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An electron is projected with an initial speed of \(3.2 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to twice the initial value?

Short Answer

Expert verified
The electron is at a distance of \( 1.01 \times 10^{-9} \, \text{m} \) from the proton.

Step by step solution

01

Identify the Initial Conditions

The initial speed of the electron is given as \( v_i = 3.2 \times 10^5 \, \text{m/s} \). The initial kinetic energy can be calculated with \( KE_i = \frac{1}{2} m v_i^2 \), where \( m \) is the mass of the electron.
02

Determine the Final Speed

We need to find when the electron's speed is twice its initial speed. Thus, the final speed is \( v_f = 2v_i = 2 \times 3.2 \times 10^5 \, \text{m/s} \).
03

Calculate the Final Kinetic Energy

The final kinetic energy \( KE_f \) is given by \( KE_f = \frac{1}{2} m v_f^2 \). Substituting \( v_f = 2v_i \), we find \( KE_f = \frac{1}{2} m (2v_i)^2 = 2 m v_i^2 \).
04

Use Conservation of Energy

Initially, the potential energy \( PE_i = 0 \) because the electron is at a great distance from the proton. At the final point, \( PE_f = -\frac{k e^2}{r} \), where \( e \) is the electron charge, and \( k \) is Coulomb's constant. Conservation of energy gives \( KE_i + PE_i = KE_f + PE_f \).
05

Substitute and Solve for Distance

Substituting the expressions from Steps 1 and 3: \( \frac{1}{2} m v_i^2 + 0 = 2 m v_i^2 - \frac{k e^2}{r} \). Simplifying gives \( \frac{k e^2}{r} = \frac{3}{2} m v_i^2 \). Solving for \( r \), the distance is \( r = \frac{2 k e^2}{3 m v_i^2} \).
06

Insert Known Values

Substitute the known values: electron charge \( e = 1.6 \times 10^{-19} \, C \), Coulomb's constant \( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), and electron mass \( m = 9.11 \times 10^{-31} \, kg \). Calculate \( r \) using these values.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
Kinetic energy depends heavily on the speed of the object, as indicated by the velocity squared term in the formula. This means that even small changes in speed result in significant changes in kinetic energy.
In the context of the original exercise, the electron has an initial kinetic energy resulting from its initial speed \( v_i = 3.2 \times 10^5 \, \text{m/s} \).
As the electron approaches the proton and speeds up, its kinetic energy increases. When its speed doubles, the kinetic energy increases by a factor of four due to the velocity squared relationship.
Potential Energy
Potential energy is the energy stored in an object due to its position relative to other objects. It commonly arises in systems where objects exert forces on each other, such as gravitational or electrostatic forces.
In this problem, potential energy becomes important as the electron moves closer to the proton. Initially, when the electron is far away, we consider its potential energy \( PE_i = 0 \) because it is at an effectively infinite distance.
As the electron nears the proton, potential energy is given by \( PE_f = -\frac{k e^2}{r} \). This expression comes from the electrostatic interaction between two charges, where \( k \) is Coulomb's constant, \( e \) is the charge of the electron, and \( r \) is the separation distance between the proton and electron.
Understanding potential energy helps explain how energy conservation works in the electron-proton interaction, enabling it to accelerate by converting potential energy into kinetic energy.
Coulomb's Constant
Coulomb's constant \( k \) is crucial when studying electrostatic forces between charged particles. It appears in Coulomb's law, which describes the force between two point charges: \[ F = \frac{k |q_1q_2|}{r^2} \].
Here, \( k \) has a value of \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \). This constant dictates how strongly charged particles interact with each other over distance.
In the exercise, Coulomb's constant is used to calculate the potential energy of the electron as it nears the proton. This is essential for applying the conservation of energy principle to the motion of the electron, determining how the potential energy converts to kinetic energy, thus influencing the electron's dynamics.
Electron Dynamics
Electron dynamics explores how electrons move, particularly under various forces. In this problem, the electron is influenced by the electrostatic force exerted by the proton.
The electron's dynamics are governed by the balance of kinetic and potential energies, linked through the conservation of energy principle. Initially, the electron moves at \( 3.2 \times 10^5 \, \text{m/s} \). As it approaches the proton, its speed—and thus its kinetic energy—increases.
By understanding electron dynamics, one can calculate how close the electron gets to the proton when its speed doubles. The exercise uses energy conservation, without needing to compute forces directly, to determine this distance.
This approach simplifies the analysis of electron motion and highlights the core principles governing charged particles' behavior.

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