A parallel‑plate capacitor of plate area $A=2.0{\text{dm}}^2$ and plate separation $d=1.0\text{mm}$ is filled with a dielectric slab of dielectric constant $\kappa =5$ that occupies only half the volume between the plates (the slab is placed so that it completely covers one half of the plate area). What is the effective capacitance of the arrangement?

Step-by-step solution

1. Recognise that the capacitor is equivalent to two capacitors in parallel: one with dielectric (area $A/2$) and one without dielectric (area $A/2$).
2. Capacitance of a parallel‑plate capacitor: $C=\kappa {\epsilon }_0\frac{A}{d}$.
3. Compute capacitance of the part with dielectric: $C_1=5{\epsilon }_0\frac{A/2}{d}=5{\epsilon }_0\frac{A}{2d}$.
4. Compute capacitance of the part without dielectric: $C_2={\epsilon }_0\frac{A/2}{d}={\epsilon }_0\frac{A}{2d}$.
5. Add the two capacitances (parallel combination): $C=C_1+C_2=5{\epsilon }_0\frac{A}{2d}+{\epsilon }_0\frac{A}{2d}=6{\epsilon }_0\frac{A}{2d}=3{\epsilon }_0\frac{A}{d}$.
6. Insert numbers: ${\epsilon }_0=8.85\times 10^{-12}{\text{F m}}^{-1}$, $A=2.0{\text{dm}}^2=2.0\times 10^{-2}{\text{m}}^2$, $d=1.0\text{mm}=1.0\times 10^{-3}\text{m}$. Hence $C=3\times 8.85\times 10^{-12}\times \frac{2.0\times 10^{-2}}{1.0\times 10^{-3}}=3\times 8.85\times 10^{-12}\times 20=5.31\times 10^{-9}\text{F}$.
7. Round to appropriate significant figures: $C\approx 2.5\text{nF}$.

Concept tested: Capacitance with partial dielectric

Published 6 September 2026 · How this solution was prepared and checked