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Two charges -q each are fixed separated by distance 2d. A third charge q of mass m placed at the midpoint is displaced slightly by x (x < < d) perpendicular to the line joining the two fixed charged as shown in figure. Show thatq will perform simpleharmonic oscillation of time period. T = [(8pi^(3) epsilon_(0)md^(3))/q^(2)]^(1//2) |
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Answer» <P> Solution :Suppose charge at A and B are - q and O is mid point of AB and PO is X.`therefore AB = AO + OB` =d+d =2d `x lt d` and `angleAPO =theta` m is mass of charge q. Attractive force by each charge A and B on charge at P, `F = (k(q)(q))/r^(2)` where r = AP = BP `F sin theta`COMPONENTS of force are of same magnitude but in opposite directions hence, their resultant is zero and Fcos6 components are in same direction, `F. = 2F cos theta` `=(2kq^(2))/r^(2) cos theta` But from figure, `r = sqrt(d^(2) + x^(2))` and `cos theta = x/r` `therefore F. = (2kq^(2))/(d^(2) + x^(2))^(2).x/(d^(2) + x^(2))^(1//2)` `=(2kq^(2)x)/(d^(2) + x^(2))^(3//2)` If `x lt lt d`, then x can be neglected, `F. = (2kq^(2)x)/(d^(3))`..........(1) `F. = Kx` where `K = (2kq^(2))/d^(3)` is constant `therefore F. prop x` Hence, q can perform SIMPLE harmonic motion. Here, force is DIRECTLY proportional to x and towards point O. `omega = sqrt(K/m)` `therefore (2pi)/T = sqrt(K/m)` `therefore T = 2pi sqrt(m/(2kq^(2)//a^(3))` By taking `k = 1/(4pi epsilon_(0))` `therefore T=[(8pi^(3)epsilon_(0)md^(3))/q^(2)]^(1//2)` |
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