Two equal negative charge –q are fixed at point (0, –a) and (0, a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will :
Answer & explanation
Correct answer: option 4
The correct answer is Option 4: execute oscillatory but not simple harmonic motion
1. Net force on the charge
At any point $x$ on the x-axis, forces due to the two charges are symmetric:
- Vertical components cancel
- Horizontal components add toward origin
So, net force is along x-axis and given by:
$F = -\frac{2kQqx}{(x^2 + a^2)^{3/2}}$
Negative sign shows restoring force toward origin
2. Nature of motion
- At $x = 2a$, charge moves toward origin
- At origin $\rightarrow$ force $= 0$ but velocity is maximum
- Due to inertia, it crosses origin
- After crossing $\rightarrow$ force reverses direction and pulls it back
Hence, motion is to and fro (oscillatory)
3. Why not SHM?
For SHM:
$F = -kx$
But here:
$F = -\frac{2kQqx}{(x^2 + a^2)^{3/2}}$
- Force is not directly proportional to $x$
- Denominator depends on $x \rightarrow$ non-linear
So, motion is not simple harmonic