Effect of Dielectric Insertion on a Capacitor: with and Without a Battery

When a dielectric slab (dielectric constant K, also called relative permittivity ε_r​) completely fills the space between capacitor plates, the outcome depends on whether the battery remains connected:

Setup: A parallel plate capacitor of capacitance C₀​ is charged by a battery to charge Q₀, potential V₀, and electric field E₀. The battery is then disconnected, and a dielectric slab of constant K is inserted, completely filling the gap.

Derivation Steps:

1. Since the capacitor is isolated, charge remains unchanged: Q = Q₀
2. The dielectric reduces the electric field by a factor of K: E = \[\frac {E_0}{K}\]
3. Since V = E ⋅ d, the potential difference also reduces: V = \[\frac {V_0}{K}\]
4. New capacitance (using C = Q/V): C = KC₀
5. New stored energy (using U = Q²/2C): U = \[\frac {U_0}{K}\]​​

Result: Capacitance increases, but stored energy decreases — the "lost" energy corresponds to work done in pulling the dielectric into the field region.

Setup: The battery remains connected while the dielectric is inserted, so the potential difference across the plates is held fixed by the battery at V₀​.

Derivation Steps:

1. Since the battery maintains the voltage, potential difference remains unchanged: V′ = V₀
2. New capacitance increases due to the dielectric: C′ = KC₀
3. New charge (using Q = CV) increases because the battery supplies more charge: Q′ = KQ₀
4. Electric field remains unchanged (since V and d are both fixed): E′ = E₀
5. New stored energy (using U = \[\frac {1}{2}\]CV²) increases: U′ = KU₀

Result: Both capacitance and stored energy increase—the battery supplies the additional energy.

Example: Two identical capacitors, each of capacitance C₀​, are connected in parallel to a battery. After they are fully charged, the battery is disconnected, and a dielectric slab of dielectric constant K = 3 is inserted into one of the capacitors, completely filling the gap. Find the ratio of the initial total electrostatic energy to the final total electrostatic energy.

Solution:

• Initial energy of the system: U = 2 × \[\frac {1}{2}C_0V_0^2\] = \[C_0V_0^2\]
• After disconnecting the battery, total charge is conserved. Redistribution occurs since one capacitor now has capacitance KC₀​ while the other remains C₀​.
• Solving using charge conservation and equivalent capacitance for the new parallel combination: \[\frac {U}{U′}\] = \[\frac {3}{5}\] = 0.6