Please answer asap with complete solution thanks a lot. I need the solution to answer other questions much alike to this one. :)
Two electrons are 10^-15 meters apart. One electron is fixed on its position. The other, originally at rest, is released. How fast is the free electron moving when it is 10^-3 in from the fixed charge. (Solve using the Law of Conservation of Energy)
To solve this problem using the Law of Conservation of Energy, we need to consider the initial and final states of the system.
Let's define the initial state as when the free electron is at rest and at a distance of 10^-15 meters from the fixed electron. The final state is when the free electron is 10^-3 meters away from the fixed electron.
In the initial state, both electrons have potential energy due to their electrostatic interaction. The potential energy is given by the equation:
PE = k * (q1 * q2) / r
Where k is the electrostatic constant, q1 and q2 are the charges of the electrons (equal to the elementary charge e), and r is the distance between the electrons.
Since one of the electrons is fixed, its position does not change, and therefore, its potential energy remains constant throughout the process. Thus, we only need to consider the change in potential energy of the free electron.
Let's denote the initial potential energy of the free electron as PE_initial and the final potential energy of the free electron as PE_final.
The change in potential energy can be calculated as:
ΔPE = PE_final - PE_initial
In the final state, when the free electron is 10^-3 meters away from the fixed electron, the potential energy can be expressed as:
PE_final = k * (e * e) / r_final
Substituting the given values into the equation, we have:
PE_final = (9 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 1.6 × 10^-19 C) / (10^-3 m)
Now, let's calculate the initial potential energy of the free electron. Since it is initially at rest, it has no kinetic energy. Hence, all its initial energy is in the form of potential energy.
PE_initial = k * (e * e) / r_initial
Substituting the given values:
PE_initial = (9 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 1.6 × 10^-19 C) / (10^-15 m)
Now, we can calculate the change in potential energy:
ΔPE = PE_final - PE_initial
Using the values of PE_final and PE_initial calculated above:
ΔPE = (9 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 1.6 × 10^-19 C) / (10^-3 m) - (9 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 1.6 × 10^-19 C) / (10^-15 m)
Now, the change in potential energy is equal to the change in kinetic energy of the free electron. According to the Law of Conservation of Energy, the total energy in the system is conserved.
In the initial state, the free electron has only potential energy, and in the final state, it has both potential and kinetic energy.
Therefore, we can write:
ΔPE = ΔKE
The kinetic energy can be expressed as:
KE = 1/2 * m * v^2
Where m is the mass of the electron, and v is its velocity.
We know that the mass of an electron is approximately 9.11 × 10^-31 kg.
Now, let's solve for the velocity v by equating the change in potential energy to the change in kinetic energy:
ΔPE = ΔKE
(9 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 1.6 × 10^-19 C) / (10^-3 m) - (9 × 10^9 Nm^2/C^2) * (1.6 × 10^-19 C * 1.6 × 10^-19 C) / (10^-15 m) = 1/2 * (9.11 × 10^-31 kg) * v^2
Solving this equation for v will give us the velocity of the free electron when it is 10^-3 meters away from the fixed electron.
Please note that you may need to simplify and rearrange the equation further to get the final solution.