A dielectric slab of thickness 1.0 cm and dielectric constant 5 is placed between the plates of a parallel plate capacitor of plate area 0.01 m² and separation 2.0 cm. Calculate the change in capacity on introduction of dielectric. What would be the change, if the dielectric slab were conducting?

Given,
Thickness of the dielectric slab, t =1 cm=10^-2 m
Dielectric constant, ε_{r =}K = 5
Area of the plates of the capacitor, A= 0.01 m²= 10^-2 m²
Distance between parallel plates of the capacitor, d =2 cm = 2 x 10^–2 m

Therefore,
Capacity with air in between the plates
              $$\begin{gathered} \boxed{{\mathrm{C}}_0 = \frac{{\in }_0\mathrm{A}}{\mathrm{d}}} \\ = \frac{8.85 \times {10}^{-12} \times {10}^{-2}}{2 \times {10}^{-2}} \\ {\mathrm{C}}_0= 4.425 \times {10}^{-12} \mathrm{Farad} \end{gathered}$$ 

Capacity with dielectric slab in between the plates
              $$\begin{gathered} \boxed{\mathrm{C} = \frac{{\in }_0\mathrm{A}}{\mathrm{d}-\mathrm{t}\left(1-{\displaystyle \frac{1}{\mathrm{K}}}\right)}} \\ = \frac{8.85 \times {10}^{-12} \times {10}^{-2}}{2 \times {10}^{-2}-{10}^{-2}\left(1-{\displaystyle \frac{1}{5}}\right)} \\ \mathrm{C} = 7.375 \times {10}^{-12} \mathrm{farad} \end{gathered}$$
 
Capacity with conducting slab in between the plates

                  $$\begin{gathered} \boxed{\mathrm{C}\text{'} = \frac{{\in }_0\mathrm{A}}{\mathrm{d}-\mathrm{t}}} \\ = \frac{8.85 \times {10}^{-12} \times {10}^{-2}}{2 \times {10}^{-2}-1\times {10}^{-2}} \end{gathered}$$
           $$\begin{gathered} = \frac{8.85 \times {10}^{-14}}{{10}^{-2}} \\ = 8.85 \times {10}^{-12} \mathrm{Farad} \end{gathered}$$ 

Increase in capacity on introduction of dielectric
               C – C₀ = 7.375 x 10^–12 – 4.425 x 10^–12
                          = 2.95 x 10^–12 farad

Increase in capacity on introduction of conducting slab
              C’ – C₀ = 8.85 x 10^-12 – 4.425 x 10^–12
                         = 4.425 x 10^–12 farad.