Problem 1
Question
A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and that has a magnitude of \(4.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?
Step-by-Step Solution
Verified Answer
(a) 0 J, (b) 0.75 J, (c) -2.06 J.
1Step 1: Understand the Relationship Between Work, Electric Force, and Electric Field
The work done by an electric force can be calculated by the formula \( W = qEd \cos \theta \), where \( q \) is the charge, \( E \) is the electric field strength, \( d \) is the displacement, and \( \theta \) is the angle between the electric field and the direction of displacement.
2Step 2: Work Calculation for 0.450 m to the Right
For part (a), since the movement is to the right while the electric field is vertical, the angle \( \theta \) between the electric field and the direction of displacement is \( 90^{\circ} \). Thus, \( \cos 90^{\circ} = 0 \). The work done is: \[ W = 28.0 \times 10^{-9} \, \text{C} \times 4.00 \times 10^4 \, \text{N/C} \times 0.450 \, \text{m} \times \cos 90^{\circ} \]\[ W = 0 \, \text{J} \].Since the force and displacement are perpendicular, no work is done.
3Step 3: Work Calculation for 0.670 m Upward
For part (b), the movement is in the same direction as the electric field (upward). Therefore, \( \theta = 0^{\circ} \). The work done is: \[ W = 28.0 \times 10^{-9} \, \text{C} \times 4.00 \times 10^4 \, \text{N/C} \times 0.670 \, \text{m} \times \cos 0^{\circ} \] \[ W = 28.0 \times 10^{-9} \times 4.00 \times 10^4 \times 0.670 \times 1 \]\[ W = 0.75 \, \text{J} \].
4Step 4: Work Calculation for 2.60 m at an Angle of 45.0° Downward
For part (c), the movement is at an angle \( 45.0^{\circ} \) downward, making the angle between the displacement and electric field \( 135^{\circ} \) (since the field is upward). The work done is: \[ W = 28.0 \times 10^{-9} \, \text{C} \times 4.00 \times 10^4 \, \text{N/C} \times 2.60 \, \text{m} \times \cos 135^{\circ} \] \[ \cos 135^{\circ} = -\frac{\sqrt{2}}{2} \], so:\[ W = 28.0 \times 10^{-9} \times 4.00 \times 10^4 \times 2.60 \times -\frac{\sqrt{2}}{2} \]\[ W \approx -2.06 \, \text{J} \].
Key Concepts
Electric ForceWork Done by Electric ForceAngle of DisplacementElectric Charge Calculation
Electric Force
Electric force is the attraction or repulsion interaction between any two charged objects. This force is a direct result of the electric field surrounding a charged particle. Think of the electric field as the force field produced by any charged object, which influences other charges present in its vicinity.
In mathematical terms, the electric force (\( F \)) can be calculated by using Coulomb's Law:
In mathematical terms, the electric force (\( F \)) can be calculated by using Coulomb's Law:
- \( F = k \frac{q_1 q_2}{r^2} \)
- where \( F \) is the force
- \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges
- and \( r \) is the distance between the two charges
Work Done by Electric Force
The work done by an electric force is critical in determining how much energy is needed or released when a charge moves in an electric field. The concept of work in physics relates the force applied to an object and the distance it moves in the direction of the force. In the context of electric fields, we use the formula:
- \( W = qEd \cos \theta \)
- where \( W \) is the work done
- \( q \) is the electric charge
- \( E \) is the electric field strength
- \( d \) is the displacement
- \( \theta \) is the angle between the electric field and the direction of displacement
Angle of Displacement
The angle of displacement is crucial when calculating the work done by electric force because it affects how effectively a force contributes to moving a charge. When a charge moves at an angle in an electric field, only a component of the force assists with this movement. This is why the angle \( \theta \) in the formula \( W = qEd \cos \theta \) is vital.
- If \( \theta = 0^{\circ} \), the movement is fully aligned with the electric field, so the cosine part becomes 1, maximizing the work done.
- If \( \theta = 90^{\circ} \), the charge moves perpendicular to the electric field, resulting in no work getting done.
- For angles greater than \( 90^{\circ} \), the work can actually become negative, meaning that the force is applied in such a way that it opposes the displacement.
Electric Charge Calculation
Electric charge calculation is fundamental when working with electric forces and fields. Understanding how to determine the values of charge is crucial for various applications in physics and engineering. A charge (\( q \)) quantifies the electricity held by an object, and its effect is observed through the electric field it generates.
- Charges are typically measured in coulombs (C), but in many practical applications, smaller units like nanocoulombs (nC) or microcoulombs (µC) are used, as they deal with relatively small quantities.
- Charge quantification helps in using formulas such as \( F = Eq \) to determine the force applied on charges in an electric field.
- In practical scenarios, electric charge calculations enable the analysis and design of circuits, understanding capacitors, and examining electrostatic scenarios and phenomena.
Other exercises in this chapter
Problem 2
Two very large charged parallel metal plates are 10.0 \(\mathrm{cm}\) apart and produce a uniform electric field of \(2.80 \times 10^{6} \mathrm{N} / \mathrm{C}
View solution Problem 3
How far from a \(-7.20 \mu \mathrm{C}\) point charge must a \(+2.30 \mu \mathrm{C}\) point charge be placed in order for the electric potential energy of the pa
View solution Problem 4
A point charge \(q_{1}=+2.40 \mu C\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu C\) moves from the point \(x=0.150 \mathrm{m}, y=
View solution