Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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
(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
(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
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -