Gokul43201 said:
Okay, so YOU prove that the slab will undergo periodic motion if released, and calculate what the period will be.
As I explained before, I claim there will be no periodic motion because the slab will slide out of the plates and fall on the floor and live there happily ever after. Please tell me how this is wrong.
Prove that your statement is true...:) The energy stored in the capacitor is higher if it is filled with a dielectric than without it. This does not mean that the energy of the system battery-capacitor is also higher with the dielectric in.
I will prove that there is an inward force acting on the slab if it was pushed to move inward at the beginning. And it will continue to move inward with constant acceleration till it reaches the end of the capacitor.
Problem:
A plane capacitor is connected to a battery that ensures constant voltage V across its plates.There are no losses, no resistance, no friction. The plates of the capacitor are of length l, width w, and they are d distance apart. The area of one plate is A = l*w. The electric field intensity inside the capacitor is E=V/d.
A dielectric slab of dimensions l x w x d and relative dielectric constant \kappa is placed so that its front face is inside the capacitor, at distance "a" from the edge. It is pushed a little inward. What happens?
Consider a small displacement \delta x of the slab inward. The intensity of the electric field, just as the voltage remains unchanged. The capacitance will increase, and so will the energy of the capacitor. At the same time, some free surface charge will disappear because of the dipole-chains built up in the dielectric. To maintain constant voltage, the battery will supply extra charges, but it has to exert work to do this. Moreover, the KE of the slab can change.
Work of the battery= change of the energy of the capacitor + change of the KE of the slab. \delta W_B=\delta W_C + \delta KE.
The work of the battery is:
\delta W_B=\delta Q*V
The change of the surface charge on the planes of the capacitor is equal to the change of the electric displacement, D, multiplied by the increament of the surface. When the dielectric replaces the vacuum and the electric field, E, stays the same D changes by (\kappa - 1)\epsilon_0*E. So
\delta W_B=\delta Q*V=(\kappa-1)\epsilon_0*E^2*w*d*\delta x
The capacitance changes by \delta C=(\kappa-1)\epsilon_0*\delta x*w/d, and the energy stored in the capacitor would change by
\delta W_C=0.5*\delta C*V^2=0.5*(\kappa-1)\epsilon_0*\delta x*w*d*E^2.
From the condion of work-energy balance we get:
(\kappa-1)\epsilon_0*E^2*w*d*\delta x =0.5*(\kappa-1)\epsilon_0*E^2*w*d*\delta x+\delta KE
Rearranging the equation:
\delta KE=0.5*(\kappa-1)\epsilon_0*E^2*w*d*\delta x.
The time rate of change for the kinetic energy is
dKE/dt=m*v*\dot v =0.5*(\kappa-1)\epsilon_0*E^2*w*d*v
According to the formulation of the problem, the slab is given a little push inward. So v is small at the beginning, but positive. The diection of the acceleration is also positive. The speed will increase, there is an inward force acting on the slab.
This little push is crucial, as we could not say if the slab starts to move or stays in its original position using the argument above. But there is really an inward force on it, coming from the "fringe effect" at the edges of the capacitor plates. The field near the edges is inhomogeneous and produces an inward force onto the slab. But to prove this is beyond the "college level".
Anyway, the slab will not "slide out of the plates and fall on the floor and live there happily ever after move out of the capacitor". If it moves out, we have 1-\kappa in the formula for dKE/dt, so it would be negative. The slab would slow down, that means an inward force again.
The slab was assumed moving inward at the beginning. We have deduced that it will continue to move inward with constant aceleration ,
\dot v=0.5*(\kappa-1)\epsilon_0*E^2*w*d/m
till its front face reaches the edge of the capacitor. From there on, the situation is reversed, new free charges appear on the plates as the slab moves outward, and the capacitor feeds back charges to the battery. Its energy decreases but to feed back the charges, additional energy is needed on the account of the KE of the slab. At the end the slab will stop. Now it is the fringe effect that will start the slab to move inward again.
The time now.
The displacement is l-a, the time needed is
t=\sqrt{\frac{2*(l-a)}{\dot v}}<br />
=2*\sqrt{\frac{(l-a)*m}{(\kappa-1)\epsilon_0*E^2*w*d}.
The time priod is four times longer:
T=4*t=8*\sqrt{\frac{(l-a)*m}{(\kappa-1)*\epsilon_0*E^2*w*d}
The solution quoted by ambuj was 8sqrt( ((l-a)lmd)/(eAE^2(K-1))) which is very similar to my result, but dimensionally not correct.
ehild
