# Find the minimum distance between 2 particles

• Helly123
In summary, the minimum distance between two point particles is when the potential energy is at its maximum and the kinetic energy is at its minimum.

## Homework Statement

A point particle of mass m and charge q(>0) approaches to a point particle Q(>0) at a fixed position. When the distance between the two particles is L, the speed of the moving particle is v. The permittivity of the vacuum is denoted as Epsilon0. Find the minimum distance between the two particles?

## The Attempt at a Solution

What i get is, the Q will repel q at the shortest distance q will ever able to reach. Because of the Force between them. So
I used
Vfinal ^2 = Vinitial ^2 -2as

S is the question
a is force/m

Force is kQq/r^2

I know i am wrong.
I don't know what epsilon for..

This is an energy problem. What is the initial potential energy and kinetic energy? And what will be the kinetic energy as the distance is minimized? Hint: What will happen after the distance reaches a minimum? They want you to use MKS units: ## F=\frac{Qq}{4 \pi \epsilon_o r^2} ##, with a related expression for the potential energy.

• Helly123
Helly123 said:
I used
Vfinal ^2 = Vinitial ^2 -2as

S is the question
a is force/m

Force is kQq/r^2
You cannot use the eqn ##v^2=u^2+2as## unless the acceleration ##a## is constant. Here the particle experiences a varying force throughout its approach towards ##Q##.
Think of a way you can employ the conservation of energy principle, or find an expression for the velocity of ##q## at an instant as a function of ##r##, the distance between ##q## & ##Q##.

• PKM said:
You cannot use the eqn ##v^2=u^2+2as## unless the acceleration ##a## is constant. Here the particle experiences a varying force throughout its approach towards ##Q##.
Think of a way you can employ the conservation of energy principle, or find an expression for the velocity of ##q## at an instant as a function of ##r##, the distance between ##q## & ##Q##.
Yes.. the problem i get is non constant of acceleration. @Charles Link also pointed out important thing for me.

Ek at min distance is zero.
Ep = kQq/r_2
r_2 is the min distance
K = 1/4pi.epsilon0
Ek1 + Ep1 = Ep2 + 0
Solve for r2

Thanks all

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