A parallel‑plate capacitor has plate area $A=50{\text{cm}}^2$ and plate separation $d=1.0\text{mm}$. A dielectric slab of dielectric constant $\kappa =4$ fills the lower half of the separation (thickness $d/2$) while the upper half is air. Determine the effective capacitance of the arrangement. Use ${\epsilon }_0=8.85\times 10^{-12}{\text{F m}}^{-1}$.

Step-by-step solution

1. Convert all dimensions to SI units: $A=50{\text{cm}}^2=5.0\times 10^{-3}{\text{m}}^2$ and $d=1.0\text{mm}=1.0\times 10^{-3}\text{m}$.
2. The capacitor can be regarded as two capacitors in series:
3. – $C_1$ (dielectric) has thickness $d/2$ and permittivity $\kappa {\epsilon }_0$, so $C_1=\frac{\kappa {\epsilon }_{0}A}{d/2}=\frac{2\kappa {\epsilon }_{0}A}{d}$.
4. – $C_2$ (air) also has thickness $d/2$, so $C_2=\frac{{\epsilon }_{0}A}{d/2}=\frac{2{\epsilon }_{0}A}{d}$.
5. For series combination, $\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}$.
6. Substitute $C_1$ and $C_2$: $\frac{1}{C_{\text{eq}}}=\frac{d}{2\kappa {\epsilon }_{0}A}+\frac{d}{2{\epsilon }_{0}A}=\frac{d(\kappa +1)}{2\kappa {\epsilon }_{0}A}$.
7. Invert to get $C_{\text{eq}}=\frac{2\kappa {\epsilon }_{0}A}{d(\kappa +1)}$.
8. Insert the numbers: $C_{\text{eq}}=\frac{2\times 4\times 8.85\times 10^{-12}\times 5.0\times 10^{-3}}{1.0\times 10^{-3}\times (4+1)}$.
9. Calculate numerator: $2\times 4\times 8.85\times 10^{-12}\times 5.0\times 10^{-3}=3.54\times 10^{-13}$.
10. Denominator: $1.0\times 10^{-3}\times 5=5.0\times 10^{-3}$.
11. Finally, $C_{\text{eq}}=\frac{3.54\times 10^{-13}}{5.0\times 10^{-3}}=7.08\times 10^{-11}\text{F}$.
12. Thus the effective capacitance is $7.08\times 10^{-11}\text{F}$ (≈ 70.8 pF).

Concept tested: Series combination of dielectrics in a capacitor

Published 11 September 2026 · How this solution was prepared and checked