Question

a parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates the quantity that remains unchanged is​

Answer 1

Answer:

Explanation:

Here,the capacitor is isolated. It is already charged or else there would be no force. But when the dielectric is inserted, the charge remains fixed. If the capacitor were connected to a voltage source, then current would flow to keep the voltage fixed and then the force would increase.The voltage goes down even though there is no current in or out.The permittivity of any material is greater than the permittivity of free space.  

Therefore,When we insert the dielectric slab the capacitance rises.

W.k.t

C=q/V

If charge, q, is fixed, and if the capacitance, C, goes up, then the voltage, V, goes down

Force between two parallel plates with charge, Q; area, S; and permittivity, epsilon, between the plates.

F=Q²/2€S  

The point in their derivation was that this equation is independent of d, the separation distance between the plates. But it also shows that for constant charge, increasing the permittivity, epsilon, will DECREASE the force