Problem 8 - Applied Physics

Question

What force is exerted on a test charge of \(4.00 \times 10^{-5} \mathrm{C}\) if it is placed in an electric field of magnitude \(3.00 \times 10^{6} \mathrm{~N} / \mathrm{C} ?\)

Step-by-Step Solution

Answer

The force exerted on the charge is 120 N.

Step 1: Understand the problem

We need to find the force exerted on a test charge in a given electric field.

Step 2: Identify the formula

The formula to calculate the force exerted on a charge in an electric field is: \[ F = qE \] where \( F \) is the force, \( q \) is the charge, and \( E \) is the electric field magnitude.

Step 3: Substitute the given values

Substitute the given values into the formula: - \( q = 4.00 \times 10^{-5} \mathrm{C} \)- \( E = 3.00 \times 10^{6} \mathrm{~N} / \mathrm{C} \)The equation becomes: \[ F = (4.00 \times 10^{-5} \mathrm{C}) \times (3.00 \times 10^{6} \mathrm{N} / \mathrm{C}) \]

Step 4: Calculate the result

Perform the multiplication to find the force:\[ F = 4.00 \times 3.00 \times 10^{-5} \times 10^{6} \]This simplifies to:\[ F = 12.0 \times 10^{1} = 1.2 \times 10^{2} \mathrm{~N} \]

Key Concepts

Electric Field

An electric field is an invisible force field surrounding electric charges. It represents how a charge would exert force on other charges around it. Imagine it as a map showing how the force acts over space. The electric field strength is denoted by the symbol \( E \) and is measured in Newtons per Coulomb (\( \mathrm{N/C} \)). The greater the electric field strength, the stronger the force that would be exerted on any charge placed within it. In our example exercise, the electric field has a magnitude of \( 3.00 \times 10^{6} \, \mathrm{N/C} \). This tells us how strong the field is at that particular point.

Test Charge

A test charge is a small positive charge used to test the electric field's effects. It helps physicists understand how strong an electric field is at a certain point without disturbing the field significantly. By convention, a test charge is always positive, making it easier to predict the direction of the force: it will move away from positive charges and towards negative charges. In our problem, the test charge has a value of \( 4.00 \times 10^{-5} \, \mathrm{C} \). This is a very small charge, ensuring it won't alter the surrounding electric fields. Understanding the test charge concept helps us appreciate how forces in electric fields are calculated without altering the system being observed.

Physics Formula

When calculating forces in electric fields, we rely on a fundamental physics formula: \[ F = qE \] This equation tells us that the force \( F \) on a charge is determined by multiplying the electric charge \( q \) and the electric field strength \( E \). Here, \( q \) is the test charge, and \( E \) is the electric field's magnitude. In this equation:

• \( q \) represents the amount of charge you place within the electric field, measured in Coulombs.
• \( E \) represents the field's strength at that location, giving you the force per unit charge.

This formula serves as a straightforward tool in physics to understand how electric fields influence charges. Understanding this equation helps us solve problems dealing with how charges interact in various contexts.

Force Calculation

To compute the force acting on a test charge within an electric field, use the equation \( F = qE \). Substituting the values from our example, we have:

• \( q = 4.00 \times 10^{-5} \, \mathrm{C} \)
• \( E = 3.00 \times 10^{6} \, \mathrm{N/C} \)

Thus, the equation becomes:\[ F = (4.00 \times 10^{-5} \, \mathrm{C}) \times (3.00 \times 10^{6} \, \mathrm{N/C}) \]Calculating it step by step:1. Multiply the numerical parts: \( 4.00 \times 3.00 = 12.00 \)2. Handle the powers of ten: \( 10^{-5} \times 10^{6} = 10^{1} \)Putting these together gives \( F = 12.00 \times 10^{1} \), which can be simplified further to \( 1.2 \times 10^{2} \, \mathrm{N} \). This result shows the force that the electric field exerts on the test charge, illustrating the practical application of the physics formula discussed.