Question

Explain in detail the effect of a dielectric placed in a parallel plate capacitor.

Answer 1

1. When the capacitor is disconnected from the battery:
   Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀. The capacitance of the capacitor without the dielectric is
   `"C"_0 = "Q"_0/"V"_0`   .....(1)
   The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by
   
   (a) Capacitor is charged with a battery
   
   (b) Dielectric is inserted after the battery is disconnected
   E = `"E"_0/ε_"r"`   .....(2)
   Here E₀ is the electric field inside the capacitors when there is no dielectric and ε_r is the relative permeability of the dielectric or simply known as the dielectric constant. Since ε_r > 1, the electric field E < E₀. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected. Hence the new potential difference is
   V = Ed = `"E"_0/ε_"r" "d" = "V"_0/ε_"r"`  ....(3)
   We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases. Thus new capacitance in the presence of a dielectric is
   C = `"Q"_0/"V" = ε_"r" "Q"_0/"V"_0 = ε_"r" "C"_0`    .....(4)
   Since ε_r > 1, we have C > C₀. Thus insertion of the dielectric constant ε_r increases the capacitance. Using equation,
   C = `(ε_0"A")/"d"`
   C = `(ε_"r"ε_0"A")/"d" = (ε"A")/"d"`......(5)
   where ε = ε_rε₀ is the permittivity of the dielectric medium. The energy stored in the capacitor before the insertion of a dielectric is given by U₀ = `1/2 "Q"_0^2/"C"_0`   .....(6)
   After the dielectric is inserted, the charge Q₀ remains constant but the capacitance is increased. As a result, the stored energy is decreased.
   U = `1/2 "Q"_0^2/(2"C") = 1/2 "Q"_0^2/(2 ε_"r""C"_0) = "U"_0/ε_"r"`
   

   Since ε_r> 1 we get U < U₀. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

2. When the battery remains connected to the capacitor:
   Let us now consider what happens when the battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
   The potential difference V₀ across the plates remains constant. But it is found experimentally (first shown by Faraday) that when dielectric is inserted, the charge stored in the capacitor is increased by a factor ε_r.
   
   (a) Capacitor is charged through a battery
   
   (b) Dielectric is inserted when the battery is connected.
   Q = ε_rQ₀ ….. (1)
   Due to this increased charge, the capacitance is also increased. The new capacitance is
   C = `"Q"_0/"V" = ε_"r" "Q"_0/"V"_0 = ε_"r" "C"_0`  ....(2)
   However the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
   Now, C₀ = `(ε_0"A")/"d"` and, C = `(ε"A")/"d"`    .....(3)
   `"U"_0 = 1/2 "C"_0 "V"_0^2`    ....(4)
   Note that here we have not used the expression
   `"U"_0 = 1/2 "V"_0^2 "C"_0`
   because here, both charge and capacitance are changed, whereas in equation 4, V₀ remains constant. After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
   U = `1/2 "CV"_0^2 = 1/2 ε_"r" "CV"_0^2 = ε_"r" "U"_0`
   

   Since e_r > 1 we have U > U₀
   It may be noted here that since voltage between the capacitor V₀ is constant, the electric field between the plates also remains constant.