Question

Consider the situation shown in figure (31-E31). The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with a length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.
Figure

Answer 1

$\Rightarrow \frac{\mathrm{dc}}{\mathrm{da}}=\frac{-\mathrm{d}}{\mathrm{da}}\left[\frac{\varepsilon_{0} \mathrm{A}}{\ell \mathrm{d}}\{\ell+\mathrm{a}(\mathrm{k}-1)\}\right]$

Explanation:

We know that Capacitance (C) = k€ A/d

Capacitance of the portion without dielectrics, C2 = ε0(l - a)A / ld

Net Capacitance C = C1 + C2

$\Rightarrow(1 / 2)(\mathrm{dc}) \mathrm{E}^{2}=(\mathrm{dc}) \mathrm{E}^{2}-\mathrm{fdx}$

$\mathrm{fdx}=(1 / 2)(\mathrm{dc}) \mathrm{E}^{2} \Rightarrow \mathrm{f}=\frac{1}{2} \frac{\mathrm{E}^{2} \mathrm{dc}}{\mathrm{dx}}$

$c=\frac{\varepsilon_{0} A}{\ell d}[\ell+a(k-1)]$      $(here x=a)$

$\frac{\mathrm{d} \mathrm{c}}{\mathrm{da}}=\frac{-\mathrm{d}}{\mathrm{da}}\left[\frac{\varepsilon_{0} \mathrm{A}}{\ell \mathrm{d}}\{\ell+\mathrm{a}(\mathrm{k}-1)\}\right]$

Where

C = Capacitance of the circuit

A = Area of the plates of the capacitor

d = distance between two capacitor plates

k = dielectric constant of the material that used between the two plates of the capacitor. This dielectric constant is acts as a medium between the plates which helps to pass the conductivity between the plates.