Effect of a dielectric on capacitance

What does a dielectric do to a capacitor?

In plain words: the dielectric is polarised by the field between the plates. Its bound charges sit right next to the plates, with the opposite sign, and partly cancel the field of the free plate charges, without removing those free charges. The field and the voltage between the plates both drop, so the same charge now needs a smaller voltage. A smaller V for the same Q means a bigger C = Q/V.

Capacitance with a dielectric between the plates

Use a homogeneous, linear isotropic dielectric in electrostatics, with fringing neglected. The comparison of fields uses the same free plate charge; it does not describe insertion while a battery fixes the voltage.

Take a parallel plate capacitor with plates of area A, a distance d apart, carrying ±Q, so the charge density is ±σ with σ = Q/A (NCERT pages 69–70).

• Vacuum: E₀ = σ/ε₀, V₀ = E₀d and C₀ = Q/V₀ = ε₀A/d (NCERT Eq. (2.46), page 70).
• Dielectric filling the gap: it is polarised and, as NCERT explains in Section 2.10, acts like two sheets with bound charge densities σ_p and −σ_p on its faces next to the plates. The field in the dielectric is then that of a net density ±(σ − σ_p) on the plates (NCERT Eq. (2.47), page 70):

• The potential difference is V = Ed (NCERT Eq. (2.48)). For a linear dielectric σ_p is proportional to E₀, that is to σ, so (σ − σ_p) is proportional to σ too. NCERT writes the constant of proportionality as 1/K (NCERT Eq. (2.49), page 70):

• K is a constant of the dielectric, and K > 1 since σ_p is positive. Then V = σd/(ε₀K) = Qd/(Aε₀K) (NCERT Eq. (2.50)), and the capacitance is (NCERT Eq. (2.51), page 70):

Dielectric constant

• The product ε₀K is the permittivity of the medium, ε = ε₀K. For vacuum K = 1 and ε = ε₀, the permittivity of vacuum (NCERT Eq. (2.52), page 70).
• The dimensionless ratio K = ε/ε₀ is the dielectric constant of the substance, and for the ordinary static dielectrics considered here it is greater than 1 (NCERT Eq. (2.53), page 70).
• Comparing NCERT Eqs. (2.46) and (2.51) gives K = C/C₀ (NCERT Eq. (2.54), page 70). This holds for any capacitor geometry when the same uniform dielectric fills its entire electric-field region, not only a parallel plate one, and can be taken as the general definition of the dielectric constant (page 71).

• Link to the susceptibility (reasoned from NCERT Section 2.10, pages 66–67, and Eqs. (2.37) and (2.49)): a slab of face area A and thickness L has dipole moment PAL, since P is the dipole moment per unit volume. Its face charges ±σ_pA, a distance L apart, give the dipole moment σ_pAL, so σ_p = P = ε₀χ_eE. With ε₀E = σ − σ_p, this gives σ_p = χ_e(σ − σ_p), so σ = (1 + χ_e)(σ − σ_p) and K = 1 + χ_e.

A slab that fills only part of the gap

NCERT Example 2.8 puts in a slab of thickness 3d/4 (page 71). The same reasoning works for any thickness t < d (reasoned from NCERT Example 2.8):

• With the charge fixed, the field is E₀ in the air gaps, of total thickness d − t, and E₀/K inside the slab.
• So V = E₀(d − t) + E₀t/K, and C = Q/V = ε₀A/(d − t + t/K).
• With t = d this gives KC₀; with t = 3d/4 it gives NCERT’s 4KC₀/(K + 3). Where the slab sits in the gap does not matter.
• If the dielectric fills the whole gap over only part of the plate area, the two parts act as capacitors in parallel. Over half the area, C = (K + 1)C₀/2 (reasoned from NCERT Eqs. (2.51) and (2.64)). A slab that fills part of the thickness, as above, acts like capacitors in series.

Battery connected or disconnected

What changes when a dielectric fills the gap depends on what is held fixed (reasoned from NCERT Eqs. (2.38) and (2.51)):

• Battery disconnected first (Q fixed): C becomes KC₀, so V = Q/C falls to V₀/K, and the field falls to E₀/K.
• Battery still connected (V fixed): C becomes KC₀, so the charge rises to KQ₀, drawn from the battery. The field V/d stays E₀.

Solved examples

Common mistakes to avoid

Quick revision summary

• A dielectric filling a capacitor is polarised; its bound charges ±σp reduce the net density to σ − σp = σ/K.
• So the field and V drop for the same Q, and C = ε_₀KA/d, equivalently C = KC_₀.
• K = ε/ε_₀ = C/C_₀ is the dielectric constant, greater than 1 for ordinary static dielectrics; ε = ε_₀K, and C/C_₀ = K requires complete filling of the electric-field region.
• With Q fixed, V and the field fall to 1/K (E = E_₀/K); with the battery connected, V stays and Q rises K times.
• A slab of thickness t: C = ε_₀A/(d − t + t/K); for t = 3d/4 this is 4KC_₀/(K + 3).

Questions students ask

Is the dielectric constant the same as relative permittivity?

Yes, the dielectric constant is the relative permittivity: K = ε/ε₀ is the permittivity of the medium relative to that of vacuum, which is where the second name comes from (NCERT Eq. (2.53), page 70). It is a pure number with no unit, being a ratio of two permittivities, or equally of two capacitances, C/C₀ (NCERT page 70; the table on page 78 lists it as dimensionless). For vacuum K = 1; beyond the NCERT text, dry air at 20 °C and 1 atm has K ≈ 1.0005 (US National Bureau of Standards, Circular 537, 1953), which is why air between the plates is treated as vacuum.

Why is the dielectric constant of a metal taken as infinite?

The dielectric constant of a metal is taken as infinite because a metal cancels an applied field completely: a dielectric cuts the field from E₀ to E₀/K, and only K → ∞ gives the zero field found inside a conductor (reasoned from NCERT Eq. (2.49), page 70, and page 62). In the same way, the induced charge on a metal’s face equals the free charge on the plate next to it, while a dielectric’s bound charge is only σ_p = σ(1 − 1/K). This is a limit, not a measured value: a metal is not a dielectric, since its charges are free, not bound (NCERT page 65).

Is a dielectric slab pulled into a charged capacitor?

Yes, a dielectric slab partly inside a charged capacitor is pulled further in, battery connected or not, because the fringing field at the plate edges acts on its bound charges (beyond the NCERT text, reasoned from NCERT Sections 2.13 and 2.15). With the battery removed, Q is fixed and U = Q^²/2C falls as C rises, the energy lost becoming work done on the slab; with the battery connected, U = ½CV^² rises, but the battery supplies twice that rise and the other half does work on the slab. For a slab as thick as the gap d, between plates of width b, the force is F = ε₀(K − 1)bV^²/2d, with V the potential difference at that moment.

How is the dielectric constant related to the refractive index?

For a transparent, approximately non-magnetic isotropic material the refractive index is the square root of the dielectric constant, n = √K, if both are taken at the same frequency (beyond the NCERT text, Maxwell’s relation, as stated in US National Bureau of Standards Circular 537; it follows from NCERT Chapter 8, Eq. (8.11), page 207, v = 1/√(με)). The static K fits for the non-polar gases of Circular 537, not for water, whose K ≈ 78.5 at 25 °C (NBS Circular 514) goes with n = 1.33 (NCERT Class 12 Physics Part II, Table 9.1, page 230), so n^² ≈ 1.77. Its polar molecules cannot turn fast enough to follow the field of light, which reverses about 10^¹⁵ times a second.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A 5.0 pF air capacitor has its gap filled with glass of dielectric constant 7.0. What is its new capacitance?

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   C = KC₀ = 7.0 × 5.0 = 35 pF (NCERT Eq. (2.54), page 70).

2. Q2. Why is K greater than 1 for the ordinary static dielectrics in this model?

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   The bound charges on the dielectric’s faces are opposite to those on the plates next to them, so σ − σ_p = σ/K is less than σ, which needs K > 1 (NCERT Eq. (2.49), page 70).

3. Q3. A charged capacitor is disconnected from its battery, and a dielectric of K = 3 is slid in to fill the gap. What happens to the charge and to the potential difference?

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   The charge stays the same; the capacitance triples, so the potential difference falls to one third (NCERT Eqs. (2.38) and (2.54)).