A capacitor has capacitance 5 $$\mu$$F when its parallel plates are separated by air medium of thickness $$d$$. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $$\frac{d}{2}$$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be _____ $$\mu$$F.

Initial capacitance with air:

$$C_0=5μF$$

When a dielectric slab of thickness $$\frac{d}{2}$$​ and dielectric constant K=1.5 is inserted, the system behaves like two capacitors in series:

• Air layer thickness $$\frac{d}{2}$$
• Dielectric layer thickness $$\frac{d}{2}$$

Effective separation:

$$d_{\text{eff}}=\frac{d}{2}+\frac{1}{K}\cdot\frac{d}{2}$$

Substitute K=1.5:

$$d_{\text{eff}}=\frac{d}{2}+\frac{d}{3}$$

$$d_{\text{eff}}=\frac{5d}{6}$$

Since capacitance is inversely proportional to effective distance,

$$C=\frac{\varepsilon_0A}{d_{\text{eff}}}$$

$$C=\frac{d}{5d/6}C_0$$

$$C=\frac{6}{5}\times5$$

C=6 μF