Question

How do i find the minimum initial speed in here?

Answer 1

`v_0 = Q/sqrt(4 pi epsilon_0 m)sqrt[1/a+2/(d-a)- (3+2sqrt2)/d]`

Explanation:

The potential energy of the particle of charge `-Q/2` at a point `(x,0)` is given by

`V(x) = 1/(4pi epsilon_0)[(Q times (-Q/2))/x+(2Q times(-Q/2))/(d-x)]`
`qquad = -Q^2/(8pi epsilon_0)[1/x+2/(d-x)]`

This potential has a maximum at a point between the two conducting spheres, as the graph below shows :

graph{-1/x-2/(6-x) [-2.17, 7.83, -5.1, -0.1]}

In order for a particle ejected from the surface of one of the spheres to reach the other, it must have enough initial kinetic energy to cross this potential maximum.

Now, to find the location of this maximum, we have to solve

`(dV)/dx = 0 implies Q^2/(8pi epsilon_0)[1/x^2-2/(d-x)^2]=0 implies (d-x)/x = sqrt 2`

Thus the maximum occurs at `x_0 = d/(1+sqrt 2)` and the maximum value for the potential energy is

`V_m = -Q^2/(8pi epsilon_0)[1/x_0+2/(d-x_0)]=-Q^2/(8pi epsilon_0) (3+2sqrt2)/d`

The total energy has to exceed this value for the charged particle to move from the surface of conductor (1) to conductor (2). Thus, the initial velocity must satisfy

`1/2mv^2 -Q^2/(8pi epsilon_0)[1/a+2/(d-a)] >= -Q^2/(8pi epsilon_0) (3+2sqrt2)/d`

so

`1/2mv^2 >= Q^2/(8pi epsilon_0)[1/a+2/(d-a)- (3+2sqrt2)/d]`

So, the minimum velocity required is

`v_0 = Q/sqrt(4 pi epsilon_0 m)sqrt[1/a+2/(d-a)- (3+2sqrt2)/d]`