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. A positively charged thin metal ring of radius R is fixed in the xy plane its centre at the origin 0.Anegatively charged particle Pis released from rest at the point (0, 0, z0) where z0> 0. Then the motion of P is (a) periodic, for all values of z0, satisfying 0 0(b) simple harmonic, for all values of zo satisfying 0 0≤ R(c) approximately simple harmonic, provided z0(d) such that P crosses O and continues to move along the negative z axis towards z = - ∞

Radhika Batra , 10 Years ago
Grade 11
anser 1 Answers
Kevin Nash

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
. (a, c) Let Q he the charge on the ring, the negative charge – q is released from point P (0, 0,Zo). The electric field at Pdue to the charged ring will be along positive z-axis and its magnitude will be
E = 1/ 4πε0 QZ0/(R2 +Z20)3./2
Therefore, force on charge P will be towards centre as shown, and its magnitude is
Fe = qE = 1/4πε0Qq/(R2 + Z20)3/2 Z0
Thanks
Kevin Nash
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