Let the min. distance between the particles be d' vel. at that point v'.
12mv2=12mv'2 +KqQd'
mvd=mv'd'....(conservation of ang. momentum about the fixed charge)v' is perp. to d' at min. seperation.
vcosθ=v'
Solving the three equation,
tanθsecθ=Qq2πe0dmv2
secθ≈1
(a)
- Sourish Ghosh Can you explain this line a bit more: vcos(theta) = v' ? And how are you saying that (theta) is the total deflection angle?Upvote·0· Reply ·2014-02-16 22:27:57
- Niraj kumar Jha oh,it should be v'cos(theta)=v.Since it says that the horizontal component does not change much.
- Sourish Ghosh Ok. But theta is the angle during min separation. How can you say that is the angle of deflection? Shouldn't the angle of deflection be the angle made by the particle when it is at x = +infinity ?
- Niraj kumar Jha my mistake,didn't read the question properly!