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Total charge - Q is uniformly spread along length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring. (a) Show that the particle executes a simple harmonic oscillation. (b) Obtain its time period. |
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Answer» Solution :(a) The force acting on a point charge q at point P at DISTANCE x <<<< R from the centre of ring of radius R on its axis by (- dQ) charge at A on ring is, `dF = (K(-dQ)q)/(SQRT(R^(2)+x^(2)))^(2) =-k.((dQ)q)/(R^(2) + x^(2))` As shown in figure, rfFsin0 are same in magnitude but opposite in direction. Hence, they cancelled the effect of each other. Hence, F will be the sum of only `dFcostheta` components which are parallel to axis and TOWARDS centre O. `F = ointdFcos theta` `=oint -k((dQ)q)/(R^(2) + x^(2)) xx x/sqrt(R^(2) + x^(2))` `therefore F = -kQq x/(R^(2) + x^(2))^(3//2)(-Q)` Here, `x^(2) lt lt lt lt lt R^(2)` hence, `R^(2) + x^(2) = R^(2)` `F = -(kQq)/R^(3)`x.........(1) `therefore F prop (-x)`........(2) The above relation shows that the oscillations of +q about axis of ring is simple harmonic motion. If force constant is K., then `k. = -F/x = (Qq)/(4piepsilon_(0)R^(3))`..........(3) From EQNS. (1) (b) Here, time period of +q charge of mass M is `T = 2pisqrt(m/k.)` `therefore T = 2pi sqrt(m/((Qq)/(4piepsilon_(0)R^(3)))` `therefore T = 2pisqrt((4piepsilon_(0)mR^(3))/(Qq))`........(4) Equation (2) and (4) show the asked results. |
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