Effect of Dielectric on Capacitance

Imagine squeezing more electrical energy into the same capacitor without changing its size — that is exactly what a dielectric does. A dielectric is an insulating material that, when inserted between the plates of a capacitor, increases its capacitance by reducing the electric field between the plates.

This concept is the physical reason real-world capacitors (in phones, laptops, and power supplies) use materials like mica, paper, or ceramic between their plates rather than air.

The product of vacuum permittivity and dielectric constant of the medium.

ε = ε₀K

The ratio of the permittivity of a medium to the permittivity of vacuum.

K = ε / ε₀

OR

Dielectric constant is the factor by which the capacitance of a capacitor increases when a dielectric is completely inserted between its plates.

A dielectric is a non-conducting (insulating) material in which charges are bound to their atoms/molecules and cannot move freely. When placed in an external electric field, the molecules of the dielectric get polarised — they develop induced dipole moments that partially oppose the external field.

The process by which the molecules of a dielectric develop induced dipole moments when placed in an external electric field. The induced dipole moments align opposite to the field, creating an opposing induced field E_P.

The maximum electric field a dielectric can withstand before it breaks down (becomes conducting). Measured in V/m. Example: Air ≈ 3 × 10⁶ V/m.

Consider a parallel plate capacitor with:

• Plate area = A
• Separation between plates = d
• Charge on plates = Q (constant; battery disconnected)

Without any dielectric (air/vacuum between plates):

Where σ = surface charge density, ε₀ = permittivity of free space = 8.85 × 10⁻¹² C²N⁻¹m⁻¹.

Physical Mechanism (Intuition First)

When a dielectric slab fills the space between the plates:

1. The external field E₀ acts on the dielectric molecules.
2. The molecules polarise — positive charges shift slightly toward the negative plate; negative charges shift toward the positive plate.
3. This creates an induced (opposing) electric field E_P​ inside the dielectric.
4. The net field is reduced: E = E₀ − E_P
5. Since V = Ed decreases, and Q is constant, capacitance C = Q/V increases.

Analogy: Think of the dielectric as a "shock absorber" for the electric field. Just as a shock absorber reduces mechanical impact, the dielectric reduces the effective electric field — allowing the capacitor to store more charge at the same voltage.

Let the dielectric constant (relative permittivity) of the inserted material = K

By definition, the dielectric reduces the field by the factor K:

Since V = E ⋅ d (for full slab, thickness = d):

Since Q remains constant (battery disconnected):

Where ε = Kε₀ is the permittivity of the medium.

If the battery remains connected (V = constant), then Q increases by K and energy increases by K instead.

Given:

• A parallel plate capacitor with plate separation = d
• A dielectric slab of thickness t = \[\frac {3d}{4}\]​ and dielectric constant = K is inserted

Formula Used:

For a partial dielectric slab of thickness t inside a gap d, the effective capacitance is:

Solution:

Substituting t = \[\frac {3d}{4}\]​:

Answer: The new capacitance is \[\frac {4K}{K+3}\] times the original capacitance C₀​.

Check: When K = 1 (no dielectric effect): C = \[\frac {4}{4}\]C₀ = C₀ ✓ (consistent)