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



area of the plates of the capacitors = A.



a = length of the dielecric slab is inside the capacitor.



Therefore, the area of the plate covered with dielectric is =



The capacitance of the portion with dielectric is given by



The capacitance of the portion without dielectric is given by



The two parts can be considered to be in parallel.



Therefore, the net capacitance is given by-





Let us consider a small displacement da of the slab towards the inward direction.


With increase in the displacement of slab, the capacitance will increase, hence the energy stored in the capacitor will also increase.


In order to maintain constant voltage, the battery will supply extra charge, and gets damage .


Therefore the battery will do work.


Now,


Work done by the battery


= Energy change of capacitor + work done by the force F on the capacitor



1)


Let’s take the differential charge dq is supplied by the battery, and the change in the capacitor be dC


We know that energy in capacitor dWB




we know q = cv


2)



And force F is given by,




From 1) and 2)






We know


2)


where m is the mass of the object


a is the acceleration


From 1) and2)-


the acceleration of the dielectric a0 is given by =



where, m is the mass.


As, the force is in inward direction, it tends to make the dielectric to completely fill the space inside the capacitors.


As, the dielectric tends to completely fills the space inside the capacitor, at this instant its velocity is not zero.


After that the dielectric slab tends to move outside the capacitor.


As the slab tends to move out, the direction of force reverses.


Hence, the dielectric slab will maintain periodic motion.


Now, the time required for moving a distance l-a) can be-





=


For completing cycle, the time taken will be four times the time taken for covering distance l-a)



hence,


T=4t


=4×


=8


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