In short

A dielectric is an insulating material placed between the plates of a capacitor. Every molecule in a dielectric, whether it carries a built-in dipole moment or not, responds to an applied field by becoming slightly polarised — the positive side of each molecule leans toward the negative plate and vice versa. The result is a thin layer of bound charge on each face of the dielectric that partly cancels the field from the free charge on the plates.

The cancellation is captured by a single dimensionless number, the dielectric constant \kappa (also written \varepsilon_r). Inside a linear dielectric,

E \;=\; \frac{E_0}{\kappa},

where E_0 is the field the same free charge would have produced in vacuum. Because the voltage across the plates scales with the field (V = Ed), and the charge on the plates is unchanged, the capacitance rises by the same factor:

\boxed{\;C = \kappa\,C_0\;} \qquad (\kappa \ge 1).

For air and vacuum \kappa \approx 1; for mica 5–7; for ceramic 10–10000; for water 80. Polar molecules (built-in dipoles, like water) give large \kappa; non-polar molecules (induced dipoles only) give small \kappa. The dielectric constant encodes how easily a material's charges can be displaced by an external field.

Take a parallel-plate capacitor made of two aluminium foils separated by an air gap and measure its capacitance. Now slide a piece of Jharkhand mica — the silvery flaky mineral India has been mining commercially since the 1900s — between the plates, without changing the plate area or the gap. Measure again. The capacitance has jumped by a factor of 6. Nothing has been added to the circuit. The battery has not been touched. No charge has come in from anywhere. And yet the capacitor now holds six times as much charge at the same voltage.

The same trick, at an industrial scale, is how the ceramic disc capacitors that sit by the thousand in every phone charger, every ceiling fan regulator, and every Jharia-based Indian ceramic factory achieve their capacitance. A ceramic disc is nothing more than a dielectric slab — titanium oxide, barium titanate, mica, polyester — with conductive paint on both faces. The paint is the "plates". Everything interesting happens in the slab. The dielectric constant of the ceramic, which can reach several thousand for specialised titanates, is what turns a 1 cm × 1 cm chip into a useful microfarad.

This chapter is the explanation of where that factor comes from. You will see why an insulator — a material with no mobile charges at all — can change the capacitance of a capacitor. You will see the molecular mechanism, the field reduction, and the way the effect rolls back up into the familiar formula C = \kappa\,C_0. And you will see why different materials give wildly different \kappa: it comes down to how easily the positive and negative charges inside each molecule can shift in response to an applied field.

What is a dielectric, anyway

An insulator is a material in which there are essentially no free electrons to carry current. Put a voltage across an insulator and no conduction happens — charges stay where they are.

A dielectric is the same material, viewed through a different lens: an insulator used for its ability to be polarised in an electric field. The word "dielectric" emphasises what the material does (respond to fields by rearranging its internal charges) rather than what it does not do (conduct). All dielectrics are insulators; the term is used when the insulating behaviour is the useful property.

The key claim of this chapter: although no charge can flow through a dielectric, the charges inside each molecule can shift slightly. That shift is small — often less than an atomic diameter per molecule — but there are about 10^{28} molecules in a cubic metre of typical material, and the cumulative effect is large.

Polarisation: the molecular picture

Molecules come in two flavours.

Non-polar molecules. The centre of positive charge and the centre of negative charge coincide — there is no built-in dipole moment. Hydrogen, nitrogen, oxygen, methane, benzene, paraffin wax, polyethylene. These are electrically symmetric.

Polar molecules. The centre of positive charge and the centre of negative charge are offset — the molecule has a permanent dipole moment \vec{p}_0. Water, hydrogen chloride, ammonia, acetone. These are electrically asymmetric.

In an applied electric field, both types respond — but by different mechanisms.

Non-polar molecules: induced dipoles

A hydrogen atom in zero field has a nucleus at the centre of a spherical electron cloud. Apply an external field \vec{E}_0. The nucleus gets pushed slightly in the direction of \vec{E}_0; the electron cloud gets pulled slightly in the opposite direction. The centres separate — and a tiny induced dipole moment \vec{p} appears, pointing along \vec{E}_0.

Non-polar molecule before and after applying an external field On the left, a neutral atom drawn as a sphere with a positive nucleus at its centre — no dipole moment. On the right, the same atom in an external field pointing right: the nucleus has shifted slightly right and the electron cloud slightly left, producing a small induced dipole pointing along the field. no external field electron cloud + no dipole: $\vec{p} = 0$ with external field $\vec{E}_0$ → $\vec{E}_0$ + − $\vec{p}$ induced dipole: $\vec{p} = \alpha \vec{E}_0$
A non-polar molecule in an external field: the nucleus shifts slightly along the field, the electron cloud against, producing a small induced dipole $\vec{p}$ proportional to $\vec{E}_0$.

For small fields, the induced dipole moment is proportional to the applied field:

\vec{p} \;=\; \alpha\,\vec{E}_0,

where \alpha is the polarisability of the molecule (units C²·m/N in SI). Why: for small displacements the restoring force inside the atom is roughly linear — like a mass on a spring. A linear restoring force plus a constant driving force gives a displacement proportional to the drive. The proportionality constant is a property of the molecule's internal structure.

The effect is small — the induced separation in a hydrogen atom in a laboratory field of 10^5 V/m is about 10^{-16} m, which is a hundred-thousandth of the atomic radius. But because there are about 3 \times 10^{25} molecules per cubic metre in a solid, the net effect adds up.

Polar molecules: permanent dipoles and their alignment

A polar molecule already has a dipole moment \vec{p}_0 even with no external field — but in a bulk sample at room temperature, the thermal motion randomises the orientations of the individual dipoles, so the net polarisation is zero. Apply an external field and the dipoles experience a torque \vec{p}_0 \times \vec{E}_0 that tries to align them along the field. Thermal motion fights back, keeping the alignment incomplete, but a net average alignment emerges.

Polar molecules before and after applying an external field On the left, many small dipole arrows pointing in random directions — zero net polarisation. On the right, the same dipoles, mostly aligned along the external field direction, showing a net polarisation along the field. no external field: random $\langle \vec{p} \rangle = 0$ with external field $\vec{E}_0$ → $\langle \vec{p} \rangle \neq 0$, pointing along $\vec{E}_0$
Polar molecules such as water: in zero field the thermal jiggle randomises their orientations, giving no net polarisation. Apply a field, and the torques on the permanent dipoles produce a partial alignment — a non-zero average dipole per molecule, pointing along the field.

The statistical mechanics of this process (Langevin's calculation) gives, for weak fields or high temperatures,

\langle \vec{p} \rangle \;=\; \frac{p_0^2}{3 k_B T}\,\vec{E}_0,

so the effective polarisability for orientation is \alpha_\text{or} = p_0^2/(3 k_B T).

Why: the field provides an energy bias of order p_0 E_0 \cos\theta for a dipole at angle \theta to the field; thermal motion smears orientations with Boltzmann weights e^{-U/k_B T}. The expansion for small p_0 E_0/(k_B T) gives the average above. The key feature: polar dipoles get weaker alignment at higher temperature, because thermal kicks undo the alignment faster than the field can establish it.

Water has p_0 \approx 6.2 \times 10^{-30} C·m, giving an orientation polarisability that is a hundred times larger than the induced polarisability of a typical non-polar molecule. This is why water has \kappa \approx 80 while non-polar materials like paraffin have \kappa \approx 2.

Polarisation density — the bulk quantity

At the molecular level you have many tiny dipoles. The useful bulk quantity is the polarisation density \vec{P} — dipole moment per unit volume:

\vec{P} \;=\; n \langle \vec{p} \rangle,

where n is the number of molecules per unit volume. \vec{P} has units C/m². For a linear dielectric — the type you meet in 99% of textbook problems — \vec{P} is proportional to the field:

\vec{P} \;=\; \varepsilon_0 \chi_e\,\vec{E},

where \chi_e (chi sub e) is the electric susceptibility, dimensionless, material-dependent. Different materials have different \chi_e for the same reasons they have different \alpha: molecular structure, bond stiffness, whether or not the molecule is polar.

Bound charges on the dielectric faces

Here is the move that turns polarisation into something with visible consequences. Inside a uniformly polarised slab of dielectric, every molecular dipole is oriented the same way. The negative end of each dipole sits close to the positive end of its neighbour — they cancel. Inside the bulk, there is no net charge density, even though every molecule is polarised.

Bound charges appearing on the faces of a uniformly polarised dielectric A rectangular slab of dielectric between two capacitor plates. Inside, rows of molecular dipoles all point the same way. The positive ends along the right face add up to a layer of bound positive charge; the negative ends along the left face form a layer of bound negative charge. Interior charges cancel neighbour by neighbour. +σ (free) −σ (free) dielectric slab (polarised) −σ_b (bound) +σ_b (bound)
Inside the slab, the positive end of one dipole sits next to the negative end of its neighbour, and they cancel. Only at the faces does the cancellation fail: the left face ends on a row of negative tails ($-\sigma_b$) and the right face ends on a row of positive heads ($+\sigma_b$). These surface layers are the **bound charge** that every real dielectric develops in an applied field.

At the two faces of the slab, the cancellation is incomplete: one face ends on negative tails (bound negative charge), the other on positive heads (bound positive charge). The surface density of this bound charge is

\sigma_b \;=\; P \qquad (\text{for a uniformly polarised slab with its polarisation perpendicular to its faces}).

Why: think of a slab of thickness d with n dipoles per unit volume, each carrying charge q displaced by a distance \delta. Per unit face area, the column holds nd dipoles, of which the positive ends near the right face add up to a surface density nq\delta — which is exactly P = n q \delta. The same argument gives a negative layer of magnitude P on the left face.

The bound charge is real — you can measure it with a sensitive electrometer — but it is not the kind of charge that flows. It is stuck to the dielectric molecules themselves. That is why the word "bound" is used.

Field reduction inside a dielectric

Now put all this together inside a parallel-plate capacitor.

Free charge on the plates. Surface densities +\sigma on the top plate and -\sigma on the bottom plate. In vacuum, this pair of charged sheets alone would produce a uniform field E_0 = \sigma/\varepsilon_0 between the plates, pointing from the positive to the negative plate (see Applications of Gauss's Law — Spheres and Planes for the derivation).

Bound charge on the dielectric faces. The bound charges, in the opposite arrangement — -\sigma_b on the face near the positive plate and +\sigma_b on the face near the negative plate — produce their own field E_b = \sigma_b/\varepsilon_0 that points opposite to E_0.

Field reduction inside a dielectric A parallel plate capacitor with a dielectric filling the gap. The field from the free charges on the plates points right. The bound charges on the dielectric faces point left. The net field inside the dielectric is smaller than E_0. +σ −σ −σ_b +σ_b $E_0$ from free charges $E_b$ from bound charges (opposes $E_0$) $E = E_0 - E_b = E_0/\kappa$ (net field inside)
The field from the free charges on the plates (top arrow) is partly cancelled by the field from the bound charges on the dielectric faces (middle, dashed, pointing opposite). The net field inside the dielectric (bottom) is smaller by the dielectric constant $\kappa$.

Derivation of E = E_0/\kappa

Assumptions: Linear dielectric (polarisation proportional to field). Uniformly filled capacitor (the dielectric occupies the entire gap). Plate dimensions much larger than the gap, so edge effects are negligible. The dielectric has no free charges of its own, only bound ones that arise in response to the field.

Step 1. Add the fields.

The two sources of field are the free charge \pm\sigma on the plates and the bound charge \mp\sigma_b on the dielectric faces. Each pair of infinite sheets produces a uniform field in the region between them:

E \;=\; E_0 - E_b \;=\; \frac{\sigma - \sigma_b}{\varepsilon_0}.

Why: a pair of infinite parallel planes with opposite charges produces a field \sigma/\varepsilon_0 between them and zero outside. The two pairs — free plates and bound faces — superpose linearly. The bound pair points opposite to the free pair (negative bound charge sits on the face near the positive plate), so the two fields subtract in magnitude.

Step 2. Use the relation \sigma_b = P (for uniform polarisation perpendicular to the slab).

E \;=\; \frac{\sigma - P}{\varepsilon_0}.

Step 3. Use the linear-dielectric law P = \varepsilon_0 \chi_e E.

E \;=\; \frac{\sigma - \varepsilon_0 \chi_e E}{\varepsilon_0} \;=\; \frac{\sigma}{\varepsilon_0} - \chi_e E.

Why: P itself depends on E, not E_0. This is the crucial subtlety: the bound-charge response is a feedback loop. The free charge produces a field, that field polarises the dielectric, the bound charges weaken the field, which reduces the polarisation — and everything settles to a self-consistent steady state. The equation above is the fixed-point equation for that steady state.

Step 4. Solve for E.

Bring all E terms to the left:

E + \chi_e E \;=\; \frac{\sigma}{\varepsilon_0}.
E\,(1 + \chi_e) \;=\; \frac{\sigma}{\varepsilon_0} \;=\; E_0.
E \;=\; \frac{E_0}{1 + \chi_e}.

Step 5. Define the dielectric constant \kappa.

The factor 1 + \chi_e appears so often that it gets its own name — the dielectric constant (or relative permittivity):

\kappa \;\equiv\; 1 + \chi_e.

Then

\boxed{\;E \;=\; \frac{E_0}{\kappa}.\;} \tag{1}

Why: \kappa is a single dimensionless number that captures how much the dielectric weakens the field. Because \chi_e \ge 0 for ordinary materials (polarisation always points with the field, not against it), \kappa \ge 1 — always. Vacuum has \chi_e = 0, so \kappa = 1. Real materials range from about 1.006 (air) to several thousand (specialised ceramic titanates).

From field reduction to C = \kappa C_0

Everything else is one line.

Step 1. Voltage between the plates. The gap is d and the field inside the dielectric is E, so

V \;=\; E\,d \;=\; \frac{E_0 d}{\kappa} \;=\; \frac{V_0}{\kappa},

where V_0 = E_0 d is the voltage that the same free charge would have produced with a vacuum gap.

Why: the voltage across the plates is just the integral of the field along a path between them. Reduce the field by a factor \kappa and the voltage drops by the same factor.

Step 2. Capacitance. The free charge on the plates is unchanged (Q = \sigma A), so

C \;=\; \frac{Q}{V} \;=\; \frac{Q}{V_0/\kappa} \;=\; \kappa \cdot \frac{Q}{V_0} \;=\; \kappa\,C_0.

Why: capacitance is charge divided by voltage. The charge stayed the same, the voltage dropped by \kappa, so the ratio — the capacitance — rose by \kappa.

For a parallel plate capacitor specifically:

\boxed{\;C \;=\; \kappa\,\frac{\varepsilon_0 A}{d} \;=\; \frac{\varepsilon_0 \kappa A}{d} \;=\; \frac{\varepsilon A}{d}\;} \tag{2}

where \varepsilon \equiv \kappa \varepsilon_0 is the permittivity of the dielectric. The formulas for cylindrical and spherical capacitors are modified by the same \kappa factor: every \varepsilon_0 becomes \kappa \varepsilon_0.

Dielectric constants of real materials

Material \kappa Notes
Vacuum 1 (exact) Definition
Air (dry, 1 atm) 1.00054 Usually treated as 1 in textbooks
Paraffin wax 2.2 Non-polar, small \kappa
Polystyrene 2.56 Used in precision capacitors
Paper 3.7 Older paper capacitors
Mica 5–7 India's Jharkhand/Bihar deposits were historically world-dominating
Glass 5–10 Depending on composition
Ceramic (low-\kappa) 10–100 Class-1 ceramic caps, stable with temperature
Ceramic (high-\kappa, BaTiO₃) 1000–10000 Class-2/3, variable, used in bulk caps at Jharia factories
Water (liquid, 20 °C) 80 Polar, huge \kappa
Ethanol 25 Polar
Teflon (PTFE) 2.1 Non-polar, very low loss

A few patterns:

Explore the effect yourself

The interactive below shows a parallel-plate capacitor with variable dielectric constant \kappa. Drag \kappa from 1 (vacuum) to 10 and watch how the field inside the dielectric, the voltage across the plates, and the capacitance respond — with the free charge on the plates held fixed.

Interactive: field, voltage, and capacitance versus dielectric constant Curves showing E/E_0 = 1/kappa falling from 1, V/V_0 = 1/kappa falling from 1, and C/C_0 = kappa rising from 1 linearly. The reader drags kappa from 1 to 10. dielectric constant $\kappa$ ratio to vacuum value 1.0 0.5 0 1 5 10 $E/E_0 = V/V_0 = 1/\kappa$ $C/C_0 = \kappa$ (rescaled by 10 to fit) drag the red point along the axis
Slide $\kappa$ from 1 (vacuum, no polarisation) to 10 (a glass or low-$\kappa$ ceramic). The field and voltage both drop as $1/\kappa$ because the induced bound charges partially cancel the free-charge field. The capacitance rises linearly in $\kappa$ for the same reason — fixed charge, smaller voltage.

Worked examples

Example 1: Sliding a mica sheet into a capacitor

A parallel-plate capacitor has plate area A = 200 cm² and gap d = 1.5 mm. In vacuum it holds Q = 1.2 nC when charged to V_0 = 100 V. A sheet of Jharkhand mica (\kappa = 6), cut to fit the gap exactly, is slid between the plates while the battery is disconnected (so the charge Q is locked in). Find the new field, new voltage, and new capacitance.

Mica inserted between plates of disconnected capacitor A 1.5 millimetre gap capacitor with plate area 200 square centimetres. On the left, vacuum gap with V0 = 100 V. On the right, the same capacitor with a mica slab filling the gap, kappa = 6, showing V = 100/6 volts. before: vacuum gap +Q −Q $V_0 = 100$ V $C_0 = 118$ pF after: mica slab ($\kappa$ = 6) mica +Q −Q $V = V_0/6 \approx 16.7$ V $C = 6 C_0 \approx 708$ pF
Before and after the mica slide: the charge stays the same (battery disconnected), but the voltage drops sixfold and the capacitance rises sixfold.

Step 1. Identify what is held fixed.

The battery is disconnected before the mica goes in, so no charge can flow on or off the plates. Charge Q is constant. What will change: the field inside (now reduced by mica), the voltage across the plates, and the capacitance.

Why: the difference between "battery connected" and "battery disconnected" is the single most common trap in dielectric problems. Connected → voltage fixed. Disconnected → charge fixed. Here the problem specifies disconnected, so Q is locked.

Step 2. Original vacuum capacitance C_0.

C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(200 \times 10^{-4})}{1.5 \times 10^{-3}}\ \text{F} = \frac{8.85 \times 10^{-12} \times 2 \times 10^{-2}}{1.5 \times 10^{-3}}\ \text{F}.
C_0 = \frac{1.77 \times 10^{-13}}{1.5 \times 10^{-3}}\ \text{F} = 1.18 \times 10^{-10}\ \text{F} = 118\ \text{pF}.

Why: start from the parallel-plate formula in SI units. Convert cm² to m² (multiply by 10^{-4}) and mm to m (multiply by 10^{-3}) so units cancel to farads.

Step 3. Original field E_0 inside the vacuum gap.

E_0 = \frac{V_0}{d} = \frac{100}{1.5 \times 10^{-3}}\ \text{V/m} = 6.67 \times 10^{4}\ \text{V/m}.

Step 4. After inserting the mica. Field, voltage, and capacitance all transform by \kappa = 6.

Field inside the dielectric:

E = \frac{E_0}{\kappa} = \frac{6.67 \times 10^4}{6}\ \text{V/m} \approx 1.11 \times 10^4\ \text{V/m}.

Voltage across the plates:

V = \frac{V_0}{\kappa} = \frac{100}{6}\ \text{V} \approx 16.7\ \text{V}.

New capacitance:

C = \kappa\,C_0 = 6 \times 118\ \text{pF} = 708\ \text{pF}.

Why: each quantity transforms by a single factor of \kappa — some up, some down. Field down, voltage down, capacitance up. The charge Q = CV stays at Q = 708\ \text{pF} \times 16.7\ \text{V} = 1.18 \times 10^{-8}\ \text{C}\cdot \text{V/V} = 1.2 nC, matching the starting charge.

Step 5. Check charge conservation.

Before: Q = C_0 V_0 = 118\ \text{pF} \times 100\ \text{V} = 1.18 \times 10^{-8}\ \text{C} = 1.18 \times 10^{-8}\ \text{C}.

After: Q = C V = 708\ \text{pF} \times 16.7\ \text{V} \approx 1.18 \times 10^{-8}\ \text{C}. ✓

Why: the charge should be exactly what was locked in at the start. The small 1% mismatch is rounding from using 16.7 instead of the exact 100/6.

Result: Field drops from 67 kV/m to 11 kV/m. Voltage drops from 100 V to 16.7 V. Capacitance rises from 118 pF to 708 pF. The charge stays at 1.2 nC.

What this shows: When the battery is disconnected, inserting a dielectric lowers the voltage (good for avoiding breakdown) and raises the capacitance. The field inside the dielectric is only a sixth of what it would have been in vacuum — that is the physical reason the voltage drops.

Example 2: Sari fabric as dielectric

A physics student in Chennai makes a homemade capacitor from two aluminium foil plates, each 20\ \text{cm} \times 20\ \text{cm}, separated by a single layer of silk sari fabric 0.25 mm thick. The silk has \kappa \approx 3.2. The student connects the capacitor to a 9 V battery and waits for it to charge fully. Find the capacitance and the charge stored.

Homemade capacitor with silk sari fabric dielectric Two aluminium foil plates of 20 cm by 20 cm sandwiching a 0.25 mm silk layer. Circuit shows a 9 V battery connected across the plates. silk, $\kappa$ = 3.2, $d$ = 0.25 mm +Q −Q 9 V $A = 20 \times 20 = 400$ cm²
Two foil plates with silk fabric as dielectric, charged by a 9 V cell.

Step 1. Vacuum-gap capacitance.

C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(400 \times 10^{-4})}{0.25 \times 10^{-3}}\ \text{F}.

Numerator: 8.85 \times 10^{-12} \times 4 \times 10^{-2} = 3.54 \times 10^{-13}. Denominator: 2.5 \times 10^{-4}.

C_0 = \frac{3.54 \times 10^{-13}}{2.5 \times 10^{-4}}\ \text{F} = 1.42 \times 10^{-9}\ \text{F} = 1.42\ \text{nF}.

Step 2. Apply the dielectric factor.

C = \kappa\,C_0 = 3.2 \times 1.42\ \text{nF} = 4.53\ \text{nF}.

Why: the silk fabric carries the same factor-\kappa multiplier as any other linear dielectric. The value \kappa = 3.2 is similar to paper — unsurprising, because silk is a natural polymer with polar amide groups in its protein chains.

Step 3. Charge stored at V = 9 V.

Q = C V = 4.53 \times 10^{-9}\ \text{F} \times 9\ \text{V} = 40.8\ \text{nC}.

Why: the battery is connected, so V is fixed at 9 V. The charge the student's capacitor accumulates is simply CV.

Step 4. Sanity check the scale.

40 nC sounds tiny, but a typical household static-shock discharge carries about 100 nC. The silk capacitor is storing charge of similar order — enough to notice with a sensitive electroscope, not enough to feel. Any larger foil area, or any thinner sari, would give proportionally more.

Result: Vacuum-gap capacitance 1.42 nF. With silk as dielectric, 4.53 nF. Charge stored: 40.8 nC.

What this shows: Ordinary household materials — sari silk, banana-leaf, paper, thin plastic — are all perfectly valid dielectrics. The only properties that matter are \kappa, the dielectric strength (before breakdown, which for silk is roughly 10^7 V/m — more than enough for this 9 V application), and the absence of mobile charges. This is the physical basis for paper and film capacitors that are still manufactured in large numbers today.

Example 3: Calculating bound charge on a ceramic disc

A ceramic disc capacitor with \kappa = 100 has circular plates of radius 5 mm, gap 0.1 mm. It is charged to 25 V. Find the free charge on each plate, the polarisation P inside the ceramic, and the bound surface charge density \sigma_b on each dielectric face.

Step 1. Plate area.

A = \pi r^2 = \pi (0.005)^2 = 7.85 \times 10^{-5}\ \text{m}^2.

Step 2. Capacitance.

C = \frac{\kappa \varepsilon_0 A}{d} = \frac{100 \times 8.85 \times 10^{-12} \times 7.85 \times 10^{-5}}{10^{-4}}\ \text{F}.
C = \frac{6.95 \times 10^{-14}}{10^{-4}}\ \text{F} = 6.95 \times 10^{-10}\ \text{F} = 0.695\ \text{nF}.

Step 3. Free charge.

Q = CV = 6.95 \times 10^{-10} \times 25 = 1.74 \times 10^{-8}\ \text{C} = 17.4\ \text{nC}.

Free surface charge density:

\sigma = \frac{Q}{A} = \frac{1.74 \times 10^{-8}}{7.85 \times 10^{-5}} = 2.22 \times 10^{-4}\ \text{C/m}^2.

Step 4. Field inside the ceramic.

E = \frac{V}{d} = \frac{25}{10^{-4}} = 2.5 \times 10^{5}\ \text{V/m}.

Step 5. Polarisation P.

P = \varepsilon_0 \chi_e E = \varepsilon_0 (\kappa - 1) E = 8.85 \times 10^{-12} \times 99 \times 2.5 \times 10^{5}.
P = 2.19 \times 10^{-4}\ \text{C/m}^2.

Why: by definition \chi_e = \kappa - 1, and the linear dielectric law gives P = \varepsilon_0 \chi_e E. For \kappa = 100, almost all of the electrostatic interaction is carried by the polarisation rather than by the field directly.

Step 6. Bound surface charge.

\sigma_b = P = 2.19 \times 10^{-4}\ \text{C/m}^2.

The ratio:

\frac{\sigma_b}{\sigma} = \frac{\kappa - 1}{\kappa} = \frac{99}{100} = 0.99.

Why: for a high-\kappa dielectric, the bound charge nearly matches the free charge. Of the 2.22 × 10⁻⁴ C/m² on each plate, 99% is "neutralised" by the bound charge on the adjacent dielectric face. Only 1% of the free charge's field actually reaches across the gap — that is the physical meaning of E = E_0/\kappa = E_0/100.

Result: Free charge Q = 17.4 nC on each plate; polarisation P = 2.19 \times 10^{-4} C/m²; bound charge density \sigma_b = 0.99\,\sigma on each dielectric face.

What this shows: In a high-\kappa ceramic, nearly all the free charge is "cancelled" by bound charge, leaving only a tiny effective field. This is why high-\kappa ceramics give such large capacitances in small packages — the bound charge does almost all the work, and the free charge (which is what you measure at the terminals) can be enormous before the voltage climbs to breakdown.

Common confusions

If you have E = E_0/\kappa, C = \kappa C_0, and the molecular picture of polarisation, you have what you need for JEE Main. What follows is the JEE Advanced and beyond — the general \vec{D} field, anisotropic dielectrics, the Clausius-Mossotti relation, ferroelectrics, and dielectric breakdown.

The electric displacement \vec{D}

In problems where free and bound charges coexist, it is useful to define the electric displacement field

\vec{D} \;\equiv\; \varepsilon_0 \vec{E} + \vec{P},

which has the property that its divergence comes only from the free charge:

\nabla \cdot \vec{D} \;=\; \rho_\text{free}.

This is Gauss's law in its most general form — it works for any medium, not just vacuum. For a linear dielectric, \vec{P} = \varepsilon_0 \chi_e \vec{E}, so \vec{D} = \varepsilon_0 (1 + \chi_e) \vec{E} = \varepsilon_0 \kappa \vec{E} = \varepsilon \vec{E}. The \vec{D} field is a book-keeping device: it lets you use Gauss's law without worrying about the bound charges, because they are built into the definition.

In the parallel-plate-with-dielectric geometry, D = \sigma_\text{free} (no reference to \kappa), while E = D/\varepsilon = \sigma_\text{free}/(\kappa \varepsilon_0).

Anisotropic dielectrics

In a general crystal, \vec{P} may not be parallel to \vec{E} — the dielectric response depends on the direction relative to the crystal axes. The susceptibility becomes a 3 \times 3 tensor \chi_{ij}, and

P_i \;=\; \varepsilon_0 \sum_j \chi_{ij}\,E_j.

For a crystal with three perpendicular axes of symmetry (most common mineral crystals), there are three principal values \chi_1, \chi_2, \chi_3 along those axes. Along each principal direction the crystal behaves like a linear isotropic dielectric with that single \chi. Off-axis, \vec{P} and \vec{E} point in different directions, and phenomena like birefringence arise.

Mica is anisotropic — its \kappa along the layer plane differs from \kappa across the layers by 20–30%. This is why precision mica capacitors specify both the thickness direction and the in-plane direction.

Clausius-Mossotti: from molecular to bulk

The relation between the molecular polarisability \alpha (units C²·m/N) and the bulk dielectric constant \kappa is not simply \kappa - 1 = n\alpha/\varepsilon_0 — because each molecule sits in the field of its neighbours as well as the applied field. Solving this self-consistency problem (the "local field" correction) gives the Clausius-Mossotti equation:

\frac{\kappa - 1}{\kappa + 2} \;=\; \frac{n \alpha}{3 \varepsilon_0}.

For gases (low n), this reduces to \kappa - 1 \approx n\alpha/\varepsilon_0. For dense materials it saturates more slowly in n. The Clausius-Mossotti relation is the bridge between molecular-scale quantum-chemistry calculations of \alpha and measured macroscopic \kappa.

Ferroelectrics and high-\kappa ceramics

Some materials — barium titanate BaTiO₃ is the classic example, and also the workhorse of Jharia's high-capacity ceramic capacitor industry — have a spontaneous polarisation even with no external field, much like ferromagnets have spontaneous magnetisation. Applying a field can flip the spontaneous \vec{P}, giving rise to a hysteresis loop in the P–E plot. The effective \kappa near the transition temperature can reach 10000 or more.

Ferroelectrics are not linear dielectrics — the formula C = \kappa C_0 is an approximation near a particular operating voltage. This is why Class-2 and Class-3 ceramic capacitors have specified tolerance curves showing how \kappa drops under bias; datasheets include a "DC bias de-rating" curve for this reason.

Dielectric strength and breakdown

Every dielectric has a maximum field E_\text{max} it can withstand before electrons are ripped off molecules, an avalanche of ionisation occurs, and the dielectric briefly behaves like a conductor. This is dielectric breakdown. Typical values:

  • Air: 3 \times 10^6 V/m (the threshold for lightning on a large scale).
  • Mica: 100 \times 10^6 V/m to 200 \times 10^6 V/m — one of the best insulators known.
  • Polyester film: 400 \times 10^6 V/m.
  • Ceramic (BaTiO₃): 20 \times 10^6 V/m — modest, because the high \kappa is bought at the cost of more defect-prone grain boundaries.

The maximum operating voltage of a capacitor is V_\text{max} = E_\text{max} d. You can often increase a capacitor's working voltage by using a thicker dielectric, but this reduces the capacitance (C = \kappa \varepsilon_0 A/d) — the trade-off between voltage rating and capacitance is fundamental.

Frequency dependence — \kappa at AC

At DC, polar molecules have time to align fully, and their full orientation polarisability contributes to \kappa. At high frequencies (GHz), they cannot keep up with the reversing field — the orientation contribution drops out, leaving only the much smaller electronic contribution. Water's \kappa drops from 80 at DC to about 2 at optical frequencies, where it is determined solely by the electronic polarisability of the atoms.

This is why microwave ovens work: water's \kappa around 2.45 GHz has a large imaginary component (absorption), and the field energy converts to heat as the water molecules try and fail to track the rapidly rotating field. The precise frequency-dependent \kappa(\omega) of a dielectric is the subject of dielectric spectroscopy, and it is how chemists identify unknown compounds in pharma labs across India.

Where this leads next