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Step-by-Step Solution
Step 1: Understand the System
We have two identical positive charges $Q$ fixed at positions separated by distance $2a$. A smaller point charge $q_{0}$, of mass $m$, is positioned exactly at the midpoint between these two fixed charges. When displaced slightly along the line joining these charges, $q_{0}$ experiences a restoring force and can perform simple harmonic motion (SHM) for small displacements.
Step 2: Express the Net Electrostatic Force
Let the midpoint be the origin of our coordinate system. If $q_{0}$ moves a small distance $x$ along the line joining the charges, one charge of $Q$ is at $(+a, 0)$ and the other is at $(-a, 0)$.
Due to the charge at $a + x$ from $q_{0}$, the repulsive force will push it in one direction; due to the charge at $a - x$, it will be pushed in the opposite direction. For small $x$ (where $x \ll a$), the net force can be approximated as proportional to $x$ but in the opposite direction, consistent with a restoring force:
$F_{\text{net}} = -\left(\frac{k\,q_{0}\,Q}{(a - x)^2} - \frac{k\,q_{0}\,Q}{(a + x)^2} \right)\,.$
Step 3: Approximation for Small Displacement ($x \ll a$)
Under the condition $x \ll a$, a binomial expansion or first-order approximation shows that the difference in the forces becomes proportional to $x$. After simplification, one obtains:
$F_{\mathrm{net}} = -\frac{4k\,q_{0}\,Q}{a^{3}}\,x\,.$
Here, the negative sign indicates that the force is directed opposite to the displacement, confirming it is a restoring force.
Step 4: Write the Equation of Motion
Using Newton’s second law, $F_{\mathrm{net}} = m\,\alpha$, we get:
$m\,\alpha = -\frac{4k\,q_{0}\,Q}{a^{3}}\,x\,.$
Hence, the acceleration is:
$\alpha = -\frac{4k\,q_{0}\,Q}{m\,a^{3}}\,x\,.$
Recall that for SHM, the general form is $\alpha = -\omega^{2} x$, where $\omega$ = angular frequency. Thus, we identify:
$\omega^{2} = \frac{4k\,q_{0}\,Q}{m\,a^{3}}\,.$
Step 5: Relate $k$ and $\varepsilon_0$
The Coulomb constant $k$ can be written as $k = \frac{1}{4\pi \varepsilon_0}$. Therefore,
$\omega^{2} = \frac{4}{m\,a^{3}} \cdot \frac{q_{0}\,Q}{4\pi \varepsilon_0} = \frac{q_{0}\,Q}{\pi \varepsilon_0 m a^{3}}\,.$
Hence,
$\omega = \sqrt{\frac{q_{0}\,Q}{\pi \varepsilon_0 m a^{3}}}\,.$
Step 6: Find the Time Period
The time period of simple harmonic motion is $T = \frac{2\pi}{\omega}$. Substituting the expression for $\omega$ gives:
$T
= 2\pi \sqrt{\frac{\pi \varepsilon_0 m a^{3}}{q_{0}\,Q}}
= \sqrt{\frac{4\pi^{3}\varepsilon_0 m a^{3}}{q_{0}\,Q}}\,.
This matches the given correct answer.
Step 7: Final Answer
Therefore, the time period of oscillation of the charge $q_{0}$ is:
$T = \sqrt{\frac{4 \pi^{3} \varepsilon_{0} m a^{3}}{q_{0} Q}}\,.
Illustration
