Force between the plates of a capacitor when a dielectric slab is added

In summary, the force of attraction between the two charged parallel plates will decrease when a solid block of dielectric is introduced between them, due to a reduction in the electric field by a factor of 1/k. This decrease in force is due to the work being done when the dielectric is introduced, and would be more significant if the plates were not connected to a voltage supply.
  • #1
avicenna
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Say there are two parallel plates separated by a small fixed distance d and charged +q, -q. The force attracting the plates is F. If a solid block of dielectric (dielectric constant k) is introduced between the plates, will the force of attraction be different from F.

My answer: The electric field will be reduce by a factor of 1/k; thus the force is lesser than F.
Is this answer correct?
 
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  • #2
avicenna said:
Say there are two parallel plates separated by a small fixed distance d and charged +q, -q. The force attracting the plates is F. If a solid block of dielectric (dielectric constant k) is introduced between the plates, will the force of attraction be different from F.
Try a related question first. Instead of introducing a dielectric, reduce the separation between the plates by a factor of 2. Does the attractive force increase, decrease or remain the same?
 
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  • #3
This thread could lead further to an interesting discussion; the Energy situation is as relevant as the Force situation.
You would need to specify whether the capacitor is connected to a voltage supply or just free-standing if you want a quantitative answer.
Work will be done when the dielectric is introduced; the dielectric would be 'pulled in', which would account for the lower PD across the plates when not connected to a supply.
 
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FAQ: Force between the plates of a capacitor when a dielectric slab is added

1. What is the equation for the force between the plates of a capacitor with a dielectric slab added?

The force between the plates of a capacitor with a dielectric slab added can be calculated using the equation F = εA(E1^2 - E2^2)/2d, where F is the force, ε is the permittivity of the dielectric material, A is the area of the plates, E1 and E2 are the electric fields in the dielectric and air, and d is the distance between the plates.

2. How does the dielectric slab affect the force between the plates of a capacitor?

The dielectric slab reduces the force between the plates of a capacitor due to the presence of the dielectric material, which decreases the electric field and thus the force. This is because the dielectric material has a higher permittivity than air, which reduces the electric field between the plates.

3. What factors affect the force between the plates of a capacitor with a dielectric slab?

The force between the plates of a capacitor with a dielectric slab is affected by the permittivity of the dielectric material, the area of the plates, the distance between the plates, and the electric field in both the dielectric and air regions. Additionally, the type and thickness of the dielectric material also play a role in determining the force.

4. How does the force between the plates change when the dielectric constant of the material is increased?

As the dielectric constant of the material increases, the force between the plates decreases due to the higher permittivity of the material. This leads to a decrease in the electric field between the plates, resulting in a smaller force. However, if the distance between the plates is also increased, the force will increase due to the larger area and electric field.

5. Can the force between the plates of a capacitor be negative when a dielectric slab is added?

No, the force between the plates of a capacitor cannot be negative. It is always positive and directed towards the opposite plate. However, the force can decrease when a dielectric slab is added due to the decrease in electric field, which can give the appearance of a negative force. In reality, the force is still positive but reduced in magnitude.

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