Capacitance change with dielectric insertion A sheet of Bakelite (a solid dielectric) is inserted fully between the parallel plates of an air capacitor without changing plate area or separation. What happens to the capacitance?

Electronics and Communication Engineering Materials and Components Difficulty: Easy
Choose an option
  • A
    Increase
  • B
    Decrease
  • C
    Remain the same
  • D
    May increase or decrease depending on frequency only
  • E
    Becomes zero because the dielectric blocks charge

Answer

Correct Answer: Increase

Explanation

Introduction / Context:Understanding how dielectrics affect capacitance is fundamental in electrical engineering and materials science. When a material like Bakelite (with relative permittivity greater than that of air) is inserted between capacitor plates, the electric field, stored energy, and charge–voltage relationship change in predictable ways.

Given Data / Assumptions:

  • Parallel-plate capacitor with fixed plate area A and fixed separation d.
  • Originally the space is air (approximately vacuum for permittivity purposes).
  • A solid dielectric sheet of Bakelite fully fills the gap.
  • Linear, homogeneous, isotropic dielectric; fringing and edge effects neglected.

Concept / Approach:The capacitance of a parallel-plate capacitor is given by C = ε * A / d, where ε = ε0 * εr. Air has εr ≈ 1. Bakelite has εr greater than 1 (typically around 4–6 depending on grade). Replacing air with a dielectric increases ε and thus increases C in direct proportion to εr.

Step-by-Step Solution:Start with air-filled value: C_air = ε0 * A / d.Insert Bakelite fully: ε = ε0 * εr → C_dielectric = ε0 * εr * A / d.Compute ratio: C_dielectric / C_air = εr.Since εr > 1 for Bakelite, C increases by a factor εr.

Verification / Alternative check:From an energy viewpoint, at fixed voltage V the stored energy U = (1/2) * C * V^2 increases because the dielectric allows more charge to be stored for the same V. At fixed charge Q, inserting the dielectric lowers the voltage by 1/εr, consistent with increased C.

Why Other Options Are Wrong:“Decrease” contradicts C ∝ ε. “Remain the same” would imply εr = 1, which is false for Bakelite. “May increase or decrease depending on frequency only” is misleading; while εr can be frequency-dependent, across normal ranges for Bakelite εr > 1, so C increases. “Becomes zero” is incorrect; dielectrics do not block static electric fields; they reduce field strength inside by polarization.

Common Pitfalls:

  • Confusing conductor insertion (which shorts plates) with dielectric insertion (which increases permittivity).
  • Ignoring that partial filling or series layers require equivalent-capacitance calculations; here the gap is fully filled.

Final Answer:Increase

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