The Polarization Formula gives the relationship between the electric polarisation vector P, the electric susceptibility χ, and the applied electric field E in a dielectric medium. Expressed as χε₀E, it is a core concept in Class 12 Physics (NCERT Chapter 2 — Electrostatic Potential and Capacitance) and appears regularly in JEE Main, JEE Advanced, and NEET examinations. This article covers the complete polarization formula, its derivation, a full formula sheet, three progressive solved examples, CBSE exam tips for 2025–26, common mistakes, and JEE/NEET application patterns.
Key Polarization Formulas at a Glance
Quick reference for the most important polarization and dielectric formulas.
- Electric Polarisation: \( \mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} \)
- Dipole Moment per unit volume: \( P = \frac{p}{V} \)
- Dielectric Constant relation: \( K = 1 + \chi_e \)
- Electric Displacement: \( \mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} \)
- Capacitance with dielectric: \( C = K C_0 \)
- Surface charge density of polarisation: \( \sigma_p = P \cos\theta \)
- Polarisation energy density: \( u = \frac{1}{2} \varepsilon_0 \chi_e E^2 \)
What is the Polarization Formula?
The Polarization Formula describes how a dielectric material responds to an external electric field. When an electric field is applied to an insulator (dielectric), the positive and negative charges inside each atom shift slightly in opposite directions. This creates tiny electric dipoles throughout the material. The collective effect of all these dipoles per unit volume is called electric polarisation, denoted by P.
In NCERT Class 12 Physics, Chapter 2 (Electrostatic Potential and Capacitance), polarisation is introduced as a vector quantity. Its SI unit is C/m² (coulombs per square metre). The formula connects polarisation to the electric susceptibility \( \chi_e \) of the material and the applied field \( \mathbf{E} \). A higher susceptibility means the material polarises more easily.
Polarisation also explains why inserting a dielectric slab between capacitor plates increases capacitance. The polarised charges on the slab surfaces create an opposing field. This reduces the net field inside the dielectric. As a result, the capacitor can store more charge for the same voltage. Understanding this mechanism is essential for both CBSE board exams and competitive entrance tests.
Polarization Formula — Expression and Variables
The primary polarization formula is:
\[ \mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} \]
Equivalently, polarisation can be defined as the electric dipole moment per unit volume:
\[ P = \frac{p}{V} \]
The electric displacement vector D relates to polarisation as:
\[ \mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} \]
And since \( \mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} \), we get:
\[ \mathbf{D} = \varepsilon_0 (1 + \chi_e) \mathbf{E} = \varepsilon \mathbf{E} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \mathbf{P} \) | Electric Polarisation (vector) | C/m² |
| \( \varepsilon_0 \) | Permittivity of free space | C²/(N·m²) ≈ 8.85 × 10⁻¹² F/m |
| \( \chi_e \) | Electric susceptibility (dimensionless) | Dimensionless |
| \( \mathbf{E} \) | Applied electric field | N/C or V/m |
| \( p \) | Total electric dipole moment | C·m |
| \( V \) | Volume of dielectric | m³ |
| \( \mathbf{D} \) | Electric displacement vector | C/m² |
| \( \varepsilon \) | Absolute permittivity of medium | F/m |
| \( K \) or \( \varepsilon_r \) | Relative permittivity (dielectric constant) | Dimensionless |
| \( \sigma_p \) | Polarisation surface charge density | C/m² |
Derivation of the Polarization Formula
Consider a dielectric placed in an external electric field \( \mathbf{E} \). Each molecule develops an induced dipole moment \( \mathbf{p}_i \) proportional to the local field. For a linear, isotropic dielectric, the molecular polarisability \( \alpha \) gives \( \mathbf{p}_i = \alpha \mathbf{E} \).
If there are \( n \) molecules per unit volume, the polarisation is:
\[ P = n \alpha E \]
Comparing with \( P = \varepsilon_0 \chi_e E \), we identify \( \chi_e = \frac{n\alpha}{\varepsilon_0} \). The dielectric constant then follows as \( K = \varepsilon_r = 1 + \chi_e \). This derivation appears in NCERT Class 12 Physics and is a standard two-mark derivation in CBSE board exams.
Complete Electrostatics and Dielectrics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Electric Polarisation | \( P = \varepsilon_0 \chi_e E \) | P=polarisation, χₑ=susceptibility, E=field | C/m² | Class 12, Ch 2 |
| Dipole Moment per Volume | \( P = p/V \) | p=dipole moment, V=volume | C/m² | Class 12, Ch 2 |
| Electric Displacement | \( D = \varepsilon_0 E + P \) | D=displacement, E=field, P=polarisation | C/m² | Class 12, Ch 2 |
| Dielectric Constant | \( K = 1 + \chi_e \) | K=dielectric constant, χₑ=susceptibility | Dimensionless | Class 12, Ch 2 |
| Capacitance with Dielectric | \( C = K C_0 \) | C=new capacitance, C₀=original capacitance | Farad (F) | Class 12, Ch 2 |
| Polarisation Surface Charge | \( \sigma_p = P \cos\theta \) | σₚ=surface charge density, θ=angle with normal | C/m² | Class 12, Ch 2 |
| Electric Dipole Moment | \( p = q \cdot d \) | p=dipole moment, q=charge, d=separation | C·m | Class 12, Ch 1 |
| Absolute Permittivity | \( \varepsilon = \varepsilon_0 K \) | ε=permittivity, K=dielectric constant | F/m | Class 12, Ch 2 |
| Energy Density in Dielectric | \( u = \frac{1}{2} \varepsilon E^2 \) | u=energy density, ε=permittivity, E=field | J/m³ | Class 12, Ch 2 |
| Polarisation Energy Density | \( u_p = \frac{1}{2} \varepsilon_0 \chi_e E^2 \) | uₚ=polarisation energy density | J/m³ | Class 12, Ch 2 |
| Clausius-Mossotti Relation | \( \frac{\varepsilon_r – 1}{\varepsilon_r + 2} = \frac{n\alpha}{3\varepsilon_0} \) | εᵣ=relative permittivity, n=molecular density, α=polarisability | Dimensionless | Class 12, Ch 2 (Advanced) |
Polarization Formula — Solved Examples
Example 1 (Class 11–12 Level)
Problem: A dielectric material has an electric susceptibility of 4.5. It is placed in a uniform electric field of magnitude 2 × 10⁵ N/C. Calculate the electric polarisation of the material.
Given:
- Electric susceptibility, \( \chi_e = 4.5 \)
- Electric field, \( E = 2 \times 10^5 \) N/C
- \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m
Step 1: Write the polarization formula: \( P = \varepsilon_0 \chi_e E \)
Step 2: Substitute values:
\[ P = 8.85 \times 10^{-12} \times 4.5 \times 2 \times 10^5 \]
Step 3: Multiply step by step:
\[ P = 8.85 \times 4.5 \times 2 \times 10^{-12+5} = 79.65 \times 10^{-7} \approx 7.97 \times 10^{-6} \text{ C/m}^2 \]
Answer
Electric Polarisation \( P \approx 7.97 \times 10^{-6} \) C/m² ≈ 7.97 μC/m²
Example 2 (Class 12 / CBSE Board Level)
Problem: A parallel plate capacitor has plate area 200 cm² and plate separation 5 mm. A dielectric slab with dielectric constant K = 6 fills the space between the plates. The applied voltage is 300 V. Find: (a) the electric field inside the dielectric, (b) the polarisation P, and (c) the polarisation surface charge density \( \sigma_p \).
Given:
- Plate area, \( A = 200 \text{ cm}^2 = 200 \times 10^{-4} \text{ m}^2 \)
- Separation, \( d = 5 \text{ mm} = 5 \times 10^{-3} \text{ m} \)
- Dielectric constant, \( K = 6 \)
- Voltage, \( V = 300 \text{ V} \)
- \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m
Step 1: Find the electric field inside the dielectric.
\[ E = \frac{V}{d} = \frac{300}{5 \times 10^{-3}} = 6 \times 10^4 \text{ V/m} \]
Step 2: Find electric susceptibility from dielectric constant.
\[ \chi_e = K – 1 = 6 – 1 = 5 \]
Step 3: Calculate polarisation using the polarization formula.
\[ P = \varepsilon_0 \chi_e E = 8.85 \times 10^{-12} \times 5 \times 6 \times 10^4 \]
\[ P = 8.85 \times 5 \times 6 \times 10^{-8} = 265.5 \times 10^{-8} \approx 2.655 \times 10^{-6} \text{ C/m}^2 \]
Step 4: For a slab with face perpendicular to the field, \( \theta = 0^\circ \), so \( \sigma_p = P \cos 0^\circ = P \).
\[ \sigma_p = 2.655 \times 10^{-6} \text{ C/m}^2 \]
Answer
(a) Electric field \( E = 6 \times 10^4 \) V/m
(b) Polarisation \( P \approx 2.655 \times 10^{-6} \) C/m²
(c) Polarisation surface charge density \( \sigma_p \approx 2.655 \times 10^{-6} \) C/m²
Example 3 (JEE/NEET Application Level)
Problem: A dielectric sphere of radius R = 0.1 m and dielectric constant K = 3 is placed in a uniform external electric field \( E_0 = 1.5 \times 10^5 \) N/C. The internal field inside the sphere is \( E_{in} = \frac{3}{K+2} E_0 \). Find the polarisation inside the sphere and the total induced dipole moment of the sphere.
Given:
- \( R = 0.1 \) m, \( K = 3 \), \( E_0 = 1.5 \times 10^5 \) N/C
- \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m
Step 1: Find the internal electric field inside the dielectric sphere.
\[ E_{in} = \frac{3}{K+2} E_0 = \frac{3}{3+2} \times 1.5 \times 10^5 = \frac{3}{5} \times 1.5 \times 10^5 = 9 \times 10^4 \text{ N/C} \]
Step 2: Find electric susceptibility.
\[ \chi_e = K – 1 = 3 – 1 = 2 \]
Step 3: Apply the polarization formula to find P.
\[ P = \varepsilon_0 \chi_e E_{in} = 8.85 \times 10^{-12} \times 2 \times 9 \times 10^4 \]
\[ P = 8.85 \times 2 \times 9 \times 10^{-8} = 159.3 \times 10^{-8} \approx 1.593 \times 10^{-6} \text{ C/m}^2 \]
Step 4: Find the volume of the sphere.
\[ V = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \pi \times (0.1)^3 \approx 4.189 \times 10^{-3} \text{ m}^3 \]
Step 5: Find the total induced dipole moment using \( p = P \times V \).
\[ p = 1.593 \times 10^{-6} \times 4.189 \times 10^{-3} \approx 6.67 \times 10^{-9} \text{ C\cdot m} \]
Answer
Polarisation inside sphere \( P \approx 1.593 \times 10^{-6} \) C/m² and total induced dipole moment \( p \approx 6.67 \times 10^{-9} \) C·m.
CBSE Exam Tips 2025-26
- Memorise the key relation: Always remember \( K = 1 + \chi_e \). Many CBSE 2025–26 questions ask you to find susceptibility when the dielectric constant is given. Rearranging gives \( \chi_e = K – 1 \) instantly.
- Units matter: Polarisation P and surface charge density \( \sigma \) both have the unit C/m². We recommend writing the unit explicitly in every step to avoid losing marks.
- Vector vs. scalar: In board exams, P is often treated as a scalar magnitude. In JEE, its vector nature (direction opposite to E for some materials) is tested. Know the distinction.
- Derivation of D = ε₀E + P: This three-line derivation carries 2 marks in CBSE. Practice writing it clearly. Our experts suggest writing each step on a separate line.
- Capacitance questions: Polarisation is always linked to the dielectric chapter. When a capacitor problem mentions a dielectric, check whether you need to use \( C = KC_0 \) or \( P = \varepsilon_0 \chi_e E \) or both.
- NCERT Examples first: Solve all NCERT in-text examples and exercises for Chapter 2 before attempting extra questions. The CBSE 2025–26 board paper heavily draws from NCERT examples.
Common Mistakes to Avoid
- Confusing K with \( \chi_e \): A very common error is substituting K directly into \( P = \varepsilon_0 K E \). The correct formula uses \( \chi_e = K – 1 \). Always subtract 1 from K before using it as susceptibility.
- Wrong unit for P: Some students write the unit of polarisation as C/m³ (volume charge density) instead of C/m². Remember, P is dipole moment per unit volume: (C·m)/m³ = C/m².
- Ignoring the direction of P: Polarisation P points in the same direction as E for most dielectrics (positive susceptibility). Forgetting this sign leads to errors in vector problems involving D = ε₀E + P.
- Using E₀ instead of E inside dielectric: In a capacitor with a dielectric, the field inside is \( E = E_0/K \), not \( E_0 \). Always use the reduced internal field when applying the polarization formula inside a dielectric medium.
- Forgetting ε₀ in the formula: Writing \( P = \chi_e E \) without \( \varepsilon_0 \) is a frequent slip. The formula is \( P = \varepsilon_0 \chi_e E \). The factor \( \varepsilon_0 \) is essential for dimensional consistency.
JEE/NEET Application of Polarization Formula
In our experience, JEE aspirants encounter the Polarization Formula in three main contexts. Understanding each pattern helps you solve problems faster under exam pressure.
Pattern 1 — Dielectric in a Capacitor (JEE Main, NEET): A capacitor is charged, then a dielectric slab is inserted. Questions ask for the new capacitance, energy stored, or the induced surface charge density. The key chain is: find K, compute \( \chi_e = K-1 \), find E inside, then apply \( P = \varepsilon_0 \chi_e E \). The polarisation surface charge \( \sigma_p = P \) (for normal incidence) then equals the bound charge on the dielectric face.
Pattern 2 — Finding Bound Charge Density (JEE Advanced): JEE Advanced problems often give a non-uniform polarisation field \( \mathbf{P}(x) \) and ask for the volume bound charge density \( \rho_b = -\nabla \cdot \mathbf{P} \). This requires differentiating the polarization formula expression. Mastering the divergence of P is essential for JEE Advanced preparation.
Pattern 3 — Energy and Force (JEE Main): The energy stored in a polarised dielectric is \( u = \frac{1}{2}\varepsilon_0 K E^2 \). Comparing this with the vacuum case \( u_0 = \frac{1}{2}\varepsilon_0 E^2 \) shows that the dielectric increases energy density by a factor of K. JEE Main frequently asks for the ratio of energies before and after inserting a dielectric at constant voltage or constant charge. At constant voltage, energy increases by K. At constant charge, energy decreases by K. Knowing the polarization formula helps you understand why this happens physically.
For NEET, the Polarization Formula appears in 1–2 MCQs per year within the Electrostatics chapter. NEET questions focus on definitions, units, and simple substitution. Knowing that P = ε₀χₑE and that K = 1 + χₑ is usually sufficient for full marks in NEET polarisation questions.
FAQs on Polarization Formula
Explore More Physics Formulas
We recommend exploring related formula articles to strengthen your understanding of electrostatics and wave physics. Visit our complete Physics Formulas hub for a full list of Class 11 and Class 12 formula guides. For wave-related polarisation concepts, read our article on the Superposition Formula, which explains how waves combine and interfere. If you are studying electromagnetic induction alongside dielectrics, our Induced Voltage Formula guide covers Faraday's law with solved examples. For thermal physics alongside electrostatics in your JEE preparation, the Heat Load Formula article is also highly recommended. All articles are aligned with NCERT 2025–26 syllabus and CBSE board exam patterns.