Chapter 20: Problem 22
Find the magnitude of the electric force on a \(2.0-\mu \mathrm{C}\) charge in a \(100-\mathrm{N} / \mathrm{C}\) electric field.
Short Answer
Expert verified
The magnitude of the electric force on a 2.0-μC charge in a 100-N/C electric field is \(0.2 N\).
Step by step solution
01
Identify given variables
The given variables are the charge (\(Q = 2.0\mu C\)) and the electric field (\(E = 100N/C\)). We must convert the charge from microcoulombs to coulombs by multiplying by \(1x10^{-6}\). So, the charge is \(2.0 \times 10^{-6} C\).
02
Use Coulomb's Law to find electric force
Coulomb's Law is stated as: \(F = QE\), where \(F\) is the electric force, \(Q\) is the charge, and \(E\) is the electric field. Plugging in the values: \(F = (2.0 \times 10^{-6} C)(100 N/C)\)
03
Compute for electric force
By multiplying the given values, the electric force on the charge is \(0.2 N\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coulomb's Law
Coulomb's Law is a fundamental principle in physics that describes the force between two charged objects. According to this law, the electric force (\( F \)) acting on charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. However, if the electric field (\( E \)) is known, the force on a single charge can be calculated using the simple expression:
This straightforward formula makes it easier to calculate the force without needing to consider the distance between charges, focusing instead on the interaction of a charge with its surrounding electric field.
- \( F = QE \)
This straightforward formula makes it easier to calculate the force without needing to consider the distance between charges, focusing instead on the interaction of a charge with its surrounding electric field.
Electric Field
The concept of an electric field is crucial in understanding how electric forces operate from a distance. An electric field is a region around a charged object where other charges experience a force. It can be visualized as lines emanating from positive charges and terminating at negative charges.
The electric field strength (\( E \)) is a vector quantity marked by both magnitude and direction. It quantifies the force per unit charge and is expressed as:
The electric field strength (\( E \)) is a vector quantity marked by both magnitude and direction. It quantifies the force per unit charge and is expressed as:
- \(E = F/Q\)
Charge Conversion
Charge conversion is a fundamental step when working with electrical problems, especially when the given charge units are not in the standard Coulomb unit. Often, charges are expressed in microcoulombs (\( \mu C \)).
To convert microcoulombs to coulombs, you need to multiply the charge by \( 1 \times 10^{-6} \). For instance, a charge of \(2.0 \mu C \) is equivalent to:
To convert microcoulombs to coulombs, you need to multiply the charge by \( 1 \times 10^{-6} \). For instance, a charge of \(2.0 \mu C \) is equivalent to:
- \( 2.0 \times 10^{-6} C \)
Physics Problem Solving
Solving physics problems involves several clear, systematic procedures to ensure accurate results. Here's a helpful strategy:
1. **Identify the Given Variables**: Understanding what is provided in the problem is the first step. Clearly note down known values and units.
1. **Identify the Given Variables**: Understanding what is provided in the problem is the first step. Clearly note down known values and units.
- Example: Charge (\(Q = 2.0 \mu C\)) and Electric Field (\(E = 100 N/C\)).
- Example: Convert \(2.0 \mu C\) to \(2.0 \times 10^{-6} C\)
- Example: Use \( F = QE \) to find the electric force.