Problem 50
Question
A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s} .\) Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 \(\mathrm{m} / \mathrm{s} ?\)
Step-by-Step Solution
Verified Answer
The acceleration of the sphere is determined by using conservation of energy to find the distance to the fixed charge and then applying Coulomb's Law and Newton's second law.
1Step 1: Understand the Forces Involved
In this problem, we are dealing with electrostatic forces between two point charges. The force \( F \) on the sphere with charge \( q_2 \) due to the fixed charge \( q_1 \) can be calculated using Coulomb's law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where \( k = 8.99 \times 10^9 \, \mathrm{N m^2/C^2} \) is the electrostatic constant, and \( r \) is the distance between the charges.
2Step 2: Calculate the Distance Where Speed is 25 m/s
At the point when the sphere's speed is 25 m/s, the sphere has moved closer to \( q_1 \). We need to find this new distance \( r \). Use energy conservation between initial and this point to calculate the change in distance:- Initial kinetic energy: \( KE_i = \frac{1}{2} m v_i^2 \)- Initial potential energy: \( PE_i = \frac{k q_1 q_2}{r_i} \)- Final kinetic energy: \( KE_f = \frac{1}{2} m v_f^2 \)- Final potential energy: \( PE_f = \frac{k q_1 q_2}{r_f} \)Apply conservation of energy: \[ KE_i + PE_i = KE_f + PE_f \]Solve for \( r_f \).
3Step 3: Calculate the Final Position r_f
Initially, \( v_i = 40.0 \, \mathrm{m/s} \), \( v_f = 25.0 \, \mathrm{m/s} \), \( r_i = 6.00 \, \mathrm{cm} = 0.06 \, \mathrm{m} \). Set the conservation of energy equation:\[ \frac{1}{2} m v_i^2 + \frac{k q_1 q_2}{r_i} = \frac{1}{2} m v_f^2 + \frac{k q_1 q_2}{r_f} \]Solve for \( r_f \). After simplification, we find:\[ r_f = \frac{k q_1 q_2}{\frac{1}{2} m (v_i^2 - v_f^2) + \frac{k q_1 q_2}{r_i}} \]
4Step 4: Determine Acceleration Using Coulomb's Force
With the value of \( r_f \) determined, apply Coulomb's law to find the force, then use Newton's second law to find acceleration:- Coulomb's force: \[ F = \frac{k |q_1 q_2|}{r_f^2} \]- Newton's second law: \[ a = \frac{F}{m} \]Substitute the computed force into the equation to find acceleration.
5Step 5: Solve for Acceleration
Plug all known values into the final form to calculate acceleration:\( k = 8.99 \times 10^9 \, \mathrm{N m^2/C^2} \), \( q_1 = 5.00 \times 10^{-6} \, \mathrm{C} \), \( q_2 = 2.00 \times 10^{-6} \, \mathrm{C} \), \( m = 4.00 \times 10^{-3} \, \mathrm{kg} \).Using the determined \( r_f \), calculate:\[ F = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-6} \times 2.00 \times 10^{-6}}{r_f^2} \]\[ a = \frac{F}{4.00 \times 10^{-3}} \]Compute the calculated value for \( a \).
Key Concepts
Electrostatic ForcesConservation of EnergyNewton's Second Law
Electrostatic Forces
Electrostatic forces are the attractive or repulsive interactions between any two charged particles. These forces are essential in understanding how charges interact in space. The fundamental principle governing electrostatic forces is Coulomb's Law. Coulomb's Law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically, this is expressed as:
- \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2} \),
- \( q_1 \) and \( q_2 \) are the values of the respective charges,
- \( r \) is the distance between the two charges.
Conservation of Energy
The concept of conservation of energy is a fundamental principle in physics. It describes how the total energy in a closed system remains constant over time, even as the system undergoes changes. In the context of moving charged particles, this principle plays a crucial role when analyzing their motion and interactions.
When a charged sphere moves through an electric field, two forms of energy are primarily considered:
When a charged sphere moves through an electric field, two forms of energy are primarily considered:
- Kinetic Energy (KE): The energy due to its motion, given by \( KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.
- Potential Energy (PE): The energy due to its position relative to another charge, given by \( PE = \frac{k q_1 q_2}{r} \).
- Initial total energy: \( KE_i + PE_i \)
- Final total energy: \( KE_f + PE_f \)
Newton's Second Law
Newton's Second Law is a key principle that links force, mass, and acceleration. It is commonly expressed with the formula:
By rearranging the equation, we can express acceleration as \( a = \frac{F}{m} \). Here, once we have the force acting on the charged sphere, which is calculated by substituting the determined distance between the charges into Coulomb's force equation, we can use this form of Newton's Second Law to solve for the acceleration. In this way, Newton's Second Law bridges the gap between the powerful concept of force and the resulting motion, quantified by acceleration.
- \( F = ma \)
- \( F \) is the force applied to an object,
- \( m \) is the object's mass,
- \( a \) is the acceleration of the object.
By rearranging the equation, we can express acceleration as \( a = \frac{F}{m} \). Here, once we have the force acting on the charged sphere, which is calculated by substituting the determined distance between the charges into Coulomb's force equation, we can use this form of Newton's Second Law to solve for the acceleration. In this way, Newton's Second Law bridges the gap between the powerful concept of force and the resulting motion, quantified by acceleration.
Other exercises in this chapter
Problem 46
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A point charge \(q_{1}=4.00 \mathrm{nC}\) is placed at the origin, and a second point charge \(q_{2}=-3.00 \mathrm{nC}\) is placed on the \(x\) -axis at \(x=+20
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A small sphere with mass \(5.00 \times 10^{-7} \mathrm{kg}\) and charge \(+3.00 \mu C\) is released from rest a distance of 0.400 \(\mathrm{m}\) above a large h
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