Problem 3 - Applied Physics

Question

An electric field exerts a force of \(2.50 \times 10^{-4} \mathrm{~N}\) on a positive test charge of \(5.00 \times 10^{-4} \mathrm{C}\). Find the magnitude of the field at the charge location.

Step-by-Step Solution

Answer

The magnitude of the electric field is 0.5 N/C.

Step 1: Identify Known Values

The problem provides the following information: the force exerted by the electric field on the charge, which is \( F = 2.50 \times 10^{-4} \; \mathrm{N} \), and the charge, which is \( q = 5.00 \times 10^{-4} \; \mathrm{C} \).

Step 2: Understand the Relationship

The electric field \( E \) at a point can be calculated using the formula \( E = \frac{F}{q} \), where \( F \) is the force on the charge and \( q \) is the magnitude of the charge. This formula relates the force to the electric field and the charge.

Step 3: Substitute Values into the Formula

Substitute the known values into the formula to find the electric field: \( E = \frac{2.50 \times 10^{-4} \; \mathrm{N}}{5.00 \times 10^{-4} \; \mathrm{C}} \).

Step 4: Calculate the Electric Field

Perform the division: \( E = \frac{2.50 \times 10^{-4}}{5.00 \times 10^{-4}} = 0.5 \; \mathrm{N/C} \).

Step 5: State the Final Result

The magnitude of the electric field at the location of the charge is \( 0.5 \; \mathrm{N/C} \).

Key Concepts

Electric Force

Electric force is a fundamental force in physics that describes the attraction or repulsion between two charged objects. This force is due to their electric charges, and it can either pull the objects together or push them apart, depending on whether the charges are of like or opposite types.
The electric force is central to determining how charged particles interact with each other.
In any scenario involving electric charges, this force acts as a defining factor. The size of the electric force depends on two main factors:

• The amount of charge on both objects.
• The distance between the objects.

In this exercise, the electric force is given directly as a force acting on a test charge, making it easier for us to calculate related quantities like the electric field.

Test Charge

A test charge is an imaginary or theoretical charge with a very small magnitude used to probe the electric field strength at a particular point without disturbing the field. Think of it like a tiny, harmless guest in the field, allowing us to explore the environment.
Typically, these charges are positive, which makes it simpler to define the direction and nature of an electric field.

• Using a test charge helps simplify calculations and conceptualize how electric fields behave.
• The impact of the test charge on the field is negligible, allowing for an accurate measurement of the field's original properties.

In this problem, the given test charge of \( 5.00 \times 10^{-4} \; \mathrm{C} \) is used to measure the effect of the electric field at its location.

Electric Field Magnitude

The concept of electric field magnitude helps us quantify how strong an electric field is at a particular point. It is defined as the force exerted per unit charge at that point.
The electric field magnitude \( E \) can be calculated using the relationship:

• \( E = \frac{F}{q} \)

Where:

• \( F \) is the electric force.
• \( q \) is the charge experiencing the force.

The formula indicates that if you know the force on a charge and the magnitude of the charge itself, you can easily calculate how strong the electric field is.
In our example, this resulted in a field of \( 0.5 \; \mathrm{N/C} \), suggesting a moderate field strength at the location of the charge.

Coulomb's Law

Coulomb's Law provides a way to calculate the electric force between two point charges. This fundamental principle explains that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. For two point charges, this can be expressed as:

• \( F = k \frac{|q_1 q_2|}{r^2} \)

Where:

• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \; \mathrm{Nm}^2/\mathrm{C}^2) \).
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the distance between the charges.

Coulomb's Law helps us understand the principles behind electric forces and fields.
Even though the law wasn't needed directly in the given exercise, it underpins the fundamental concepts we explored, explaining why charges exert forces as they do.
This knowledge is instrumental when dealing with electric fields and forces across various problems.