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Total charge -Q is uniformly spread alon...

Total charge `-Q` is uniformly spread along length of a ring of radius R. A small test `+q` of mass m is kept at the center of the ring .
(a) Show that the particle executes a single harmonic oscillation.
(b) Obtain its time period.

Text Solution

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In fig, we have shown `-Q` charge distributed uniformly over a ring of radius R with center O. A and B are two pionts on the ring at the ends of a diameter.
If we take line elements of charge at A and B, having unit length, then charge on each element `= - (Q)/(2pi R)`
A small test charge `+q` of mass m kept at the center O of the ring is given a gentle push to P where `OP = z`
If `AP = BP = r = (z^(2) + R^(2))^(1//2) and /_ APO = theta`, fig, then force on q to the two line elements
`dF = 2 (- (Q)/(2pi R)) q xx (1)/(4pi in_(0)) (1)/(r^(2)) cos theta`
Total force on the charge q, due to entire ring
`F = -(Qq)/(pi R) (pi R) (1)/(4pi in_(0)) (1)/(r^(2)) , (z)/(r) = (Qqz)/(4pi in_(0) (z^(2) + R^(2))^(3//2))`
When `z lt lt R, F = - (Qqz)/(4pi in_(0) R^(3)) = -Kz where K = (Qq)/(4pi in_(0) R^(3))` = constant.
Clearly force on q is propertional to negative of its displacement. Therefore, motion of q is simple harmonic.
`omega = sqrt((K)/(m)) and T = (2pi)/(omega) = 2pi sqrt((m)/(K)) = 2pi sqrt((m 4pi in_(0) R^(3))/(Qq)) = 2pi sqrt((4pi in_(0) m R^(3))/(Qq))`
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