InterviewSolution
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InterviewSolution
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A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point `(0, 0, z_0)` where `z_0gt0`. Then the motion of P isA. (a) periodic, for all values of `z_0` satisfying `0ltz_0ltoo`B. (b) simple harmonic, for all values of `z_0` satisfying `0ltz_0leR`C. (c) approximately simple harmonic, provided `z_0lt ltR`D. (d) such that P crosses O and continues to move along the negative z axis towards `z=-oo` |
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Answer» Correct Answer - A::C Let Q be the charge on the ring, the negative charge `-q` is released from point `P(0, 0, Z_0)`. The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be `E=(1)/(4piepsilon_0)*(QZ_0)/((R^2+Z_0^2)^(3//2)` Therefore, force on charge P will be towards centre as shown, and its magnitude is `F_e=qE=(1)/(4piepsilon_0)*(Qq)/((R^2+Z_0^2)^(3//2)).Z_0` ...(1) Similarly, when it crosses the origin, the force is again towards centre O. Thus the motion of the particle is periodic for all values of `Z_0` lying between 0 and `oo`. Secondly if `Z_0lt lt R, (R^2+Z_0^2)^(3//2)~~R^3` `F_e=(1)/(4piepsilon_0)*(Qq)/(R^3).Z_0` [From equation 1] i.e. the restoring force `F_e prop -Z_0`. Hence the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position). |
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