A parallel-plate capacitor has its plate length and width each doubled (area ×4) and the plate separation doubled (d ×2). Neglecting fringing, to keep the capacitance unchanged, how must the dielectric constant (relative permittivity ε_r) be.

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:Capacitance in a parallel-plate capacitor scales directly with plate area and dielectric constant, and inversely with plate separation. When geometry is modified, compensating changes in the dielectric constant can maintain the same capacitance, an important idea in sensor design and capacitor packaging. Given Data / Assumptions: Initial capacitance: C = ε_0 ε_r A / d. New dimensions: area A' = 4 A (both length and width doubled); separation d' = 2 d. Fringing fields are neglected (valid when plate dimensions ≫ separation). Concept / Approach:Write the new capacitance in terms of ε_r' and set it equal to the original. Solve for the required ε_r' as a multiple of ε_r. This relies purely on proportionality; no numerical values are necessary. Step-by-Step Solution: Original: C = ε_0 ε_r A / d.Modified: C' = ε_0 ε_r' (4 A) / (2 d) = ε_0 ε_r' (2 A / d).Set C' = C ⇒ ε_0 ε_r' (2 A / d) = ε_0 ε_r (A / d).Cancel common factors ⇒ 2 ε_r' = ε_r ⇒ ε_r' = ε_r / 2. Verification / Alternative check: Insert sample values (e.g., ε_r = 4): original C ∝ 4 A / d; new with ε_r' = 2 gives C' ∝ 2 * 2 A / d = 4 A / d, matching. Why Other Options Are Wrong: Keeping ε_r the same doubles C; doubling or quadrupling ε_r increases C beyond target; reducing to one quarter under-compensates. Common Pitfalls: Forgetting that area increased by factor 4 (both dimensions doubled) and separation doubled by factor 2. Final Answer: Halved (ε_r → ε_r/2)

A parallel-plate capacitor has its plate length and width each doubled (area ×4) and the plate separation doubled (d ×2). Neglecting fringing, to keep the capacitance unchanged, how must the dielectric constant (relative permittivity ε_r) be adjusted?

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