(sorry Fred I had missed this bit:) >> > Since the potential V at a distance r from a charge = k * q/2r volts >> >> V=k * q/r (cf same post, come on you had it right two days ago) >> > That is total energy,
No, that is voltage! Easy to check by yourself: take the derivative of this formula wrt r to get the electric field (reminder: derivative of 1/x is -1/x^2), which if you multiply it by your electron charge -q yields the coulombic force. If force formula works out as k*(-q)*(+q)/r^2 as you expect (negative i.e. towards the proton) then V formula is right, agreed? > Potential plus Kinetic.A 0.5 kilogram raindrop > (hailstones like those > that trashed my roof in October 2004) formed at 1 Km or more has a > potential energy of mgh, > but while it is falling it's gaining kinetic energy too. No? Yes, precisely: E = P.E. + K.E. is constant so while one decreases the other increases. In the case of the H electron at 0.53 10^-10m of the proton: P.E. = -27.2 eV K.E. = +13.6 eV => E = -27.2 + 13.6 = -13.6 If you could let the electron (hailstone) fall towards the proton (the earth) it's P.E. would indeed decrease, say by 10 eV to -37.2 eV, so it's K.E. would increase to 13.6+10 = 23.6 eV. Michel
