Problem 5 - Applied Physics

Question

An electric field of magnitude \(0.450 \mathrm{~N} / \mathrm{C}\) exerts a force of \(8.00 \times 10^{-4} \mathrm{~N}\) on a test charge placed in the field. What is the magnitude of the test charge?

Step-by-Step Solution

Answer

The magnitude of the test charge is \(1.78 \times 10^{-3} \mathrm{~C}\).

Step 1: Identify Given Information

An electric field has a magnitude of \(0.450 \mathrm{~N} / \mathrm{C}\), and it exerts a force of \(8.00 \times 10^{-4} \mathrm{~N}\) on a test charge. These will be our known values for the calculation.

Step 2: Understand the Formula

The relationship between the electric field (\(E\)), force (\(F\)), and charge (\(q\)) is given by the formula \(F = qE\). We will use this formula to find \(q\).

Step 3: Rearrange the Formula

Solve for the charge \(q\) by rearranging the formula: \(q = \frac{F}{E}\). This allows us to isolate \(q\) and use the given values to calculate it.

Step 4: Substitute Values

Substitute the known values into the rearranged formula: \(q = \frac{8.00 \times 10^{-4} \mathrm{~N}}{0.450 \mathrm{~N} / \mathrm{C}}\).

Step 5: Calculate the Charge

Apply the calculation: \(q = \frac{8.00 \times 10^{-4}}{0.450} = 1.78 \times 10^{-3} \mathrm{~C}\). Thus, the test charge has a magnitude of \(1.78 \times 10^{-3} \mathrm{~C}\).

Key Concepts

Understanding the Test Charge

In the world of electromagnetism, the concept of a test charge helps us explore and understand electric fields. A test charge is a small, hypothetical positive charge. It is used to measure the strength and direction of an electric field at a specific point. By convention, test charges are positive, and thus, allow us to observe their interaction with the electric field without altering the field itself. When we place a test charge within an electric field, it experiences a force. This force is due to the interaction between the charge of the test charge and the electric field. By analyzing the magnitude and direction of this force, we can get detailed information about the characteristics of the electric field. A key aspect of using test charges is that they are assumed to be infinitesimally small. This means they don't influence the electric field they're introduced into, ensuring that our measurements reflect the field's original conditions. This simplification is crucial for accurate calculations and predictions in physics.

Force and Electric Field Relationship

The interaction between a test charge and an electric field can be expressed through a simple yet powerful relationship. This relationship is given by the formula: \[ F = qE \]Here, \( F \) is the force exerted on the charge, \( q \) is the magnitude of the charge, and \( E \) is the electric field strength. This equation tells us that the force a test charge experiences is directly proportional to both the magnitude of the charge and the strength of the electric field.

• When \( q \) is larger, the force is stronger.
• When \( E \) is larger, the force is stronger.

Since electric fields can exert forces at a distance, understanding this relationship is fundamental to predicting how charged objects will behave in an electric field. This formula simplifies the complex interactions into manageable calculations.For example, if you know two of the three variables (force, test charge, or electric field strength), you can easily calculate the third. This relationship also helps clarify why test charges always move in the direction of the electric field lines when the charge is positive.

The Magnitude of Charge

The magnitude of a charge, denoted as \( q \), is a fundamental quantity in determining the behavior of electric forces. Calculating the magnitude of a charge involves using the relationship between the electric field (\( E \)), force (\( F \)), and the charge itself. In our problem, we rearranged the formula:\[ q = \frac{F}{E} \]This rearranged equation allows us to solve for \( q \), the magnitude of the charge, when the force and the electric field values are known.

• This step-by-step logic helps us derive the charge's magnitude accurately.
• After identifying known values, substituting and solving becomes straightforward.

In the given exercise, by substituting the values for the force (\( 8.00 \times 10^{-4} \mathrm{~N} \)) and the electric field (\( 0.450 \mathrm{~N} / \mathrm{C} \)), we found \( q \) to be \( 1.78 \times 10^{-3} \mathrm{~C} \). This process highlights how the magnitude of the charge directly influences the force experienced by it in an electric field, underscoring the vital role that charge magnitude plays in electromagnetic interactions.