Four identical particles each have charge q and mass m. They are released from rest at the vertices of a square of side L. How fast is each charge moving when their distance from the center of the square doubles?
To find the speed of each charge when their distance from the center of the square doubles, we can use the law of conservation of energy.
First, let's consider the initial situation when the charges are at the vertices of the square. At this point, the charges are at rest, so their initial kinetic energy is zero.
The charges repel each other due to their identical charges, therefore, work is done to move them away from each other. As the charges move to double the distance from the center, their potential energy increases. Let's denote the initial distance from the center as R.
The potential energy U between two charges can be given by Coulomb's law: U = k * (q^2) / r, where k is the electrostatic constant, q is the charge, and r is the distance between the charges.
Since the charges are at the vertices of a square, the distance r between them is L. Hence, the potential energy between two charges at the vertices is U_initial = k * (q^2) / L.
When the charges move to double the distance from the center, their new distance from each other becomes √2L (using Pythagorean theorem).
Now, the potential energy between two charges at this new distance can be calculated as U_final = k * (q^2) / (√2L).
According to the law of conservation of energy, the decrease in potential energy is converted into an increase in kinetic energy.
The change in potential energy ΔU is given by: ΔU = U_initial - U_final.
The change in kinetic energy is equal to the change in potential energy: ΔK = ΔU.
The initial kinetic energy is zero, and the final kinetic energy is given by: K_final = (1/2) * m * v^2, where m is the mass of each charge and v is their final velocity.
Therefore, we can equate the change in kinetic energy to the change in potential energy: ΔK = ΔU.
(1/2) * m * v^2 = ΔU = U_initial - U_final.
Substituting the expressions for U_initial and U_final, we have:
(1/2) * m * v^2 = k * (q^2) / L - k * (q^2) / (√2L).
Simplifying further:
(1/2) * m * v^2 = k * (q^2) / L * (1 - 1/√2).
Now, solving for v:
v^2 = (2 * k * (q^2) / L) * (1 - 1/√2).
Taking the square root of both sides:
v = √[(2 * k * (q^2) / L) * (1 - 1/√2)].
Hence, the speed of each charge when their distance from the center of the square doubles is given by √[(2 * k * (q^2) / L) * (1 - 1/√2)].