Force needed to hold together a capacitor

[phantomvommand]

Homework Statement

The inner surfaces of 2 *insulating* plates are given a charge Q. What is the force required to hold them together?

Relevant Equations

F = 1/2 (E outside + E inside) * Q

I have 2 methods, which give 2 different solutions:
Let sigma = charge per unit area
Let plate 1 be the left plate, plate 2 = right plate.
Method 1:
Because they are insulating, consider the electric field at 3 regions; region 1 to the left of plate 1, region 2 between the plates, and region 3 to the right of plate 2.

Because the plates are insulating, Electric field in the regions are superpositions of the field due to each plate.
Region 1 has field = sigma/e0 + 0 = sigma/e0 directed leftwards,
Region 2 has field = sigma/e0 - sigma/e0 = 0,
Region 1 has field = sigma/e0 + 0 = sigma/e0 directed rightwards,

Average force on plate 1 = sigma/2e0 *Q (Average E field * Q)= Asigma^2/2e0 = Q^2/2Ae0, where A is area of plate. (force directed leftwards)
Similarly, force on plate 2 = Q^2/2Ae0. (force directed rightwards)

Thus, total force needed to hold the plates together is Q^2/Ae0.

Method 2:
Consider the capacitance of the system.
C = e0A/x, where x is plate separation.
Energy = Q^2x/2Ae0.
F = dE/dx = Q^2/2A e0.

Method 2 gives the correct answer. Why is method 1 wrong?

 

[phantomvommand]

I have since realized why my 2 methods don't seem to "reconcile". Please do tell me if I'm right on this.

They in fact are the same thing. The force given by -dU/dx refers to the force experienced by each capacitor plate, which is exactly the same as what method 1 gives.

 

[haruspex]

Region 1 has field = sigma/e0 + 0 = sigma/e0 directed leftwards,
Region 2 has field = sigma/e0 - sigma/e0 = 0,

Not sure of your reasoning there. I would have said Region 1 has field = the sum of the fields from the two plates = sigma/(2e0) + sigma/(2e0) = sigma/e0 directed leftwards,
Likewise Region 2 has field = sigma/(2e0) - sigma/(2e0) = 0,

Average force on plate 1 = sigma/2e0 *Q

Again, I would have argued that the field from plate 2 is sigma/(2e0), so the force is Q sigma/(2e0). There is no force on plate 1 from plate 1's field.
But I suppose your method works because plate 1 necessarily exerts equal and opposite fields each side, so on average exerts no field where it is.

Thus, total force needed to hold the plates together is Q^2/Ae0.

Arguably, the question is unclear, but what they mean is, suppose you tied the plates together with a cord; what would the tension be in the cord?

 

[phantomvommand]

Not sure of your reasoning there. I would have said Region 1 has field = the sum of the fields from the two plates = sigma/(2e0) + sigma/(2e0) = sigma/e0 directed leftwards,
Likewise Region 2 has field = sigma/(2e0) - sigma/(2e0) = 0,

Again, I would have argued that the field from plate 2 is sigma/(2e0), so the force is Q sigma/(2e0). There is no force on plate 1 from plate 1's field.
But I suppose your method works because plate 1 necessarily exerts equal and opposite fields each side, so on average exerts no field where it is.

Arguably, the question is unclear, but what they mean is, suppose you tied the plates together with a cord; what would the tension be in the cord?

Thank you for your reply. Why would it be sigma/2e0 though? The plates are insulating here, so the charge of Q remains solely on 1 side of the plate. The other side has 0 charge. Thus, the E-field in the region in between is 0 (as the inner surface of the other plate creates a similar field in opp direction.)

However, the E-field in the region to the left of plate 1 is due to the E-field of plate 2, which is sigma/e0 leftwards.
Thus, the average E-field around plate 1 is given by (0 + sigma/e0) /2. The force on plate 1 is in fact due to the field of plate 2.

 

[haruspex]

The plates are insulating here, so the charge of Q remains solely on 1 side of the plate. The other side has 0 charge.

The charge doesn't know it is on the surface of a plate. Half the field lines will go each side. Insulators don't block fields.