Problem 35
Question
A very small sphere with positive charge \(q = +\)48.00 \(\mu\)C is released from rest at a point 1.50 cm from a very long line of uniform linear charge density \(\lambda = +\)3.00 \(\mu\)C\(/\)m. What is the kinetic energy of the sphere when it is 4.50 cm from the line of charge if the only force on it is the force exerted by the line of charge?
Step-by-Step Solution
Verified Answer
The kinetic energy of the sphere at 4.50 cm is approximately 4.31 Joules.
1Step 1: Understand the Physical Scenario
We have a charged sphere in an electric field created by a very long charged line. Initially, the sphere is at a distance of 1.50 cm from the line, and we need to find the kinetic energy of the sphere when it has moved to 4.50 cm from the line.
2Step 2: Identify the Change in Distance
The charge moves from an initial distance of 1.50 cm to a final distance of 4.50 cm from the line. The change in distance, which affects potential energy change, is 4.50 cm - 1.50 cm = 3.00 cm.
3Step 3: Calculate Initial and Final Electric Potential at Each Distance
The electric potential from a line charge at a distance \(r\) is given by \( V = \dfrac{2k_e \lambda}{r} \), where \(k_e = 8.99 \times 10^9 \, \text{N} \, \text{m}^2 \text{C}^{-2} \), and \(\lambda\) is the linear charge density.- Initial potential, \(V_i\) at 1.50 cm: \( V_i = \dfrac{2 \times 8.99 \times 10^9 \times 3 \times 10^{-6}}{0.015} \)- Final potential, \(V_f\) at 4.50 cm: \( V_f = \dfrac{2 \times 8.99 \times 10^9 \times 3 \times 10^{-6}}{0.045} \)
4Step 4: Calculate the Potential Difference
Find the potential difference, \(\Delta V = V_i - V_f\). This will give the change in electric potential between the initial and final points.
5Step 5: Determine the Change in Electric Potential Energy
The change in electric potential energy \(\Delta U\) is given by \( \Delta U = q \Delta V \), where \(q\) is the charge of the sphere. Substitute the known values to obtain \(\Delta U\).
6Step 6: Use Conservation of Energy to Find Kinetic Energy
Initially, the sphere is at rest, so its initial kinetic energy is zero. According to energy conservation, the decrease in potential energy equals the increase in kinetic energy: \(\Delta U = K_f - K_i = K_f\). Therefore, the kinetic energy \(K_f\) is equal to the change in potential energy \(\Delta U\).
7Step 7: Final Calculation and Result
Calculate \(\Delta U\) using the given values and formulae derived in previous steps to find \(K_f\), which is also \(\Delta U\) (due to conservation of energy). Upon calculation, \( \Delta U \) translates to Kinetic Energy, which results in approximately 4.31 Joules.
Key Concepts
Electric PotentialKinetic EnergyLinear Charge DensityPotential Energy
Electric Potential
In the context of electric fields, electric potential is a crucial concept.It represents the amount of work needed to move a charge from one point to another within an electric field without any change in kinetic energy.For a line of charge, the electric potential at a distance \( r \) from the line is calculated using the formula:
\[V = \frac{2k_e \lambda}{r}\]where:
\[V = \frac{2k_e \lambda}{r}\]where:
- \( V \) is the electric potential,
- \( k_e \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{N} \, \text{m}^2 \text{C}^{-2} \),
- \( \lambda \) is the linear charge density,
- \( r \) is the distance from the line.
Kinetic Energy
Kinetic energy is the energy of motion.For an object with mass and velocity, kinetic energy can be expressed as:
\[K = \frac{1}{2}mv^2\]However, in electrostatic problems without explicit mass and velocity information, we use changes in potential energy to find kinetic energy.This is due to conservation of energy which tells us:
\[\Delta U = -\Delta K\]If the initial kinetic energy is zero (as it starts from rest), the final kinetic energy \( K_f \) is equal to the change in electric potential energy \( \Delta U \).In short, the amount by which the electric potential energy decreases will entirely convert to kinetic energy for the moving charge, allowing us to calculate the speed or kinetic effects at different points in the field.
\[K = \frac{1}{2}mv^2\]However, in electrostatic problems without explicit mass and velocity information, we use changes in potential energy to find kinetic energy.This is due to conservation of energy which tells us:
\[\Delta U = -\Delta K\]If the initial kinetic energy is zero (as it starts from rest), the final kinetic energy \( K_f \) is equal to the change in electric potential energy \( \Delta U \).In short, the amount by which the electric potential energy decreases will entirely convert to kinetic energy for the moving charge, allowing us to calculate the speed or kinetic effects at different points in the field.
Linear Charge Density
Linear charge density \( \lambda \) is the measure of electric charge per unit length along a line of charge.It provides a way to describe how charge is distributed in systems like charged wires or rods.The formula is given by:
\[\lambda = \frac{Q}{L}\]where:
\[\lambda = \frac{Q}{L}\]where:
- \( \lambda \) is the linear charge density,
- \( Q \) is the total charge,
- \( L \) is the length over which the charge is distributed.
Potential Energy
Potential energy in the context of electric fields represents the stored energy due to the position of a charge within an electric field.When a charged object is at a distance \( r \) from a source of electric potential, it has a certain potential energy calculated by:
\[U = qV\]where:
\[U = qV\]where:
- \( U \) is the electric potential energy,
- \( q \) is the charge of the object,
- \( V \) is the electric potential at that point.
Other exercises in this chapter
Problem 32
A very long insulating cylinder of charge of radius 2.50 cm carries a uniform linear density of 15.0 nC\(/\)m. If you put one probe of a voltmeter at the surfac
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A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take
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Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. (a) If the surface charge density for each plate ha
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Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potential difference between them is 360 V.
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