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?

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