Problem 15 - University Physics with Modern Physics

Question

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{kg}\) . It moves from point \(A\) , where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B\) , where the electric potential is \(V_{B}=+800 \mathrm{V}\) . The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

Step-by-Step Solution

Answer

The particle's speed at point B is 7.42 m/s. It moves faster at B than at A.

Step 1: Understand the Given Values

The problem provides the following values:- Charge of the particle, \( q = -5.00 \mu C = -5.00 \times 10^{-6} C \)- Mass of the particle, \( m = 2.00 \times 10^{-4} \text{ kg} \)- Electric potential at point \( A, V_A = +200 V \)- Electric potential at point \( B, V_B = +800 V \) - Speed at point \( A, v_A = 5.00 \text{ m/s} \).

Step 2: Calculate the Change in Electric Potential Energy

The change in electric potential energy, \( \Delta U \), as the particle moves from \( A \) to \( B \) is given by: \[ \Delta U = q(V_B - V_A) \]Substitute the given values:\[ \Delta U = (-5.00 \times 10^{-6} \text{ C})(800 V - 200 V) = -3 \times 10^{-3} \text{ J} \].

Step 3: Calculate the Change in Kinetic Energy

Since electric force is the only force acting, the change in kinetic energy, \( \Delta K \), is equal in magnitude and opposite in sign to the change in electric potential energy:\[ \Delta K = -\Delta U \]\[ \Delta K = 3 \times 10^{-3} \text{ J} \].

Step 4: Use the Kinetic Energy Formula

The initial kinetic energy at point \( A \) is:\( K_A = \frac{1}{2}mv_A^2 \)\[ K_A = \frac{1}{2}(2.00 \times 10^{-4} \text{ kg})(5.00 \text{ m/s})^2 = 2.5 \times 10^{-3} \text{ J} \].

Step 5: Calculate the Final Kinetic Energy at Point B

The final kinetic energy at point \( B \) is:\[ K_B = K_A + \Delta K \]\[ K_B = 2.5 \times 10^{-3} \text{ J} + 3 \times 10^{-3} \text{ J} = 5.5 \times 10^{-3} \text{ J} \].

Step 6: Calculate the Speed at Point B

Using the formula for kinetic energy, solve for the speed at point \( B \):\[ K_B = \frac{1}{2}mv_B^2 \]\[ 5.5 \times 10^{-3} = \frac{1}{2}(2.00 \times 10^{-4})v_B^2 \]Solve for \( v_B^2 \):\[ v_B^2 = \frac{5.5 \times 10^{-3} \times 2}{2.00 \times 10^{-4}} = 55 \]\[ v_B = \sqrt{55} \approx 7.42 \text{ m/s} \].

Step 7: Determine if the Particle is Faster or Slower at B

Since \( v_B = 7.42 \text{ m/s} \) and \( v_A = 5.00 \text{ m/s} \), the particle is moving faster at point \( B \) compared to point \( A \).

Key Concepts

Kinetic Energy

Kinetic energy is the energy that a particle possesses due to its motion. It depends on the mass of the particle and its speed (or velocity). The formula for kinetic energy (\( K \)) is given by: \[ K = \frac{1}{2}mv^2 \] Here, \( m \) is the mass of the particle, and \( v \) is the speed of the particle.

• If a particle speeds up, its kinetic energy increases.
• If it slows down, its kinetic energy decreases.

In the context of the exercise, when the particle moves from point \( A \) to point \( B \), its kinetic energy changes due to the work done by the electric force. By calculating the initial and final kinetic energy, we can determine how much faster the particle is moving at point \( B \).

Electric Charge

Electric charge is a fundamental property of particles, determining how they interact with electric fields. Charges can be positive or negative, with like charges repelling each other and opposite charges attracting.

• In this exercise, the charge of the particle is negative, \( q = -5.00 \mu C \).
• This negative charge indicates the directionality of the electric force acting on it.

This electric charge results in an electric force, which when experiencing different potentials can lead to a change in kinetic energy as seen in the problem. This change manifests when the particle transitions between regions of different electric potential.

Particle Motion

The motion of a particle in an electric field is influenced by the interplay of its kinetic and electric potential energies. As the particle moves from point \( A \) to point \( B \), its speed changes due to this energy transformation.

• In the beginning, the particle's speed is \( 5.00 \text{ m/s} \).
• As it moves through an electric field, this speed changes, reaching \( 7.42 \text{ m/s} \) at point \( B \), indicating acceleration.

The only force acting on the particle is due to the electric field, meaning mechanical energy conservation principles play a crucial role in understanding its motion. The conversion between kinetic and potential energies governs the changes in speed.

Electric Force

Electric force is a vital factor which causes a charge to experience acceleration. This force arises due to interaction with the electric field and is given by the equation \[ F = qE \] where \( F \) is the electric force, \( q \) is the charge, and \( E \) represents the electric field strength.

• In this exercise, the electric force is the only influencing factor on the particle's motion.
• This leads to a conversion of electric potential energy to kinetic energy, enhancing the particle's speed.

Because the particle starts with a specific potential energy and moves to a region with different potential, this force is responsible for altering its kinetic energy, thereby changing its velocity during the transition from \( A \) to \( B \).