# Very hard capacitor problem

#### Hercuflea

1. Homework Statement

Two square plates of sides L are placed parallel to each other with separation d as suggested in the Figure. You may assume d is much less than L. The plates carry uniformly distributed static charges +Q naught and -Q naught. A block of metal has width L, length L, and thickness slightly less than d. It is inserted a distancex into the space between the plates. The charges on the plates remain uniformly distributed as the block slides in. In a static situation, a metal prevents an electric field from penetrating inside it. The metal can be thought of as a perfect dielectric, with K--> infinity.

a) Calculate the stored energy in the system as a function of x.
b) Find the direction and magnitude of the force that acts on the metallic block.
c) The area of the advancing front face of the block is essentially equal to L*d. Considering the force on the block as acting on this face, find the stress (force per area) on it.
d) Express the energy density in the electric field between the charged plates in terms of Qnaught, L, d, and εnaught.
e) Explain how the answers to part c and d compare with each other

(The figure referred to is a horizontal positive Q plate on top, a horizontal negative Q plate on bottom, and a metallic block being inserted in the middle a distance of x measured from the left.)

2. Homework Equations

Capacitance of a square PP capacitor: KεL2 / d

Energy Density: (1/2)εE2

Potential Energy of capacitor: Q2/2C = (1/2)QΔV = (1/2)C(V)2

3. The Attempt at a Solution

a)Wow. Not sure how to start on this. It asks for the stored energy so I guess I'll start with potential energy of a Cap. However, as the problem states K--> infinity, how can I even find the capacitance of the plates if KεL2/d already goes to infinity? Does it have infinite capacitance? I need to find the capacitance to use the formula (1/2)Q2/C.

Also, for the potential energy formula, does the Q represent just the charge Q of one of the plates or does the Q in the formula represent (+Q + (-Q)) for both of the plates, which would be zero? I'm confused.

b) I suppose I will use Gauss's Law. I guess they are asking for the force "after" the block has been inserted. E*A = Qenc / ε. I'll use a cylinder cut in half parallel to the positive plate.

E = (σ*π*r2)/ (ε*(2πr*L)) (am I jumping the gun by assuming σ?)

E = (σ*r) / (2*ε*L)

c) I have no idea. He has never taught us how to do something like that. (isn't the electric field perpendicular to the block face anyways?)

d) U = (1/2)(ε)[(σ*r) / (2*ε*L)]2 from part b

e) don't know

Halp?

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#### voko

What is the field like in the portion of the capacitor filled with the block? What is the energy of that field? What about the other portion? The total energy?

#### Hercuflea

What is the field like in the portion of the capacitor filled with the block? What is the energy of that field? What about the other portion? The total energy?
So...I will make a Gaussian cylinder perpendicular to the positive plate, with diameter x (same x as the length the metal is inserted) and height h.

Therefore ∫E dA = Qenc / ε ====> E * (2πxh) = (+Qnaught)[(πx2 / 4) / x2] / ε

I am using the ratio of the area of the cross section of the cylinder to the area of x2 to find the charge enclosed.

===> E(h/2) = (+Qnaught)(π/4) / (2πεxh)

===> E(h/2) = (+Qnaught) / 8εxh

#### voko

The field, because we have a perfect dielectric, is zero. So the energy is zero as well. You could find it more formally by taking the value of capacitance - infinite - and using the formula for stored energy.

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