Figure shows a charge +Q clamped at a point in free space. From a large distance another charge particle of charge -q an.d mass m is thrown toward +Q with an impact parameter d as shown with speed v. Find the distance of closest approach of the two particles.

Explanation

Here we can see that as +q moves toward +Q, a repulsive force acts on –q radially outward +Q. Here as the line of action of force passes through the fix charge, no torque act on +q relative to the fix point charge +Q, thus here we can say that with respect to +Q, the angular momentum of +q must remain constant. Here we can say that +q will be closest to +Q when it is moving perpendicularly to the line joining the two charges as shown.
If the closest separation in the two charges is `r_(min)`, from conservation of angular momentum we can write
`mvd=mv_0r_(min)" "...(1)`
Now from energy conservation, we have
`1/2mv^2=1/2mv_0^2+(KqQ)/(r_(min))`
Here we use from equation (1) `v_0=(vd)/(r_(min))`
`1/2mv^2=1/2mv^2(d^2)/(r_(min)^2)+(KqQ)/(r_(min))" "...(2)`
Solving equation (2) we'll get the value of `r_(min)` .