A parallel‑plate capacitor has plate area $A=80{\text{cm}}^2$ 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 left slab has dielectric constant ${\kappa }_1=3$ and the right slab 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 and separation to SI units: $A=80{\text{cm}}^2=80\times 10^{-4}{\text{m}}^2=8.0\times 10^{-3}{\text{m}}^2$, $d=1\text{mm}=1\times 10^{-3}\text{m}$.
2. Since the dielectrics are side by side, the capacitors are in parallel: $C_{\text{eff}}=C_1+C_2$.
3. Each slab occupies half the area, so $A_1=A_2=\frac{A}{2}=4.0\times 10^{-3}{\text{m}}^2$.
4. Compute $C_1=\frac{{\kappa }_1{\epsilon }_0{A}_1}{d}=\frac{3\times 8.85\times 10^{-12}\times 4.0\times 10^{-3}}{1\times 10^{-3}}$.
5. Calculate $C_1=3\times 8.85\times 10^{-12}\times 4.0=1.062\times 10^{-10}\text{F}$.
6. Similarly, $C_2=\frac{{\kappa }_2{\epsilon }_0{A}_2}{d}=\frac{5\times 8.85\times 10^{-12}\times 4.0\times 10^{-3}}{1\times 10^{-3}}=1.77\times 10^{-10}\text{F}$.
7. Add the two capacitances: $C_{\text{eff}}=1.062\times 10^{-10}+1.77\times 10^{-10}=2.832\times 10^{-10}\text{F}$.
8. Express in scientific notation with appropriate significant figures: $C_{\text{eff}}\approx 2.83\times 10^{-10}\text{F}=28.3\text{pF}$.
9. If the answer is required in farads, it is $2.83\times 10^{-10}\text{F}$.

Concept tested: Capacitance with dielectrics (parallel arrangement)

Published 22 September 2026 · How this solution was prepared and checked