Choosing dielectrics for capacitors: which permittivity is preferable? For maximizing the capacitance of a given capacitor geometry, it is better to use a dielectric material having:

Introduction / Context:Capacitance depends on geometry and on the dielectric constant (relative permittivity εr) of the material between electrodes. Selecting the appropriate dielectric directly affects energy storage, device size, and performance across frequency.

Given Data / Assumptions:

• Parallel-plate or fixed geometry capacitor concept.
• Linear, isotropic dielectrics considered for simplicity.
• Breakdown strength and loss tangent are not the focus of this single-factor question.

Concept / Approach:The capacitance of a parallel-plate capacitor is C = ε0 εr A / d, where ε0 is vacuum permittivity, εr is the relative permittivity, A is plate area, and d is separation. For fixed A and d, increasing εr increases C proportionally, allowing higher energy storage (E = 0.5 C V^2) at the same voltage or a smaller component for the same C.

Step-by-Step Solution:Identify C dependence: C ∝ εr for fixed geometry.For higher C, choose larger εr materials.Therefore, “high permittivity” is preferable when maximizing capacitance alone.

Verification / Alternative check:Practical high-εr materials include ferroelectric ceramics (e.g., barium titanate-based). Conversely, PTFE or air has εr near 2–1 and produces smaller C for the same geometry.

Why Other Options Are Wrong:Low or air-like εr: Lowers capacitance. Neither high nor low: Offers no advantage in C. Negative ε: Not suitable for ordinary static capacitors; appears in metamaterials and plasmonic contexts, not conventional dielectrics.

Common Pitfalls:Ignoring trade-offs: high εr materials may have higher loss or lower breakdown; however, the question asks only about maximizing C.

Final Answer:high permittivity