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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?

Short Answer

Expert verified
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.

Step by step solution

01

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.
02

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 \(U_i = U_{i1} + U_{i2}\).
03

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 \(U_f = U_{f1} + U_{f2}\).
04

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 \(\Delta U\).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • \( F = \frac{k_e \cdot |q_1 \cdot q_2|}{r^2} \)
where \( F \) is the magnitude of the force, \( k_e \) is the Coulomb constant \((8.988 \times 10^9 \text{ Nm}^2/\text{C}^2 )\), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
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:
  • \( U = \frac{k_e \cdot Q \cdot q}{r} \)
where \( r \) is the distance between the charges.
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:
  • \( \Delta U = U_i - U_f = K \)
  • \( K = \frac{1}{2}mv^2 \)
where \( \Delta U \) represents the change in electric potential energy, \( m \) is the mass of the electron, and \( v \) is its velocity. Solving such problems requires equating the change in potential energy to the kinetic energy to find the speed of the electron.
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.
  • Knowing the distance and energies involved lets us predict the path and speed the electron attains.
This analysis is crucial for understanding not just electrostatic exercises, but also the behavior of particles in natural and technological processes.

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Most popular questions from this chapter

A point charge \(q_1\) is held stationary at the origin. A second charge \(q_2\) is placed at point a, and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \)J. When the second charge is moved to point \(b\), the electric force on the charge does \(-1.9 \times 10^{-8}\) J of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\)?

(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 3.75 kV? (b) What is the potential of the sphere 's surface relative to infinity?

Charge \(Q = +\)4.00 \(\mu\)C is distributed uniformly over the volume of an insulating sphere that has radius \(R =\) 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere?

When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.79 MeV when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has 88 protons. (a) What was the electric potential energy of the alpha\(-\)radon combination just before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus.

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\)-direction. The potential difference between point \(a\) (at \(x =\) 0.60 m) and point \(b\) (at \(x =\) 0.90 m) is 240 V. (a) Which point, \(a\) or \(b\), is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q = -\)0.200 \(\mu\)C is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

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