A parallel plate capacitor with plate area A and plate separation d, with a vacuum between the plates, has a capacitance C_0. 

The space between the plates is filled with two slabs of the same area but thicknesses (d/4) and (3d/4) made of material of dielectric constant 2 and 4, respectively. The capacitance of the capacitor is now:

Concept:

• Capacitor: 
  • ​A capacitor is a device that stores electrical energy in an electric field by virtue of accumulating electric charges on two close surfaces insulated from each other.
  • It is a passive electronic component with two terminals.
• ​Parallel plate capacitor: 
  • ​When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.
  • 
  • The capacitance of the parallel plate capacitor, \(C=\frac{A\epsilon_0}{d}\) where, A = area, d = distance between two plates
  • When a dielectric slab for dielectric constant k, then the capacitance of the parallel plate capacitor, \(C=\frac{A\epsilon_0k}{d}\)
• Capacitance in parallel combination:
  • ​Equivalent capacitance, C_eq = C₁ + C₂
• Capacitance in series combination:
  • ​Equivalent capacitance,  \(\frac{1}{C_{eq}} =\frac{1}{C_1}+\frac{1}{C_2}\)

Calculation:

In parallel plate capacitors, the capacitance in air, \(C_0=\frac{A\epsilon_0}{d}\) . . . . . . . . . .(1)

When the dielectric slab is inserted, the capacitance, \(C_0=\frac{A\epsilon_0k}{d}\)

Now, the space between the plates is filled with two slabs of the same area but thicknesses (d/4) and (3d/4) made of material of dielectric constant 2 and 4, respectively. The capacitors are in a series combination.

So, Equivalent capacitance, \(\frac{1}{C_{eq}} =\frac{1}{C_1}+\frac{1}{C_2}\)

\(C_{eq}=\frac{C_1C_2}{C_1+C_2} \)

\(C_{eq}=\frac{\frac{A\epsilon_0k_1}{d_1}\times \frac{A\epsilon_0k_2}{d_2}}{\frac{A\epsilon_0k_1}{d_1}+\frac{A\epsilon_0k_2}{d_2}} \)

\(C_{eq}=\frac{\frac{2A\epsilon_0}{d/4}\times \frac{4A\epsilon_0}{3d/4}}{\frac{2A\epsilon_0}{d/4}+\frac{4A\epsilon_0}{3d/4}} \)

\(C_{eq}=\frac{16}{5}\frac{A\epsilon_0}{d} \)

Substitute equation (1) in the above, we get

\(C_{eq}=\frac{16}{5} C_0\)