A particle of mass m and charge q moves at a high speed along the x-axis.It is initially near x = -∞ and it ends up near x = +∞.
A second charge Q is fixed at the point x=0,y = -d.As the moving charge passes the stationary charge,its x component of velocity does not change appreciably,but it acquires a small velocity in y-direction.Determine the angle through which the moving charge is deflected.
(A)θ=tan-1(qQ2πε0dmv2)
(B)θ=sin-1(qQ2πε0dmv2)
(C)θ=tan-1(qQ4πε0dmv2)
θ=sin-1(qQ4πε0dmv2)
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2 Answers
486Let 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!
1133(A)
Aditya Agarwal how did you solve it?- Sourish Ghosh First I assumed that the particle would not go much above the x-axis, i.e. the final y coordinate of the particle will be very small since it has a very high v(x). Using this assumption, I found out a relation between v(x), v(y) and x. Then tan(theta) = v(y)/v(x) at x = infinity.
- Aditya Agarwal ok. thanks. :)
