 # A charge +Q is fixed. Another charge +2q and mass M is projected from a distance R from the fixed charge at and angle 30 with the horizontal towards the fixed charge (like a projectile) at a speed of vFind the minimum separation between the two charges if the velocity becomes 1/√3 times of the projected velocity at this moment (assume gravity to be absent)ANSWER : sqrt(3) R/2, i have been trying energy conservation but unable to get this answer plz.. help ..thnx in advance

9 years ago

distance will be minimum when there will be no component ofthe velcty of the movable particle towards the centre  (i.e the fxed particle)

in this case you get a right andled triangle and the asnwer comes root 3 R/2

no need to conserver energy,

use basic concepts!!!

9 years ago

Ya,that''s fine but why should we conserve angular momentum ,because net torque acting on the charge 2q is not zero(due to force exerted at an angle by charge Q),so angular momentum should not remain conserve,plz correct me if iam wrong..

4 years ago
Angular momentum of 2q charge about Q is constant . ThereforemvRsin300=mv3–√rminsinθmvRsin300=mv3rminsinθWhen the two charges will be at minimum distance their relative velocity along the line joining them is zero . So θ=900θ=900mvR2=mvrmin×13–√mvR2=mvrmin×13rmin=3–√R2.