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Question: Two equal negative charges are fixed at a point (0,a) and (0,-a) on the y-axis. A positive charge q ...

Two equal negative charges are fixed at a point (0,a) and (0,-a) on the y-axis. A positive charge q is released from rest at the point (22a,0) on the x-axis. The charge will
a) Execute SHM about the origin
b)Move to the origin and remain at rest
c) Move to infinity
d) Execute oscillatory but not SHM

Explanation

Solution

Hint : If a particle or a body, experiences a restoring force about a mean position then the object oscillates about a fixed position. If there exists a point on either side of the mean position, such that the restoring force is the same but in the opposite direction within the confined limits then the particle will undergo SHM.

Complete solution:
In the above diagram φ is the angle between F‾\overline{F} and the x-axis.
Now let us understand the above diagram,
The charge Q initially at rest, accelerates towards the origin from 2a as it experiences a force 2F‾Cos⁡Φ.2\overline{F}\operatorname{Cos}\Phi . When it reaches the origin, the force on the charge q becomes equal to zero since the force due to both negative charges are equal and in opposite direction.
Now the charge Q, will possess some velocity at the origin and hence it will move towards negative x-axis. As the charge moves further, it will experience a force −2F‾Cos⁡Φ-2\overline{F}\operatorname{Cos}\Phi in the opposite direction of motion of the charge as seen in the diagram. Now it is to be noted that the rate of acceleration in moving from 2a to origin is equal to rate of deceleration in moving from origin to -2a. Hence the velocity of the charge q will be zero at x=-2a and again it will move towards the origin due to the force −2F‾Cos⁡Φ-2\overline{F}\operatorname{Cos}\Phi . This process will keep on continuing as long as the negative charges have the same magnitude. WE can clearly see in the above situation that the charge Q will move two and fro within a fixed interval of time and hence it will execute SHM.
Now let us understand the situation mathematically
Using Coulomb's law for force between two charges and superposition principle force on charge Q at x=2a is given by 2F‾Cos⁡Φ.2\overline{F}\operatorname{Cos}\Phi .
F‾\overline{F} is the force due to each negative charge I.e.
F‾=qQ4π∈∘r2N\overline{F}=\dfrac{qQ}{4\pi {{\in }_{\circ }}{{r}^{2}}}N
r is the distance of separation between the charges.
Let us take 14π∈∘=9×109\dfrac{1}{4\pi {{\in }_{\circ }}}=9\times {{10}^{9}} as k for simplicity.
Fore at x=-2a
−2F‾Cos⁡Φ.-2\overline{F}\operatorname{Cos}\Phi .
−2qQ4π∈∘r2Cos⁡Φ-2\dfrac{qQ}{4\pi {{\in }_{\circ }}{{r}^{2}}}\operatorname{Cos}\Phi
Since Cos⁡Φ=2ar\operatorname{Cos}\Phi =\dfrac{2a}{r} after substituting in the above equation we get,
−2qQ4π∈∘r2×2ar-2\dfrac{qQ}{4\pi {{\in }_{\circ }}{{r}^{2}}}\times \dfrac{2a}{r}
[−2qQ4π∈∘r3]2aN\left[ -2\dfrac{qQ}{4\pi {{\in }_{\circ }}{{r}^{3}}} \right]2aN is the required restoring force
Let us call this as equation 1
The above equation obtained is analogous to F‾=−kx\overline{F}=-kx….(2) which is the restoring force in SHM of a spring,
Comparing equation 1 and 2 we have
X=2a and k=2qQ4π∈∘r32\dfrac{qQ}{4\pi {{\in }_{\circ }}{{r}^{3}}}

Hence, the correct option to the question is a.

Note:
The time period of oscillation i.e. T=2πmk2\pi \sqrt{\dfrac{m}{k}}
T= 2πm2qQ4π∈∘r32\pi \sqrt{\dfrac{m}{2\dfrac{qQ}{4\pi {{\in }_{\circ }}{{r}^{3}}}}}
T=2πmqQ2π∈∘r3sec⁡2\pi \sqrt{\dfrac{m}{\dfrac{qQ}{2\pi {{\in }_{\circ }}{{r}^{3}}}}}\sec
The restoring force acts towards the origin when the charge tries to move away from the origin. Hence minus sign is to be considered next to the restoring force on charge q.