A parallel‑plate capacitor has plate area $A=40{\text{cm}}^2$ and plate separation $d=0.5\text{mm}$. 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 ${\kappa }_1=2$, and the slab on the right has dielectric constant ${\kappa }_2=6$. Find 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: $A=40{\text{cm}}^2=40\times 10^{-4}{\text{m}}^2=4.0\times 10^{-3}{\text{m}}^2$. $d=0.5\text{mm}=5.0\times 10^{-4}\text{m}$.
2. Because the dielectrics are side by side, the capacitor behaves as two capacitors in parallel, each having plate area $A/2$ and the same separation $d$.
3. Capacitance of left part: $C_1={\kappa }_1{\epsilon }_0\frac{A/2}{d}=2\times 8.85\times 10^{-12}\times \frac{2.0\times 10^{-3}}{5.0\times 10^{-4}}$.
4. Evaluate $C_1$: numerator $2\times 8.85\times 10^{-12}=1.77\times 10^{-11}$, fraction $2.0\times 10^{-3}/5.0\times 10^{-4}=4$. Thus $C_1=1.77\times 10^{-11}\times 4=7.08\times 10^{-11}\text{F}$.
5. Capacitance of right part: $C_2={\kappa }_2{\epsilon }_0\frac{A/2}{d}=6\times 8.85\times 10^{-12}\times 4=2.13\times 10^{-10}\text{F}$.
6. Effective capacitance is sum of the parallel branches: $C=C_1+C_2=7.08\times 10^{-11}+2.13\times 10^{-10}=2.84\times 10^{-10}\text{F}$. Wait, this is $2.84\times 10^{-10}$ F which equals $2.84\times 10^{-10}\text{F}=2.84\times 10^{-10}$ F. The answer line should be corrected to $2.84\times 10^{-10}\text{F}$.

Concept tested: Capacitor with mixed dielectrics (parallel arrangement)

Published 13 September 2026 · How this solution was prepared and checked