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Two identical particles of mass M carrying a charge Q each initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed V the closet distance of the approach be?
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? Two identical particles of mass M carrying a charge Q each initially...
At the distance of closest approach , d the kinetic energy K of moving mass, m is converted in to potential energy, U of system of two particles.
Therefore ,
(1/2)mv^2=kQ^2/d. Therefore ,
d=(kQ^2)(2/mv^2)=(2kQ^2/mv^2).
As gravitational interactions are about 10^-39 times the electrostatic interaction, we have neglected gravitational potential energy of interaction of the particles.
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? Two identical particles of mass M carrying a charge Q each initially...
Problem Statement:
Two identical particles of mass M carrying a charge Q each are initially placed on a smooth horizontal plane. One particle is at rest, while the other is projected along the plane directly towards the first particle from a large distance with speed V. We need to determine the closest distance of approach between the two particles.

Solution:

1. Analysis of the System:
To solve this problem, we need to analyze the forces acting on the particles and understand their motion.

2. Coulomb's Law:
The force between two charged particles is given by Coulomb's law:
F = k * (Q1 * Q2) / r^2
where F is the force, k is the electrostatic constant, Q1 and Q2 are the charges of the particles, and r is the distance between them.

3. Motion of the Projected Particle:
The projected particle is initially at a large distance from the stationary particle and is projected towards it with a certain velocity V. As there are no other forces acting on the particles, the only force acting on the projected particle is the electrostatic force from the stationary particle.

4. Conservation of Energy:
Since the only force acting on the projected particle is the electrostatic force, we can use the conservation of energy to determine its motion. The initial kinetic energy of the projected particle is given by:
KE_initial = (1/2) * M * V^2
The final kinetic energy of the projected particle at the closest distance of approach is zero since it comes to rest.

5. Work-Energy Theorem:
Using the work-energy theorem, we can equate the initial kinetic energy to the work done by the electrostatic force to bring the particle to rest:
KE_initial = Work done by electrostatic force
(1/2) * M * V^2 = k * (Q1 * Q2) / r

6. Calculation of Closest Distance of Approach:
We can rearrange the equation to solve for the closest distance of approach (r):
r = k * (Q1 * Q2) / [(1/2) * M * V^2]

7. Substituting Values:
Substituting the given values of Q, M, and V into the equation, we can calculate the closest distance of approach.

8. Final Answer:
The closest distance of approach between the two particles is calculated using the equation:
r = k * (Q^2) / [(1/2) * M * V^2]

Summary:
The closest distance of approach between the two identical particles of mass M carrying a charge Q each, where one is at rest and the other is projected towards it with speed V, can be calculated using the equation r = k * (Q^2) / [(1/2) * M * V^2]. Substituting the given values of Q, M, and V will provide the final answer.
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? Two identical particles of mass M carrying a charge Q each initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed V the closet distance of the approach be?
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