Problem 22
Question
An electron is acted on by an electric force of \(3.2 \times 10^{-14} \mathrm{~N}\). What is the magnitude of the electric field at the electron's location?
Step-by-Step Solution
Verified Answer
The magnitude of the electric field is \( 2.0 \times 10^{5} \, \mathrm{N/C} \).
1Step 1: Identify the Relationship
The force experienced by a charged particle in an electric field is given by the formula \( F = qE \), where \( F \) is the force, \( q \) is the charge, and \( E \) is the electric field.
2Step 2: Know the Charge of an Electron
The charge of an electron is \( q = -1.6 imes 10^{-19} \, \mathrm{C} \).
3Step 3: Rearrange the Formula
We need to find the magnitude of the electric field, so rearrange the formula to \( E = \frac{F}{|q|} \). Here, we take the absolute value of \( q \) because we are only finding the magnitude, not the direction.
4Step 4: Substitute the Values
Substitute the given force and the charge into the equation: \[ E = \frac{3.2 imes 10^{-14} \, \mathrm{N}}{1.6 imes 10^{-19} \, \mathrm{C}} \].
5Step 5: Calculate the Electric Field
Performing the division gives \[ E = 2.0 imes 10^{5} \, \mathrm{N/C} \]. This is the magnitude of the electric field.
Key Concepts
Force on Charged ParticleElectric Force FormulaElectron ChargeElectric Field Magnitude
Force on Charged Particle
When a charged particle, such as an electron, is placed in an electric field, it experiences a force. This force can cause the particle to move, and its magnitude depends on two main factors: the size of the charge on the particle and the strength of the electric field it is in. The direction of the force is determined by the nature of the charge:
- A positive charge experiences a force in the direction of the electric field.
- A negative charge, like an electron, experiences a force opposite to the direction of the electric field.
Electric Force Formula
The electric force acting on a charged particle is calculated using the formula \( F = qE \):
- \( F \) is the force measured in newtons \( \mathrm{N} \).
- \( q \) is the charge in coulombs \( \mathrm{C} \).
- \( E \) is the electric field in newtons per coulomb \( \mathrm{N/C} \).
Electron Charge
The charge of an electron is a fundamental constant in physics. It is denoted as \( q = -1.6 \times 10^{-19} \, \mathrm{C} \). This negative sign indicates that electrons are negatively charged particles. In calculations involving magnitude, we often use the absolute value, ignoring the negative sign, to focus solely on how much charge there is.
- This constant value helps in calculating forces in any scenario involving electrons.
- In practical applications, this charge is crucial for determining behaviors in electrical and electronic systems, such as in semiconductor devices.
Electric Field Magnitude
The magnitude of an electric field is a measurement of how strong the field is at a particular point in space. It is calculated once we rearrange the formula for electric force, focusing on the electric field:\[ E = \frac{F}{|q|} \]Where:
- \( E \) is the electric field magnitude in \( \mathrm{N/C} \).
- \( F \) is the force in newtons \( \mathrm{N} \).
- \(|q|\) is the absolute value of the charge \( \mathrm{C} \).
Other exercises in this chapter
Problem 17
An electron is placed on a line connecting two fixed point charges of equal charge but opposite sign. The distance between the charges is \(30.0 \mathrm{~cm}\)
View solution Problem 21
If the distance from a charge is doubled, is the magnitude of the electric field (1) increased, (2) decreased, or (3) the same compared to the initial value? (b
View solution Problem 23
An electron is acted on by two electric forces, one of \(2.7 \times 10^{-14} \mathrm{~N}\) acting upward and a second of \(3.8 \times 10^{-14} \mathrm{~N}\) act
View solution Problem 24
What are the magnitude and direction of the electric field at a point \(0.75 \mathrm{~cm}\) away from a point charge of \(+2.0 \mathrm{pC} ?\)
View solution