91Ó°ÊÓ

Coulomb's Law is a fundamental principle in electrostatics that describes the force between two charged particles. The law is given by the formula:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where the force \( F \) depends on:

• \( k \), the Coulomb's constant, approximately \( 8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2 \).
• \( q_1 \) and \( q_2 \), the magnitudes of the point charges involved.
• \( r \), the distance between the two charges.

In the exercise, the fixed charge is \(1 \, \mathrm{mC} = 1 \times 10^{-3}\,\mathrm{C} \) and the particle charge is \(1 \, \mu \mathrm{C} = 1 \times 10^{-6}\,\mathrm{C} \). The force is inversely proportional to the square of the distance between the charges, meaning as the distance increases, the force decreases significantly. This principle is crucial in determining the interaction and motion of charged particles.

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our exercise, this principle is applied by equating the change in electric potential energy to the change in kinetic energy. Initially, the particle possesses electric potential energy due to the distance from the fixed charge. As it moves away, this potential energy decreases and is converted into kinetic energy, making the particle accelerate.The relationship between potential and kinetic energy in this scenario is expressed through:\[ \Delta KE = PE_i - PE_f \]where:

• \( PE_i \) is the initial potential energy at \( 1 \mathrm{~m} \)
• \( PE_f \) is the final potential energy at \( 10 \mathrm{~m} \)
• \( \Delta KE \) is the change in kinetic energy

Understanding this transformation helps us calculate the change in speed as the particle moves under the influence of electric forces.

Electric potential energy is the energy that a charged particle possesses due to its position in an electric field. In the exercise, this energy is calculated using:\[ PE = \frac{k \cdot q_1 \cdot q_2}{r} \]From when the particle was at a distance of \( 1 \mathrm{~m} \) from the fixed charge, its potential energy was:\[ PE_i = \frac{8.99 \times 10^9 \cdot 1 \times 10^{-3} \cdot 1 \times 10^{-6}}{1} = 8.99 \times 10^{-3} \mathrm{~J} \]When the particle moves to \( 10 \mathrm{~m} \), the potential energy reduces to:\[ PE_f = \frac{8.99 \times 10^9 \cdot 1 \times 10^{-3} \cdot 1 \times 10^{-6}}{10} = 8.99 \times 10^{-4} \mathrm{~J} \]The decrease in potential energy corresponds with the increase in kinetic energy, hence driving the particle's motion. Electric potential energy is a key to understanding how charged objects interact in fields.

Kinetic energy is the energy of an object due to its motion. In the context of this exercise, it is derived from the transformation of potential energy as the charged particle moves away from the fixed charge.The formula for kinetic energy is:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the particle and \( v \) is its velocity. With potential energy transforming into kinetic energy, we use the equation:\[ \Delta KE = PE_i - PE_f \]Given the mass of the particle is \( 2 \, \mathrm{g} = 2 \times 10^{-3} \, \mathrm{kg} \), the transformed energy results in the particle’s speed increasing as:\[ v^2 = \frac{8.091 \times 10^{-3}}{10^{-3}} = 8.091 \times 10^1 \]\[ v = \sqrt{80.91} \approx 9 \, \mathrm{m/s} \]This allows us to determine that the speed of the particle when it is 10 meters away is approximately \( 90 \, \mathrm{m/s} \), as calculated through the changes in potential and kinetic energy.