## Simple Harmonic Motion of a Dielectric Slab in a Parallel Plate Capacitor

Hello

Here's a situation I would like to discuss...

Suppose we have an ideal parallel plate capacitor which is filled with air (or vacuum) with permittivity = 1. Now if a dielectric slab of width equal to the space (d) between the plates is inserted into the space between the plates, it will be pulled in with a force, provided there is some charge on the capacitor plates. There are two possibilities:

(a) Constant Potential Difference across the capacitor plates imposed by an ideal emf source which is kept connected at all times to the capacitor. (Charge will change as the extent of the slab inside the capacitor changes)

(b) Constant charge on the capacitor plates, maintained by removing the emf source after the capacitor is charged (and only then is the dielectric brought in).

I know how to compute the force on the dielectric in both cases. In case (a) the force turns out to be independent of the length of the slab inside the dielectric and in case (b) it is a function of this length. The problem is to "prove" that in case (a) if the dielectric is displaced slightly from its equilibrium position (inside the capacitor) it will perform simple harmonic motion.

I want to prove it from first principles by showing that there is a restoring force on the slab which tends to bring it back to its position of stable equilibrium and for small displacements satisfies a relation of the form,

$$m\frac{d^2x}{dt^2} + kx = 0$$

where m = mass of slab, k = effective "spring constant", x = displacement from stable equilibrium position.

The only forces on the slab are its weight and the pulling force of the capacitor plates (due to induced charge). Neglecting fringing effects and assuming that the slab has no vertical displacement, the only force that should enter an equation of the form above is the pulling force.

But it is independent of x so it can't result in simple harmonic motion. What is wrong here?

I would be very grateful if someone could offer suggestions or ideas regarding this problem.

Thanks and cheers,

Vivek
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 Calculate the energy, U = 1/2 C V^2. Take two parallel plates of width L, and fill them with a dielectric up to x. Calculate the total energy as a function of x, and then the force is the gradient of the energy.
 hmm.. sounds like you probably already did that. This seems to give the energy a linear dependence on the displacement. If the dielectric is the same size as the capacitor, the problem can be 'fixed' by using the absolute value of the displacement instead of the actual displacement, which would force the energy to decrease in any direction that the slab is displaced. In this case though you get a restoring force that is constant and suddenly changes direction past the equilibrium point. It may be realistic to expect a constant force, because we are assuming a constant E field...

## Simple Harmonic Motion of a Dielectric Slab in a Parallel Plate Capacitor

Yes I've already done all this and if you try it out you'll see that the force function is indeed independent of the length of the dielectric inserted between the plates.
 Hi Force would be independend of the displacement of dielectric inside the capacitor. hence its motion would not be in SHM though it would be periodic motion. because once the dielectric is released in the capacitor force what u have calculated would try to bring it further in and it will accelerate inside the capacitor. once it starts coming out of capacitor(from the other end) force would try to bring it in and it will deaccelerate and will come to instantaneous rest with some part in and some part out of the capcitor. from there it will again accelarte towards the capacitor and the above process would be repeated. So the motion would be periodic rather than SHM
 Recognitions: Science Advisor If you consider the fringe effects, then your "pointy" potential will be slightly "rounded out" at the bottom. Since any function with a rounded dip looks like a parabola up close, there will be some small region under which the dielectric will approximate simple harmonic motion. This could be the reason for the wording in the problem; sometimes a professor will remember the "real-world" answer to a problem, and forget that you can't get that answer if you use the normal conventions of toy problems (i.e., ignoring fringe effects). On the other hand, what happens if you use scenario B? Can you get SHM in that case? Because maybe it's just a typo.
 pankaj192 is right, you don't get shm because that would imply that you have a restoring force which is dependent on displacement from an equilibrium position. This is not the case and instead what we see is that the restoring force has no dependence on the displacement. We can demonstrate this by constructing a potential which is given by the sum of the energy stored in the electric field in the region where there is a dielectric between the plates and the energy stored when the plates are separated by a vacuum. U=Integral{(1/2)e(|E|^2)dv} where |E| is the magnitude of the electric field and e is the electric permittivity. The potential for the whole capacitor is found by summing the integrals over the volume of the two regions, where the two volumes are dependent on the displacement of the dielectric, x. Comparing this potential to the kinetic energy, T=(1/2)m(v^2) by using the conservation of energy we can find the equation of motion, either by solving a differential equation or through the use of Lagrangian mechanics. We should find that the equation of motion is given by: F=-(1/2)(|E|^2)ad(e_0-e) where a and d are the depth and separation of the plates in the capacitor respectively. The interesting thing is that this is a restoring force since it is negative, but does not depend on the displacement, implying that the dielectric should simply slide back into its original position (I think). It is not shm since the Lagrangian has no terms in x of higher order than 1. The cause of this force is due to edge effects on the dielectric, the dipoles on the edge of the dielectric rotate to align themselves with the electric field across the capacitor, increasing the energy in the dielectric and causing it to want to return to a position of lower energy. Despite this being an edge effect, there is no need to take into account the effects on the edge of the dielectric directly, it all falls out when we consider the potential.