Question

9. Two equal positive charges are fixed at the points[-a, 0] and [a, O] on the x-axis. A negative chargeis released from rest at the points [0, 2a] on they-axis. The charge will1(1) Move to the origin and remain at rest(2) Move to infinity(3) Execute oscillatory but not SHM7(4) Execute SHM​

Answer 1

Answer:

as both charges are positively charged so they will not form dipole and no oscillatory motion will be there now both of the last options are eliminated.

now the released charged is negatively charged so it will be attracted towards origin ...

hence it wil move towards origin

hope it help you

Answer 2

Explanation:

By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O. Net force on charge Q.

Fnet => 2F cosx = 2(1/4περ)*(-qQ/(a²+x²)*(x/(a²+x²)^1/2

i.e., Fnet = (1/4περ)*2qQx/(a²-x²)^3/2

As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic.

Therefore Your answer is :-

(3) Execute oscillatory but not SHM