A vacuum-insulated parallel-plate capacitor with plate separation d has capacitance C₀. What is the capacitance if an insulator with dielectric constant κ and thickness d/2 is slipped between the electrodes without changing the plate separation?

Verified step by step guidance

1. Start by recalling the formula for the capacitance of a parallel-plate capacitor: $C=\frac{\epsilon }{d}$, where $\epsilon$ is the permittivity of the material between the plates and $d$ is the plate separation.
2. When a dielectric material with dielectric constant $\kappa$ is introduced, the permittivity becomes $\epsilon$₀κ, where $\epsilon$₀ is the permittivity of free space. For this problem, the dielectric occupies half the separation distance ($\frac{d}{2}$), while the other half remains vacuum.
3. Treat the system as two capacitors in series: one with the dielectric (capacitance $C$₁) and one with vacuum (capacitance $C$₂). The equivalent capacitance for capacitors in series is given by $\frac{1}{C}=\frac{1}{C}$₁ + 1C₂.
4. Calculate $C$₁ for the region with the dielectric: $C$₁ = κε₀A2d. Similarly, calculate $C$₂ for the vacuum region: $C$₂ = ε₀A2d.
5. Substitute $C$₁ and $C$₂ into the series capacitance formula to find the equivalent capacitance: $C=\frac{\epsilon }{}$₀Ad2κκ

Key Concepts

Capacitance

Capacitance is the ability of a capacitor to store electric charge per unit voltage. It is defined as C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the plates. The capacitance of a parallel-plate capacitor in a vacuum is given by C₀ = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between them.

Dielectric Material

A dielectric material is an insulating substance that can be polarized by an electric field, which affects the capacitor's ability to store charge. When a dielectric with a dielectric constant κ is introduced into a capacitor, it increases the capacitance by a factor of κ, as it reduces the electric field within the capacitor. The thickness of the dielectric also plays a role in determining the effective capacitance.

Effective Capacitance with Dielectric

When a dielectric is partially inserted into a capacitor, the effective capacitance can be calculated by considering the capacitor as two capacitors in series: one with the dielectric and one without. In this case, if the dielectric occupies half the distance (d/2), the capacitance can be expressed as a combination of the capacitance of the dielectric-filled section and the air-filled section, leading to a new effective capacitance that reflects the contributions of both regions.