Question

A parallel plate capacitor with air as dielectric is connected to a power supply and charged to a potential difference Vo. After disconnecting from power supply, a sheet of insulating material is inserted between the plates completely filling the space between them. How will its
(i) charge on the plates
(ii) electric field between the plates
(iii) potential difference between the plates
(iv) capacitance
(v) energy stored in the capacitor.

Answer 1

V = power supply voltage
V0 = potential diff across capacitor
C = capacitor
Q0 = initial charge on capacitor = C V0
dielectric constant of insulator = K > 1

i)  capacitance C ' after disconnection & with insulator = K C
     charge on the plates remains same.
    but  on the sides of insulator slab facing the plates, charge Q' and -Q' are induced....   Q' =  Q0 (1 - 1/K)  respectively near  -Q0 and +Q0 plates.

ii)   E0 = Q0/(εA) = σ / ε  = electric field between plates
      E' = field after insulator is introduced = Q0/(KεA)    = E0/ K   < E0

iii)   Potential diff between plates V0
       V' after insulator introduction = Q0/(K C) = V0/K
         becomes 1/K times
iv)   C' = K C
v)     U0 = 1/2 C V0^2    or  1/2  Q0^2 / C
        U' = 1/2  (KC) V'^2  = 1/2 C V0^2/K     = U0 / K
        Energy decreases as there is work done in introducing the insulator between the plates of capacitor.