Force needed to hold together a capacitor

In summary, the conversation discusses two methods for calculating the force needed to hold two insulating plates together. Method 1 considers the electric field in three different regions and calculates the average force on each plate. However, it is argued that this method is incorrect because it does not take into account the force due to the other plate's electric field. Method 2 uses the capacitance of the system to calculate the force, which is argued to be the correct method. The conversation also touches on the question of why method 1 does not give the same answer as method 2 and clarifies that the force on each plate is due to the electric field of the other plate.
  • #1
phantomvommand
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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?
 
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  • #2
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.
 
  • #3
phantomvommand said:
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,
phantomvommand said:
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.
phantomvommand said:
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?
 
  • #4
haruspex said:
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.
 
  • #5
phantomvommand said:
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.
 
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Related to Force needed to hold together a capacitor

1. What is the force needed to hold together a capacitor?

The force needed to hold together a capacitor is dependent on several factors, including the size and shape of the capacitor, the distance between the plates, and the dielectric material used. In general, the force is directly proportional to the voltage applied and inversely proportional to the distance between the plates.

2. How does the dielectric material affect the force needed to hold together a capacitor?

The dielectric material used between the plates of a capacitor can greatly impact the force needed to hold it together. A higher dielectric constant material will result in a stronger force, while a lower dielectric constant material will result in a weaker force. This is because the dielectric material affects the electric field between the plates, which in turn affects the force.

3. Does the shape of a capacitor affect the force needed to hold it together?

Yes, the shape of a capacitor can affect the force needed to hold it together. For example, a cylindrical capacitor will have a different force requirement compared to a parallel plate capacitor with the same surface area and distance between the plates. This is because the electric field and thus the force, is distributed differently in different shapes.

4. What is the relationship between the force needed to hold together a capacitor and the voltage applied?

The force needed to hold together a capacitor is directly proportional to the voltage applied. This means that as the voltage increases, the force required to hold the plates together also increases. This is because a higher voltage results in a stronger electric field between the plates, resulting in a stronger force.

5. Can the force needed to hold together a capacitor be calculated?

Yes, the force needed to hold together a capacitor can be calculated using the formula F = Q^2 / (2 * A * d), where F is the force in newtons, Q is the charge on the plates in coulombs, A is the surface area of the plates in square meters, and d is the distance between the plates in meters. This formula assumes a parallel plate capacitor with a vacuum as the dielectric material. For different shapes and dielectric materials, different formulas may be needed.

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