- #1
photon_mass
- 28
- 0
potential = 0
particle of mass m charge q is projected with kinetic energy K at a nucleus mass M charge Q that is at rest. it is shot with 'perfect aim' (along the x axis).
find the distance x to the nucleus when dx/dt of particle is zero.
i know that F = dK/dt = cqQ/x^2 where c = 1/4pi(epsillon naught)
.5m(d/dt)v^2 = cqQ/x^2 (1)
the coulomb force is position dependent force, so:
.5m(dv^2/dx)(dx/dt) = cqQ/x^2 (2)
.5m(dv^3) = cqQ(dx/x^2) (3)
this doesn't make any sense to me. it makes more sense to take the time derivative in (1) just giving Newtons 2nd, and then doing the position dependence.
ma = cqQ/x^2
mvdv = cqQ(dx/x^2)
.5mv^2 = -cqQ/x
and then ?
i got nothing. any suggestions, clarifications or anything of the like would be much appreciated.
particle of mass m charge q is projected with kinetic energy K at a nucleus mass M charge Q that is at rest. it is shot with 'perfect aim' (along the x axis).
find the distance x to the nucleus when dx/dt of particle is zero.
i know that F = dK/dt = cqQ/x^2 where c = 1/4pi(epsillon naught)
.5m(d/dt)v^2 = cqQ/x^2 (1)
the coulomb force is position dependent force, so:
.5m(dv^2/dx)(dx/dt) = cqQ/x^2 (2)
.5m(dv^3) = cqQ(dx/x^2) (3)
this doesn't make any sense to me. it makes more sense to take the time derivative in (1) just giving Newtons 2nd, and then doing the position dependence.
ma = cqQ/x^2
mvdv = cqQ(dx/x^2)
.5mv^2 = -cqQ/x
and then ?
i got nothing. any suggestions, clarifications or anything of the like would be much appreciated.