A parallel plate capacitor (capacitance C) is charged with a battery of emf V volts. A dielectric slab of dielectric constant K is placed between the plates to fully occupy the space. The battery remains connected.
What are the changes in-1)C 2)Q (charge on capacitor) 3)E(E. field) 4)V(potential difference) 5)U (energy of capacitor) 6)F(Force of attraction between the plates). Symbols have usual meanings
All applicable to capacitors
The Attempt at a Solution
I know that when a dielectric is inserted C2=CK (by adding 2, I'm indicating the new value). Since battery is still connected, V remains V(?). Here comes my first confusion, since E2=E/K, shouldn't V change as well. But if we assume for now that V remains V, then Q must have become KQ to make capacitance KC, either that or V became V/K..which one is happening here?
also, I don't know how to approach the 6th question ,as to how the force will change because of dielectric. I'd appreciate some help.