Problem 13 - University Physics with Modern Physics

Question

A small particle has charge \(-5.00\) \(\mu\)C and mass \(2.00 \times 10^{-4}\) kg. It moves from point \(A\), where the electric potential is \(V_A = +\)200 V, to point \(B\), where the electric potential is \(V_B = +\)800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m\(/\)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 approximately 7.42 m/s. It moves faster at B than at A, due to increased kinetic energy.

Step 1: Understand the Concepts

To find the speed of the particle at point B, we need to account for changes in kinetic energy resulting from changes in potential energy from A to B. The equation for the change in kinetic energy will be derived from the work-energy principle which states that the change in kinetic energy is equal to the work done by the electric force.

Step 2: Calculate the Change in Electric Potential Energy

The change in electric potential energy (ΔU) as the particle moves from A to B can be calculated using the formula: \[ΔU = q(V_B - V_A)\]Where \(q = -5.00 \mu C = -5.00 \times 10^{-6} C\), \(V_A = 200 V\), and \(V_B = 800 V\). Substitute these values in to get:\[ΔU = (-5.00 \times 10^{-6})(800 - 200)\]\[ΔU = (-5.00 \times 10^{-6})(600)\]\[ΔU = -3.00 \times 10^{-3} J\]

Step 3: Relate Potential Energy Change to Kinetic Energy Change

According to the work-energy principle, the change in kinetic energy \(ΔK\) of the particle is equal to the negative of the change in electric potential energy:\[ΔK = - ΔU\]Thus,\[ΔK = 3.00 \times 10^{-3} J\]

Step 4: Calculate Initial Kinetic Energy

The initial kinetic energy \(K_A\) at point A is given by:\[K_A = \frac{1}{2}mv_A^2\]Substitute \(m = 2.00 \times 10^{-4} kg\) and \(v_A = 5.00 m/s\):\[K_A = \frac{1}{2}(2.00 \times 10^{-4})(5.00)^2\]\[K_A = \frac{1}{2}(2.00 \times 10^{-4})(25)\]\[K_A = 2.50 \times 10^{-3} J\]

Step 5: Find Final Kinetic Energy and Speed at B

Given \(ΔK\), the final kinetic energy \(K_B\) at point B is:\[K_B = K_A + ΔK = 2.50 \times 10^{-3} J + 3.00 \times 10^{-3} J = 5.50 \times 10^{-3} J\]The speed \(v_B\) at point B can be found by rearranging the kinetic energy formula:\[K_B = \frac{1}{2}mv_B^2 \]\[v_B^2 = \frac{2K_B}{m} = \frac{2(5.50 \times 10^{-3})}{2.00 \times 10^{-4}}\]\[v_B^2 = 55 \]\[v_B = \sqrt{55} \]\[v_B \approx 7.42 \text{ m/s}\]

Step 6: Analyze the Speed Change

Since \(v_B = 7.42 \, m/s\) is greater than \(v_A = 5.00 \, m/s\), the particle is moving faster at point B than at point A. The increase in speed is due to the particle moving to a higher potential and the negative charge causing it to gain kinetic energy.

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It can be calculated with the formula: \[ K = \frac{1}{2}mv^2 \]where:

• \( K \) = kinetic energy (in Joules)
• \( m \) = mass (in kilograms)
• \( v \) = velocity (in meters per second)

The unit of kinetic energy is Joules (J).
In the problem, the particle's initial kinetic energy was found using its mass and speed at point A. As the particle moves from point A to B, changes in electric potential energy occur, influencing its kinetic energy and consequently, its speed.
Remember, a faster particle has higher kinetic energy, while a slower one has less. This is vital to determining the speed relationship between two points.

Work-Energy Principle

The work-energy principle is a fundamental concept in physics that relates the work done on an object to its change in kinetic energy.
According to this principle, the work done by forces on an object results in a change in its kinetic energy.
The mathematical expression for this principle is:\[ \Delta K = W \]where:

• \( \Delta K \) = change in kinetic energy
• \( W \) = work done by forces (also in Joules)

In the case of electric forces acting on a charged particle, the work done is equivalent to the change in electric potential energy: \[ W = -\Delta U \]Thus, the work-energy principle states: \[ \Delta K = -\Delta U \]This relationship helps us understand how the increase in kinetic energy at point B is due to the work done by electric forces overcoming the potential energy change from point A to B.

Electric Charge

Electric charge is a property that causes particles to experience a force when placed in an electric field.
Charges can be positive or negative and are measured in Coulombs (C).
In this problem, the particle has a charge of \(-5.00\) \(\mu C\), which is a negative charge.
This negative charge interacts with the electric field between the points A and B.

• The potential at point A is \(+200\) V, and at point B, it's \(+800\) V.
• The particle, being negatively charged, naturally moves from a region of lower potential to a higher potential within an electric field, which influences its kinetic energy.

The interaction of the charge with the electric field results in the change from potential energy to kinetic energy. These interactions are crucial to understanding how the particle speeds up as it moves.

Particle Motion

Particle motion concerns the movement of particles through space influenced by various forces.
In this exercise, the sole force influencing the particle is the electrical force due to its charge and the differing potentials at points A and B.
The movement of the particle is dependent on its initial velocity and the net work done by the electrical force, which alters its kinetic energy:

• At point A, the particle begins with an initial speed of 5 m/s.
• As it travels towards point B, its kinetic energy increases owing to the work done by the electric force.
• This results in an increase in speed, reaching approximately 7.42 m/s at point B.

Particle motion in an electric field showcases how forces transform potential energy into kinetic energy, demonstrating fundamental principles of motion physics.