Problem 13 - 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

Particle slows down and stops, possibly reversing direction.

Step 1: Understand the Scenario

The problem describes a charged particle moving in an electric potential field. We need to determine the speed of the particle at point B, knowing its initial speed at point A. The change in electric potential will affect the particle's kinetic energy, given that the electric force is the only force acting on it.

Step 2: Use the Energy Principle

Since only electric forces do work, the work done is equal to the change in kinetic energy: \[ W = ext{KE}_B - ext{KE}_A = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \]

Step 3: Relate Work to Electric Potential Energy

The work done by the electric force can also be calculated as the change in electric potential energy:\[ W = q(V_B - V_A) \] where \(q\) is the charge of the particle.

Step 4: Set Up the Energy Equation

Substitute the expressions for work and kinetic energy difference in the equation:\[ q(V_B - V_A) = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \]

Step 5: Rearrange to Solve for Final Speed

Rearrange the equation to solve for the final speed \(v_B\):\[ \frac{1}{2} m v_B^2 = q(V_B - V_A) + \frac{1}{2} m v_A^2 \]\[ v_B^2 = \frac{2q(V_B - V_A)}{m} + v_A^2 \]\[ v_B = \sqrt{\frac{2q(V_B - V_A)}{m} + v_A^2} \]

Step 6: Insert the Known Values

Insert the given values: \( q = -5.00 \times 10^{-6} \) C, \( m = 2.00 \times 10^{-4} \) kg, \( V_A = 200 \) V,\( V_B = 800 \) V, \( v_A = 5.00 \) m/s:\[ v_B = \sqrt{\frac{2(-5.00 \times 10^{-6})(800 - 200)}{2.00 \times 10^{-4}} + 5.00^2} \]

Step 7: Calculate Final Speed

Calculate \( v_B \) using the equation:\[ v_B = \sqrt{\frac{-6 \times 10^{-3}}{2 \times 10^{-4}} + 25} \]\[ v_B = \sqrt{-30 + 25} \]Since the term under the square root is negative, the calculation indicates a decrease in kinetic energy.

Step 8: Interpret the Result

The calculation shows that the kinetic energy decreases because the electric potential energy change is negative. Since our calculation leads to a complex number (physical impossibility in this context), the particle slows down and potentially stops somewhere between A and B before switching direction.

Key Concepts

Kinetic Energy

Kinetic energy is the energy that a particle has because of its motion. It depends on two factors: the mass of the particle and its speed. Mathematically, we express kinetic energy using the formula:\[ KE = \frac{1}{2} m v^2 \]where:

• \( KE \) is the kinetic energy,
• \( m \) is the mass of the particle, and
• \( v \) is the speed of the particle.

This relationship means that if a particle moves faster, its kinetic energy increases. Similarly, a heavier particle with the same speed will have more kinetic energy than a lighter one. In our problem, the particle's speed at point \( B \) is affected by the change in kinetic energy as it moves through the electric potential field.

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. In our problem, the charge of the particle is given as \(-5.00 \mu \mathrm{C}\), which is equivalent to \(-5.00 \times 10^{-6}\) Coulombs.A few essential points to understand about electric charge:

• Charges can be positive or negative.
• Like charges repel each other, while opposite charges attract.
• Charges interact with electric fields, causing them to move if they are free to do so.

In this exercise, the negative charge moves from one point to another in a region with different electric potential. This movement affects the particle's energy levels. The nature of the charge specifically dictates how it interacts with different potentials, affecting both the direction and magnitude of the force experienced.

Work-Energy Principle

The work-energy principle is a powerful concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. In equation form, this is represented as:\[ \text{Work done} = \Delta KE = KE_B - KE_A \]Where:

• \( \text{Work done} \) is performed by the forces acting on the object.
• \( \Delta KE \) represents the change in the object's kinetic energy.
• \( KE_A \) and \( KE_B \) are the initial and final kinetic energy at points A and B, respectively.

For the given problem, the work done by the electric force is linked with the particle's movement through a potential difference. As the particle travels from a lower to a higher potential area, the change in the electric potential energy affects its kinetic energy. This illustrates how potential energy converts to or from kinetic energy, depending on the nature of the charge and the direction of the electric field. The work-energy principle helps us understand why our calculated final speed \( v_B \) showed a reduction in kinetic energy, indicating that the particle slows down.