ElectrostaticsHard
Question
Two equal negative charges -q each are fixed at the points (0, a) and (0, - a) on the y-axis .A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will :
Options
A.Execute simple harmonic motion about the origin
B.At origin velocity of particle is maximum.
C.Move to infinity
D.Execute oscillatory but not simple harmonic motion.
Solution
(i) From diagram, force on Q at general position x, is given by
Fnet = − 2F cos θ = −
(Towards origin)(ii) When charge moves from (2a, 0) to origin O, force keeps on acting on Q and becomes zero at O.
∴ Velocity of Q is max. at O.
∴ Velocity of Q is max. at O.
(iii) ∵ Motion is SHM for very small displacements. & 2a is not very small is motion is periodic but not SHM.
Create a free account to view solution
View Solution FreeMore Electrostatics Questions
Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius...The molarity of a solution obtained by mixing750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be :-...A non-conducting ring of radius 0.5 m carries a total charge of 1.11 × 10-10 C disturbed non-uniformaly on its circ...A ball of mass 1g and charge 10-8 C moves from a point A (VA = 600 V) to the point B whose potential is zero. Velocity o...Four charges of 1μC, 2μC, 3μC, and - 6μC are placed one at each corner of the square of side 1m.The ...