A parallel-plate capacitor is connected to a power supply that maintains a fixed potential difference between the plates. (a) If a sheet of dielectric is then slid between the plates, what happens to (i) the electric field between the plates, (ii) the magnitude of charge on each plate, and (iii) the energy stored in the capacitor? (b) Now suppose that before the dielectric is inserted, the charged capacitor is disconnected from the power supply. In this case, what happens to (i) the electric field between the plates, (ii) the magnitude of charge on each plate, and (iii) the energy stored in the capacitor? Explain any differences between the two situations.

When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.

The potential difference between the two plates is expressed as $V=ED$, the electric field strength times the distance between the plates.

(a) Here the potential is constant
(i) since the potential isconstantbetween the plates then the electricfieldis constant

(ii) The magnitude of electric chargeisn'tchange

(iii) The energy is constant because

Therefore (i) Constant (ii) Constant (iii) Constant

Letdielectricconstant be K

(i) After the source is disconnecting anddielectricis inserted theinitialpotential andbythe factor K so the electricfieldsoreducedby factor K

(ii)TheMagnitudeof chargeisn'tchange

(iii)Theenergywillreduceby factor Kbecauseandthe potentialisreducedbyconstantK

Therefore the (i) reduced by factor K (ii) Constant (iii) Reduced by factor K

The maindifferencebetween first and second case is voltage isconstantin part a so thefieldisconstantand the charge isconstantbut in secondaccessthefieldisreducedbecausethe potential is reduced

Therefore in First case field is constant and in second case the files is reduced.