{mech_f} = U_f + KE_f\). Applying the conservation of mechanical energy, we get: \(E_{mech_i} = E_{mech_f}\), or \(U_i = U_f + KE_f\).

Calculate the final kinetic energy

Solve for the final kinetic energy: \(KE_f = U_i - U_f\). Calculate the final kinetic energy: \(KE_f = 4.044\times 10^{-3} J - 2.022\times 10^{-3} J\). \(KE_f = 2.022\times 10^{-3} J\). The kinetic energy of the particle at the instant it passes the point \(x = 0.20 m\) is \(2.022 \times 10^{-3} J\).

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that helps us understand the interaction between charged particles. It states that the force between two charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The formula is:
\[ F = k \frac{|q_1 q_2|}{r^2} \] where

• \( F \) is the force between the charges,
• \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) is the electrostatic constant,
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• and \( r \) is the distance between the charges.

This law shows us how charges interact, and explains why the particle moves in the given exercise. The positive charges repel each other, causing the particle to experience a force that pushes it away, which ultimately turns potential energy into kinetic energy as it moves.

Potential Energy

Potential energy in electrostatics refers to the energy stored due to the position of charged particles. For two point charges, it can be calculated using the formula:
\[ U = \frac{k q_1 q_2}{r} \] In this formula,

• \( U \) is the potential energy,
• \( k \) is the electrostatic constant,
• \( q_1 \) and \( q_2 \) are the charges,
• and \( r \) is the distance between the charges.

In our exercise:

• The initial potential energy \( U_i \) at \( x = 0.10 \, \text{m} \) is \( 4.044 \times 10^{-3} \, \text{J} \).
• The final potential energy \( U_f \) at \( x = 0.20 \, \text{m} \) is \( 2.022 \times 10^{-3} \, \text{J} \).

As the particle moves further away from the fixed charge, the potential energy decreases, converting into kinetic energy.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another. In our scenario, the conservation of mechanical energy tells us that the sum of kinetic and potential energy remains constant as the particle moves. Initially, the kinetic energy is zero, so the total mechanical energy is just the initial potential energy:
\[ E_{\text{mech}_i} = U_i \]
At the final position, the mechanical energy is the sum of the final potential energy and kinetic energy. By applying conservation of energy, we have:
\[ E_{\text{mech}_i} = E_{\text{mech}_f} = U_f + KE_f \]
Solving for the final kinetic energy, we get:
\[ KE_f = U_i - U_f \]
The energy that was once potential has partly turned into kinetic energy, as the particle moves away. In the problem, this kinetic energy is calculated to be \( 2.022 \times 10^{-3} \, \text{J} \), which demonstrates how energy is conserved in the system.