Question

the plates of a 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.​

Answer 1

Answer:

Explanation:

Area of the plates of the capacitors is A and since, a is the  length of the dielectric slab hence, area = (A*l*a).  

We know that the formula of capacitence, C1=K∈0Aald  

when the capacitence is without dielectric then  C1=∈0A(l-a)ld are in parallel.

Hence, equivalent C=C1+C2

                ⇒C=∈0AldKa+(l-a)  

                ⇒C=∈0Aldl+a(K-1)

If there is a small displacement inside the capacitor then capacitance will increase.

Thus, energy stored within the capacitor increase.  

Work by the battery to maintain a constant voltage is change capacitors energy + work done on the capacitor (dWB = dU + dWF).

Charge dq supplied to battery dC dWB = (dq).V  

                             = (dC).V2 dU  

                             = 12(dC).V2 (dC).V2  

                             = 12(dC).V2 + F.da 12(dC).V2  

                             = F.da ⇒F=12dCdaV2  

                             ⇒F=12dda∈0Aldl+a(K-1) V2

                              = F=12∈0Ald(K-1).

Acceleration = 12∈0*A*l*dm*(K-1).

Since, force inside direction to make a periodic motion.  

Now, to move (l-a)distance  

(l-a)=12*a*0*t2t

=2l-a*a*0t

=2l-a×2ldm∈0AV2(K-1)t

=4ml-ald∈0AV2(K-1)

Hence to complete the cycle 4*0*A*V*2*(k-1)