# question____15

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in Physics A positively charged thin metal ring of radius R is fixed in x-y plane with its centre at the origin O, A negatively charged particle P is released from rest at the point (0,0,z0). Then the motion of P is

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Let Q be the charge on the ring, the negative charge -q is released from point ( 0, 0, z0 ). The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be .

E=Qz0/14πε0(R2+z20)3/2
E = 0 at centre of the ring because z0 = 0
Force on charge at P will be towards centre as shown, and its magnitude is
F0=qE= Qqz0/14πε0(R2+z20)3/2     ..........(i)
Similarly, when it crosses the origin, the force is again towards centre O.
Thus, the motion of the particle is peroidic for all values of z0 lying between 0 and ∞.

Secondly, if z0<<R,(R2+z02)3/2≈R3
F0≈Qqz0/14πε0⋅R3               [From Equation. ( i)]

i.e., the restoring force Fe∝−z0. Hence, the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position.)