Determine the Dielectric constant of slab:

An air capacitor contains two parallel plates 10 cm² in area and 0.5 cm apart. While a dielectric slab of area 10 cm2 and thickness 0.4 cm was inserted among the plates, one of the plate ought to be shifted by 0.4 cm to attain the same value of capacitance. What is the dielectric constant of slab?

Solution

Capacitance without dielectric slab :

C  = ε₀ A / d₁= (8.85 × 10 ^-12 × 10 × 10^{- 4} ) /0.5 × 10 ^{- 2}

= (8.85 × 10^-12 ) / 5

Capacitance with dielectric slab :

C₂   = (ε₀ ε_r A) /∑ d₂     

Figure

C₂ = (ε0 × 10 × 10- 4)  /((0.5/ ε_r)+0.4) × 10^-2  = ε₀/(((5/ ε_r)+ 4)= (8.85 × 10^-12 )/((5/ ε_r)+ 4)

But

∴ C₁ = C₂

(8.85 × 10^-12 )/((5/ ε_r)+ 4)= (8.85 × 10^-12 )/ 5

∴          ε_r = 5

Figure