# Dielectrics and Polarisation

Physics II · Electromagnetism · https://tryals.app/learn/physics-ii/dielectrics-and-polarisation

## The Field Inside Matter

Put a dielectric in a field and its molecules acquire aligned dipole moments. The **polarisation** $\mathbf{P}$ is the dipole moment per unit volume, and it is equivalent to a set of **bound charge densities**:

$$\sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}}, \qquad \rho_b = -\nabla \cdot \mathbf{P}$$

These are real charges, they are simply not free to move through the material.

Because bound charge is awkward to track, electrostatics introduces the **displacement field**

$$\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}$$

whose divergence counts only the **free** charge: $\nabla \cdot \mathbf{D} = \rho_f$. That is Gauss's law in a form you can actually apply, since free charge is what you put there deliberately.

For a **linear isotropic** medium the response is proportional to the field:

$$\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}, \qquad \mathbf{D} = \varepsilon_0 \varepsilon_r \mathbf{E}$$

with $\varepsilon_r = 1 + \chi_e$ the **relative permittivity**. Since $\chi_e > 0$ for every ordinary dielectric, $\varepsilon_r > 1$ always.

The practical consequence is that a dielectric **weakens** the field inside it. Insert one into a charged capacitor at fixed charge and the field drops by $\varepsilon_r$, so the voltage drops and the capacitance rises by the same factor. The bound surface charge sits antiparallel to the free charge on the plates and partly cancels it.

| Material | $\varepsilon_r$ |
|---|---|
| Vacuum | 1 (exactly) |
| Air | 1.0006 |
| Paper | 3.7 |
| Water | 80 |

> **Common pitfall:** treating $\mathbf{D}$ as "the real field". $\mathbf{E}$ is what exerts force on a charge; $\mathbf{D}$ is a bookkeeping device that hides bound charge so Gauss's law stays usable. Neither is more fundamental, they answer different questions.

## Practice questions

8 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. A dielectric has electric susceptibility $\chi_e = 2.7$. Compute its relative permittivity.

**Answer:** 3.7 (within ±0.05)

**Why:** $\varepsilon_r = 1 + \chi_e = 1 + 2.7 = 3.7$, the value for paper. The 1 is the vacuum contribution, which is why $\varepsilon_r$ can never fall below 1 for an ordinary dielectric.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/a-dielectric-has-electric-susceptibility-e-2-7-compute-its

### 2. A capacitor of relative permittivity 1 is filled with a dielectric of $\varepsilon_r = 3$ at fixed charge. Set the factor by which its capacitance changes.

**Answer:** 3 (within ±0.3)

**Why:** The field drops by $\varepsilon_r = 3$, so $V$ drops threefold while $Q$ is unchanged, and $C = Q/V$ therefore **triples**. This is why practical capacitors are filled rather than left empty.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/a-capacitor-of-relative-permittivity-1-is-filled-with-a-dielectric-of

### 3. Match each quantity to what it represents.

**Answer:**

- Polarisation $\mathbf{P}$ → Dipole moment per unit volume
- Displacement field $\mathbf{D}$ → Lets Gauss’s law be applied to free charge alone
- Susceptibility $\chi_e$ → How strongly the medium polarises per unit field
- Relative permittivity $\varepsilon_r$ → The factor by which the field is weakened

**Why:** $\mathbf{P}$ measures the aligned dipoles, $\mathbf{D}$ hides them from Gauss’s law, $\chi_e$ sets how readily they align, and $\varepsilon_r = 1 + \chi_e$ is the resulting field reduction.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/match-each-quantity-to-what-it-represents

### 4. Which statements about bound charge are correct?

A. Bound charge contributes nothing to the electric field
B. Bound charges are real charges that cannot move freely through the material
C. Volume bound charge is minus the divergence of the polarisation
D. Surface bound charge equals the polarisation component along the outward normal

**Answer:** B. Bound charges are real charges that cannot move freely through the material; C. Volume bound charge is minus the divergence of the polarisation; D. Surface bound charge equals the polarisation component along the outward normal

**Why:** Bound charges are genuinely there and genuinely produce field, that is exactly how a dielectric reduces the internal field. They differ from free charge only in being unable to migrate through the material.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/which-statements-about-bound-charge-are-correct

### 5. A capacitor of capacitance 4.0 pF in vacuum is completely filled with a dielectric of relative permittivity 2.5. Compute its new capacitance in pF.

**Answer:** 10 (within ±0.2)

**Why:** $C = \varepsilon_r C_0 = 2.5 \times 4.0 = 10$ pF. Every geometric factor is unchanged; only the medium’s response has altered.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/a-capacitor-of-capacitance-4-0-pf-in-vacuum-is-completely-filled-with

### 6. Sort each statement by whether it is true of $\mathbf{E}$ or of $\mathbf{D}$.

**Answer:**

- True of $\mathbf{E}$: Exerts the force on a test charge, Is reduced inside a dielectric at fixed charge
- True of $\mathbf{D}$: Responds to deliberately placed charge alone, Is unchanged by inserting a dielectric at fixed free charge

**Why:** $\mathbf{E}$ responds to every charge, bound included, so it falls inside a dielectric, and it is what pushes a test charge. $\mathbf{D}$ tracks only free charge, so at fixed free charge it is untouched by the slab.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/sort-each-statement-by-whether-it-is-true-of-e-or-of-d

### 7. The displacement field D satisfies Gauss's law using only free charge, whereas the electric field E responds to all charge. What follows from this distinction when determining the actual electrostatic force experienced by a test charge inside a polarised dielectric?

A. The force depends on D because bound charges screen out the field E
B. The force depends on D because free charges determine the divergence
C. The force depends on E because bound charges exert physical forces
D. The force depends on E only if no polarisation charges are present

**Answer:** C. The force depends on E because bound charges exert physical forces

**Why:** Confusing D with the force-mediating field overlooks that bound charge is physically real. D is merely an auxiliary construct that absorbs polarisation into Gauss's law; test charges experience the net Coulomb force from both free and bound sources combined.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/the-displacement-field-d-satisfies-gausss-law-using-only-free

### 8. Complete the account of what a dielectric does to a field.

**Answer:** A dielectric **weakens** the field inside it, because its **bound** charge sits antiparallel to the free charge and partly cancels it. The reduction factor is the **relative permittivity**, which for an ordinary material is always greater than **one**.

**Why:** Polarisation puts bound charge on the slab faces opposing the free charge, so the net field falls by $\varepsilon_r = 1 + \chi_e > 1$. This is also why filling a capacitor raises its capacitance by the same factor.

Page: https://tryals.app/practice/physics-ii/dielectrics-and-polarisation/complete-the-account-of-what-a-dielectric-does-to-a-field