Question 26 Solution
Step 1: Determine the Force on the Plate
The system is isolated (Charge $q$ is constant).
The capacitance depends on the overlap length $(l-x)$.
$$C(x) = \frac{\epsilon_0 \epsilon_r l (l-x)}{d}$$
Energy stored: $U = \frac{q^2}{2C(x)}$.
Force $F = -\frac{dU}{dx}$.
$$F = -\frac{d}{dx} \left( \frac{q^2 d}{2\epsilon_0 \epsilon_r l (l-x)} \right)$$
$$F = -\frac{q^2 d}{2\epsilon_0 \epsilon_r l} \cdot \frac{d}{dx}\left( \frac{1}{l-x} \right)$$
$$F = -\frac{q^2 d}{2\epsilon_0 \epsilon_r l} \cdot \frac{1}{(l-x)^2} (\text{approx constant for small } x \text{? No, force is constant for capacitor insertion})$$
Correction on Force: The force pulling a dielectric slab (or plate) into a capacitor is constant if the edge effects are ignored.
Magnitude of Force $F = \frac{q^2 d}{2\epsilon_0 \epsilon_r l^3}$. (Derived from standard formula for force on dielectric slab with constant Q).
Step 2: Type of Motion
Since the Force $F$ is constant and independent of $x$, the acceleration $a = F/m$ is constant.
The plate does not undergo Simple Harmonic Motion (SHM). Instead, it accelerates towards the center, crosses it, decelerates, stops, and returns. This is periodic motion.
Step 3: Calculate Time Period
Time to travel distance $x$ starting from rest ($u=0$) to the center:
$$x = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2x}{a}} = \sqrt{\frac{2xm}{F}}$$
Total Period $T = 4t$ (In, Out, Return In, Return Out).
$$T = 4 \sqrt{\frac{2xm}{F}}$$
Substituting $F = \frac{q^2 d}{2\epsilon_0 \epsilon_r l^3}$:
$$T = 4 \sqrt{ \frac{2xm \cdot 2\epsilon_0 \epsilon_r l^3}{q^2 d} } = 4 \sqrt{ \frac{4 m x \epsilon_0 \epsilon_r l^3}{q^2 d} }$$
$$T = \frac{8l}{q} \sqrt{ \frac{m x \epsilon_0 \epsilon_r l}{d} }$$