Very hard capacitor problem

[Hercuflea]

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.)

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

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?

 

[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.

 

[Nickie]

I would respond to this content by first acknowledging that this is indeed a very challenging capacitor problem. The problem statement is very clear and provides all the necessary information to solve the problem, but it requires a strong understanding of electrostatics and capacitance.

To start, it would be helpful to review the formula for the capacitance of a square parallel plate capacitor, which is C = KεL^2/d. In this case, as K approaches infinity, the capacitance of the plates would also approach infinity, indicating that the plates have an extremely high ability to store charge. This makes sense as the problem states that the metal block can be thought of as a perfect dielectric with K approaching infinity.

For part a, the potential energy of the capacitor can be calculated using the formula U = Q^2/2C. In this case, Q represents the charge on one of the plates, which would be +Q or -Q depending on which plate is being considered. The formula for the capacitance can be used to find the value of C, which can then be plugged into the potential energy formula.

For part b, you are correct in using Gauss's Law to calculate the electric field between the plates. However, it is not necessary to assume a surface charge density (σ) as the problem states that the charges on the plates remain uniformly distributed. Instead, you can use the formula E = Q/εA, where Q is the charge on one of the plates and A is the area of the plates. This will give you the electric field between the plates, which can then be used to calculate the force on the block using the formula F = QE.

For part c, the stress on the front face of the block can be calculated using the formula stress = force/area. In this case, the force can be found from part b and the area is given as L*d.

For part d, the energy density in the electric field can be calculated using the formula (1/2)εE^2, where E is the electric field calculated in part b.

Lastly, for part e, the answers to part c and d should be equal to each other as both are measures of the energy stored in the system. The stress on the block is a measure of the force applied to the block per unit area, while the energy density is a measure of the energy stored in the electric field per unit volume.