Problem 4 - College Physics

Question

A point charge \(q_{1}=+2.40 \mu C\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu C\) moves from the point \(x=0.150 \mathrm{m}, y=0,\) to the point \(x=0.250 \mathrm{m},\) \(y=0.250 \mathrm{m} .\) How much work is done by the electric forceon \(q_{2} ?\)

Step-by-Step Solution

Answer

The work done by the electric force on \( q_2 \) is \(-0.333 \, \text{J}\).

Step 1: Understand the Problem

We need to calculate the work done by the electric force as a point charge, \( q_2 \), moves in the electric field created by another point charge, \( q_1 \). The work done can be determined by the change in electric potential energy.

Step 2: Recall the Formula for Electric Potential Energy

The electric potential energy \( U \) between two point charges is given by: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.

Step 3: Calculate Initial Distance and Initial Potential Energy

Initially, the distance between \( q_1 \) and \( q_2 \) is \( r_i = 0.150 \, \text{m} \). Calculate the initial potential energy: \[ U_i = \frac{8.99 \times 10^9 \, (2.40 \times 10^{-6}) (-4.30 \times 10^{-6})}{0.150} \]

Step 4: Calculate Final Distance and Final Potential Energy

The final position of \( q_2 \) is \( (0.250 \text{ m}, 0.250 \text{ m}) \). The distance from \( q_1 \) is \( r_f = \sqrt{0.250^2 + 0.250^2} \). Calculate the final potential energy: \[ U_f = \frac{8.99 \times 10^9 \, (2.40 \times 10^{-6}) (-4.30 \times 10^{-6})}{r_f} \] where \( r_f = 0.354 \text{ m} \).

Step 5: Calculate the Work Done by the Electric Force

The work done by the electric force is equal to the change in electric potential energy: \[ W = U_i - U_f \] Substitute the values calculated in Steps 3 and 4 to find \( W \).

Step 6: Solve and Conclude

First, calculate \( U_i \approx -0.576 \, \text{J} \) and \( U_f \approx -0.243 \, \text{J} \). Therefore, the work done is \( W = -0.576 - (-0.243) = -0.333 \, \text{J} \).

Key Concepts

Coulomb's Law

Coulomb's law is a fundamental principle that describes the force between two charged objects. It tells us that the force (\( F \)) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. The equation is given by:

• \( F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \)

where

• \( k = 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2 \)
• \( q_1 \) and \( q_2 \) are the point charges
• \( r \) is the distance between the charges

This law helps us understand how charges interact at both microscopic and macroscopic levels. The force described can either be attractive or repulsive depending on whether the charges are of opposite or same signs, respectively. Understanding this force is crucial for calculating work done by electric forces, as it influences energy levels.

Work Done by Electric Force

The work done by an electric force occurs when a charge is moved within an electric field. Work (\( W \)) is defined as the energy transferred to or from an object via the application of force along a displacement. It's calculated using the change in electric potential energy:

• \( W = \Delta U = U_i - U_f \)

where

• \( U_i \) is the initial potential energy
• \( U_f \) is the final potential energy

Essentially, work done by the electric force is equal to the reduction in the electric potential energy of a system as the charge moves. If the potential energy decreases, positive work is done on the charge as it moves through the field. This concept is a manifestation of the conservation of energy and appears frequently in scenarios involving point charges.

Point Charges

Point charges are an idealized concept used to simplify calculations in electrostatics. A point charge is an electric charge that is assumed to occupy a single point in space. This approximation is very helpful because:

• It allows us to ignore the effects of the physical size of the charges, focusing solely on their electric effects.
• It simplifies the mathematics of evaluating forces and potentials.
• It is a practical model for understanding interactions at small distances or when other physical dimensions are much larger than the size of the charge itself.

Physically, real objects contain distributed charges, but many scenarios can effectively be modeled as point charges to make calculations manageable, without significant loss of accuracy. Point charges are instrumental in applying Coulomb's law, calculating electric fields, and understanding energy changes during charge displacement.

Electric Field

An electric field represents the area or region around a charged object where other charged objects experience a force. The electric field (\( E \)) is a vector quantity, indicating both magnitude and direction, defined at each point in space. It is related to a charge by:

• \( E = \frac{F}{q} \)

where:

• \( F \) is the force experienced by a small positive test charge \( q \)

Another expression shows how the field appears around a point charge:

• \( E = \frac{k \cdot q}{r^2} \)

The electric field is crucial because it provides a way to visualize and calculate the influence a charge will have on other charges around it. It helps us predict how charges will move and how much work the electric force will perform as charges are displaced. Understanding electric fields is foundational for mastering concepts in electricity and magnetism.