Problem 21 - College 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 approximately 7.42 m/s, faster than at point A.

Step 1: Understand the Problem

We are given a charged particle with an initial and final electric potential at points \( A \) and \( B \), respectively. We need to determine the speed of the particle at point \( B \) based on the change in electric potential energy and kinetic energy.

Step 2: Calculate the Electric Potential Energy Change

The change in electric potential energy \(( \Delta U )\) as the particle moves from \( A \) to \( B \) can be calculated using \( \Delta U = q (V_B - V_A) \), where \( q = -5.00 \times 10^{-6} \ \text{C} \). Here, \( V_A = +200 \ \text{V} \) and \( V_B = +800 \ \text{V} \).

Step 3: Substitute Values into Electric Potential Energy Equation

Substitute the values into the formula to find \( \Delta U \):\[ \Delta U = (-5.00 \times 10^{-6} \, \text{C}) \times (800 \, \text{V} - 200 \, \text{V}) = (-5.00 \times 10^{-6} \, \text{C}) \times 600 \, \text{V} = -3.00 \times 10^{-3} \, \text{J} \]

Step 4: Apply Conservation of Energy

Since the only force acting is electric, the total mechanical energy is conserved. The relationship is given by \( K_A + U_A = K_B + U_B \), where \( K \) is the kinetic energy \(( \frac{1}{2}mv^2 )\).

Step 5: Set Up Energy Conservation Equation

Reformulating the energy conservation equation, we get:\[ \frac{1}{2} m v_A^2 + q V_A = \frac{1}{2} m v_B^2 + q V_B \]

Step 6: Substitute Known Values into Energy Equation

Given \( v_A = 5.00 \, \text{m/s} \), \( m = 2.00 \times 10^{-4} \, \text{kg} \), \( q = -5.00 \times 10^{-6} \, \text{C} \), \( V_A = 200 \, \text{V} \), and \( V_B = 800 \, \text{V} \), substitute these into the energy equation:\[ \frac{1}{2} (2.00 \times 10^{-4} \, \text{kg}) (5.00 \, \text{m/s})^2 - 5.00 \times 10^{-6} \, \text{C} \times 200 \, \text{V} = \frac{1}{2} (2.00 \times 10^{-4} \, \text{kg}) v_B^2 - 5.00 \times 10^{-6} \, \text{C} \times 800 \, \text{V} \]

Step 7: Simplify and Solve for \( v_B \)

Calculate the terms:- Kinetic energy at \( A \): \( \frac{1}{2} (2.00 \times 10^{-4} \text{kg}) (5.00 \text{m/s})^2 = 2.50 \times 10^{-3} \text{J} \)- Potential part at \( A \): \(-1.00 \times 10^{-3} \text{J} \)- Potential part at \( B \): \(-4.00 \times 10^{-3} \text{J} \)Setting up the equation:\[ 2.50 \times 10^{-3} \text{J} - 1.00 \times 10^{-3} \text{J} = \frac{1}{2} (2.00 \times 10^{-4} \text{kg}) v_B^2 - 4.00 \times 10^{-3} \text{J} \]Solve for \( v_B \):\[ \frac{1}{2} (2.00 \times 10^{-4}) v_B^2 = 5.50 \times 10^{-3} \]\[ v_B^2 = \frac{5.50 \times 10^{-3}}{1.00 \times 10^{-4}} = 55 \]\[ v_B = \sqrt{55} \approx 7.42 \text{ m/s} \]

Step 8: Conclude the Speed Difference

Since \( v_B = 7.42 \, \text{m/s} \) is greater than \( v_A = 5.00 \, \text{m/s} \), the particle moves faster at point \( B \) than at point \( A \).

Key Concepts

Conservation of Energy

When discussing the movement of charged particles in an electric field, the principle of conservation of energy is essential. It states that energy cannot be created or destroyed, only transformed from one form to another. In the context of electric fields and moving charges, this typically involves the conversion between electric potential energy and kinetic energy.

Here’s how it applies to our exercise: As the particle travels from point A to point B, its total mechanical energy, comprised of both kinetic energy and electric potential energy, remains constant. The formula used to represent this scenario is:

• Initial mechanical energy at point A: kinetic energy at A (K_A = \frac{1}{2}mv_A^2) plus electric potential energy (U_A = qV_A)
• Final mechanical energy at point B: kinetic energy at B (K_B = \frac{1}{2}mv_B^2) plus electric potential energy (U_B = qV_B)

When we equate both, \( K_A + U_A = K_B + U_B \), it confirms that any decrease in electric potential energy must correspond to an increase in kinetic energy, resulting in a higher speed at point B.

Kinetic Energy

Kinetic energy, the energy of motion, is vital in analyzing the dynamics of our moving charged particle. It’s fundamental to recognize that kinetic energy depends heavily on both the speed of the particle and its mass. The standard equation for kinetic energy is:\[K = \frac{1}{2} mv^2\]where \(m\) is the mass and \(v\) is the speed of the particle.

Initially at point A, the particle possesses kinetic energy based on its given speed of 5.00 m/s. As we calculated, that kinetic energy is 2.50 \times 10^{-3} \ ext{J}. At point B, the energy equation showed us that the decrease in electric potential energy was offset by an increase in kinetic energy, causing the particle to speed up to 7.42 m/s. In this case, the electric force converts potential energy into kinetic energy, emphasizing that a charged particle in an electric field naturally accelerates, accumulating kinetic energy as it goes.

Electric Force

Electric force is the only force acting on the charged particle in this exercise, playing a crucial role in accelerating it from point A to point B. This kind of force results from the electric field generated by differences in voltage or electrical potential between points. The relationship is described by:\[ F_e = qE \]where \( F_e \) is the electric force, \( q \) is the charge of the particle, and \( E \) is the electric field strength.

In our scenario, the fact that the electric potential increases from 200 V at point A to 800 V at point B indicates a stronger electric field, effectively pulling the negatively charged particle forward. The negative sign of the particle's charge ensures that it moves in the direction of increasing potential, with the electric force doing work that increases the particle's kinetic energy as it loses potential energy. This relationship beautifully illustrates the transformation of energy facilitated by the electric force acting upon the charged particle.