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Grade: 6
        
A positively charged thin metal ring of radius R is fixed in the xy plane its centre at the origin 0.A negatively charged particle P  is 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 < z0 < 8
(b) simple harmonic, for all values of zo satisfying 0 < z0 = R
(c) approximately simple harmonic, provided z0 << R
(d) such that P crosses O and continues to move along the negative z axis towards z = - 8
4 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							

(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 P due to the charged ring will be along positive z-axis and its magnitude will be

E = 1/ 4pe0 QZ0/(R2 +Z20)3./2

Therefore, force on charge P will be towards centre as shown, and its magnitude is

Fe = qE = 1/4pe0 Qq/(R2 + Z20)3/2 Z0
4 years ago
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