A parallel‑plate capacitor has plate area $A=70,{\text{cm}}^2$ and plate separation $d=2,\text{mm}$. 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 $k_1=4$, and the slab on the right has dielectric constant $k_2=6$. 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 the area and separation to SI units: $A=70\times 10^{-4}=7\times 10^{-3},{\text{m}}^2$, $d=2\times 10^{-3},\text{m}$.
2. Find the capacitance of each half of the plate: $C_1=\frac{{\epsilon }_0{k}_1(A/2)}{d}$ and $C_2=\frac{{\epsilon }_0{k}_2(A/2)}{d}$.
3. Add the two capacitances because they are in parallel: $C_{\text{eff}}=C_1+C_2$.
4. Simplify algebraically to $C_{\text{eff}}=\frac{{\epsilon }_{0}A(k_1+k_2)}{2d}$.
5. Substitute the numerical values: $C_{\text{eff}}=\frac{(8.85\times 10^{-12})(7\times 10^{-3})(10)}{2\times 2\times 10^{-3}}$.
6. Evaluate: $C_{\text{eff}}\approx 1.55\times 10^{-10},\text{F}$.

Concept tested: Capacitance of parallel‑plate with side‑by‑side dielectrics

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