Deflection of the Particle

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)

2 Answers

486
Niraj kumar Jha ·

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)

1133
Sourish Ghosh ·

(A)

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