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

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SUMMARY

The introduction of a dielectric slab with dielectric constant k between two charged parallel plates reduces the electric field by a factor of 1/k, resulting in a decrease in the force of attraction between the plates compared to the original force F. This conclusion is based on the relationship between electric field strength and force in capacitors. Additionally, the discussion highlights that the behavior of the capacitor changes depending on whether it is connected to a voltage supply or is free-standing, affecting the potential difference across the plates.

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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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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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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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