Question

Find the capacitance of a parallel plate capacitor with a dielectric slab between the plates. 

Answer 1

1. Consider a parallel plate capacitor with the two plates each of area A separated by a distance d. The capacitance of the capacitor is given by `"C"_0 = ("A"ε_0)/"d"`
2. Let E₀ be the electric field intensity between the plates before the introduction of the dielectric slab. Then the potential difference between the plates is given by V₀ = E₀d, where E₀ = `σ/ε_0 = "Q"/("A"ε_0),` and σ is the surface charge density on the plates.
3. Let a dielectric slab of thickness t (t < d) be introduced between the plates of the capacitor as shown in the figure below.
   
   A dielectric slab in the capacitor
4. The field E₀ polarizes the dielectric, inducing charge –Q_p on the left side and +Q_p on the right side of the dielectric. 
5. These induced charges set up a field E_p inside the dielectric in the opposite direction of E₀. The induced field is given by 
   `"E"_"p" = sigma_"p"/ε_0 = "Q"_"p"/("A"ε_0)` .................`(∵ sigma_"p" = "Q"_"p"/"A")`
6. The net field (E) inside the dielectric reduces to E₀ – E_p. 
   ∴ E = E₀ - E_p = `"E"_0/"k"` ...............`(∵ "E"_0/("E"_0 - "E"_"p")= "k")`
   where k is a constant called the dielectric constant.
   ∴ E = `"Q"/("A""ε"_0"k")` or Q = Akε₀E ….(2)
7. The field E_p exists over a distance of t and E₀ over the remaining distance (d – t) between the capacitor plates. Hence the potential difference between the capacitor plates is V = E₀(d – t) + E(t) 
   = E₀(d – t) + `"E"_0/"k"`(t) ...................`(∵ "E" = "E"_0/"k")`
   = `"E"_0[("d" - "t") + "t"/"k"]`
   = `"Q"/("A"ε_0)["d" - "t" + "t"/"k"]`
8. The capacitance of the capacitor on the introduction of dielectric slab becomes
   C = `"Q"/"V" = "Q"/("Q"/("A"ε_0)("d " - "t" + "d"/"k")) = ("A"ε_0)/(("d" - "t" + "t"/"k"))` 
   This is the required expression.