Problem 16
Question
Two stationary point charges \(+\)3.00 nC and \(+\)2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the \(+\)3.00-nC charge?
Step-by-Step Solution
Verified Answer
The speed of the electron is approximately \(4.2 \times 10^5\,\text{m/s}\) when it is 10.0 cm from the \(+3.00\,\text{nC}\) charge.
1Step 1: Identify Known Values
First, note down all known values we have from the problem:\[ q_1 = +3.00 \text{ nC} = 3.00 \times 10^{-9} \text{ C} \], \[ q_2 = +2.00 \text{ nC} = 2.00 \times 10^{-9} \text{ C} \], initial position of the electron is midway (25.0 cm) between the charges, final position is 10.0 cm from \(q_1\) which is 40.0 cm from \(q_2\). The initial potential energy must convert to kinetic energy because the electron is released from rest.
2Step 2: Setup Coulomb's Law
Using Coulomb's Law, calculate the initial electric potential energy at the midpoint between the charges and the final electric potential energy when the electron is 10.0 cm from the charge \(q_1\).The Coulombic potential energy between each charge and the electron when it's at the midpoint:\[U_{i1} = \frac{k_e \cdot q_1 \cdot e}{r_{i1}} = \frac{(8.988 \times 10^9 \text{ Nm}^2/\text{C}^2)(3.00 \times 10^{-9} \text{ C})(1.60 \times 10^{-19} \text{ C})}{0.25 \text{ m}}\]\[U_{i2} = \frac{k_e \cdot q_2 \cdot e}{r_{i2}} = \frac{(8.988 \times 10^9 \text{ Nm}^2/\text{C}^2)(2.00 \times 10^{-9} \text{ C})(1.60 \times 10^{-19} \text{ C})}{0.25 \text{ m}}\]Summing these gives \(U_i = U_{i1} + U_{i2}\).
3Step 3: Calculate Final Electric Potential Energy
Now calculate the final electric potential energy when the electron is 10.0 cm from \(q_1\) and 40.0 cm from \(q_2\):\[U_{f1} = \frac{k_e \cdot q_1 \cdot e}{r_{f1}} = \frac{(8.988 \times 10^9 \text{ Nm}^2/\text{C}^2)(3.00 \times 10^{-9} \text{ C})(1.60 \times 10^{-19} \text{ C})}{0.10 \text{ m}}\]\[U_{f2} = \frac{k_e \cdot q_2 \cdot e}{r_{f2}} = \frac{(8.988 \times 10^9 \text{ Nm}^2/\text{C}^2)(2.00 \times 10^{-9} \text{ C})(1.60 \times 10^{-19} \text{ C})}{0.40 \text{ m}}\]Summing these gives \(U_f = U_{f1} + U_{f2}\).
4Step 4: Use Energy Conservation
Since energy is conserved, the difference in the electric potential energy is converted into kinetic energy:\[\Delta U = U_i - U_f = K = \frac{1}{2}mv^2\]Solve for the speed \(v\) of the electron, using the mass \(m = 9.109 \times 10^{-31} \text{ kg}\) of the electron.Arrange to obtain:\[v = \sqrt{\frac{2 \Delta U}{m}}\]Calculate \(v\) by substituting \(\Delta U\).
5Step 5: Calculate the Speed
Plug in all the calculated values for \(\Delta U\) into the equation for \(v\) obtained in the previous step to find the speed of the electron when it is 10.0 cm from the \(3.00\,\text{nC}\) charge.
Key Concepts
Coulomb's LawElectric Potential EnergyConservation of EnergyMotion of Charged Particles
Coulomb's Law
In the realm of electrostatics, Coulomb's Law is a fundamental principle that helps us understand the force between charged particles. It states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is given by:
Understanding this law is key to solving problems concerning the interaction of charged particles, allowing us to calculate the force and potential energy between them.
- \( F = \frac{k_e \cdot |q_1 \cdot q_2|}{r^2} \)
Understanding this law is key to solving problems concerning the interaction of charged particles, allowing us to calculate the force and potential energy between them.
Electric Potential Energy
Electric potential energy is the energy a charged particle possesses due to its position in an electric field. It arises from the interactions between charges and depends on the configuration of those charges.
For a point charge \( q \) in the presence of another charge \( Q \), the electric potential energy \( U \) can be calculated using:
In problems like the current one, we assess the initial and final potential energy when a charged particle moves between two points. The change in electric potential energy is crucial for understanding how kinetic energy varies as the particle moves.
For a point charge \( q \) in the presence of another charge \( Q \), the electric potential energy \( U \) can be calculated using:
- \( U = \frac{k_e \cdot Q \cdot q}{r} \)
In problems like the current one, we assess the initial and final potential energy when a charged particle moves between two points. The change in electric potential energy is crucial for understanding how kinetic energy varies as the particle moves.
Conservation of Energy
The conservation of energy principle is a cornerstone of physics that applies to various physical systems, including charged particles in electric fields. It states that the total energy of an isolated system remains constant over time.
In this context, when an electron is released from rest, the electric potential energy it initially possesses transforms into kinetic energy as it gains speed moving through the electric field. The mathematical expression for this conservation is:
In this context, when an electron is released from rest, the electric potential energy it initially possesses transforms into kinetic energy as it gains speed moving through the electric field. The mathematical expression for this conservation is:
- \( \Delta U = U_i - U_f = K \)
- \( K = \frac{1}{2}mv^2 \)
Motion of Charged Particles
The motion of charged particles is heavily influenced by the electric fields they encounter due to other charges. When an electron is placed in the vicinity of larger charged bodies, it experiences forces that govern its acceleration and trajectory.
In this exercise, when the electron is released, it accelerates due to the attractive forces from the positively charged particles.
Analyzing the motion involves understanding how the initial potential energy landscape changes as the electron moves. By calculating the shift in energy, we determine the velocity the electron acquires as it travels to a new position.
In this exercise, when the electron is released, it accelerates due to the attractive forces from the positively charged particles.
Analyzing the motion involves understanding how the initial potential energy landscape changes as the electron moves. By calculating the shift in energy, we determine the velocity the electron acquires as it travels to a new position.
- Knowing the distance and energies involved lets us predict the path and speed the electron attains.
Other exercises in this chapter
Problem 14
A particle with charge \(+\)4.20 nC is in a uniform electric field \(\overrightarrow{E}\) directed to the left. The charge is released from rest and moves to th
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Point charges \(q_1 = +\)2.00 \(\mu\)C and \(q_2 = -\)2.00 \(\mu\)C are placed at adjacent corners of a square for which the length of each side is 3.00 cm. Poi
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Two point charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charge
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