Consider the situation shown in figure. 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.
Given that the area of the plates of the capacitors is A.
As ''a'' length of the dielecric slab is inside the capacitor.
Therefore, the area of the plate covered with dielectric is `= A/la`
The capacitance of the portion with dielectric is given by `C_1 = (K∈_0Aa)/(ld)`
The capacitance of the portion without dielectric is given by `C_1 = (∈_0A(l-a))/(ld)`
The two parts can be considered to be in parallel.
Therefore, the net capacitance is given by
`C = C_1 + C_2`
`⇒ C = (∈_0A)/(ld)[Ka + (l-a)]`
`⇒ C = (∈_0A)/(ld)[l + a(K-1)]`
Let us consider a small displacement da of the slab in the inward direction. The capacitance will increase, therefore the energy of the capacitor will also increase. In order to maintain constant voltage, the battery will supply extra charges, therefore the battery will do work.
Work done by the battery = change in energy of capacitor + work done by the force F on the capacitor
`dW_B = dU + dW_F`
Let the charge dq is supplied by the battery, and the change in the capacitor be dC
`dW_B = (dq).V = (dC).V^2`
`dU = 1/2(dC).V^2`
`(dC).V^2 = 1/2(dC).V^2 + F.da`
`1/2(dC).V^2 = F.da`
`⇒ F = 1/2 (dC)/(da)V^2`
`⇒ F = 1/2 d/(da) ((∈_0A)/(ld)[l + a(K-1)])V^2`
`⇒ F = 1/2 (∈_0A)/(ld) (K - 1)`
The acceleration of the dielectric is given by `a_0 = 1/2 (∈_0A)/(ldm)(K-1)`
As, the force is in inward direction, it will tend to make the dielectric to completely fill the space inside the capacitors. As, the dielectric completely fills the space inside the capacitor at this instant its velocity is not zero. The dielectric slab tends to move outside the capacitor. As the slab tends to move out, the direction of the force due to the capacitor will reverse its direction. Thus, the dielectric slab will have a periodic motion.
The time taken to move distance `(l-a)` can be calculated as :-
`(l - a) = 1/2 a_0t^2`
`t = sqrt ((2(l-a))/(a_0))`
`t = sqrt (2(l-a) xx (2ldm)/(∈_0AV^2(K-1))`
`t = sqrt ((4m(l-a)ld)/(∈_0AV^2(K-1))`
For the complete cycle the time period will be four times the time taken for covering distance `(l-a)`.
It is given by :-
`T = 4t = 4 xx 2 sqrt((m(l-a)ld)/(∈_0AV^2(K-1))) = 8 sqrt ((m(l-a)ld)/(∈_0AV^2(K-1))`