A parallel‑plate capacitor has plate area $A=60,{\text{cm}}^2$ and plate separation $d=1,\text{mm}$. The space between the plates is filled with two dielectric slabs placed side by side, each occupying half the plate area. The slab on the left has dielectric constant ${\kappa }_1=3$, and the slab on the right has dielectric constant ${\kappa }_2=5$. What is the effective capacitance of the arrangement? Use ${\epsilon }_0=8.85\times 10^{-12},{\text{F m}}^{-1}$.

Step-by-step solution

1. Convert area to square metres. $A=60,{\text{cm}}^2=6\times 10^{-3},{\text{m}}^2$
2. Calculate the area of each slab. $A_{\text{slab}}=A/2=3\times 10^{-3},{\text{m}}^2$
3. Use the formula for a parallel‑plate capacitor with a dielectric. $C=\frac{{\epsilon }_0\kappa A_{\text{slab}}}{d}$
4. Compute the capacitance of the left slab. $C_1=\frac{8.85\times 10^{-12}\times 3\times 3\times 10^{-3}}{1\times 10^{-3}}=7.965\times 10^{-11},\text{F}$
5. Compute the capacitance of the right slab. $C_2=\frac{8.85\times 10^{-12}\times 5\times 3\times 10^{-3}}{1\times 10^{-3}}=1.3275\times 10^{-10},\text{F}$
6. Since the slabs are side by side, their capacitors are in parallel. $C_{\text{eff}}=C_1+C_2=2.124\times 10^{-10},\text{F}\approx 2.1\times 10^{-10},\text{F}$

Concept tested: Capacitance of composite dielectrics

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked