1E · Dielectrics — D, P, E + Solved PYQs

Write the relation between $\vec{D}$, $\vec{P}$ and $\vec{E}$ in a dielectric; define them and show $\vec{D} = \varepsilon_0\vec{E} + \vec{P}$.

Recall

Steps

1. Define each quantity with units: $\vec{D}$ — displacement linked to free charge on the plates, $D = \sigma_{\text{free}}$, unit C/m². $\vec{P}$ — dipole moment per unit volume of the dielectric, unit C/m². $\vec{E}$ — net field a test charge feels, unit N/C.
2. Place a slab between the plates: bound charges $\pm\sigma_p$ appear on its two faces. Net field: $E = (\sigma_{\text{free}} - \sigma_p)/\varepsilon_0$ (reason: superposition of the two sheet fields).
3. Identify terms: $D = \sigma_{\text{free}}$ and $P = \sigma_p$ (proved in Q3). So $\varepsilon_0 E = D - P$.
4. Rearrange: $\boxed{\vec{D} = \varepsilon_0\vec{E} + \vec{P}}$. Add the linear case $\vec{D} = \varepsilon\vec{E}$ and the free-charge Gauss law $\oint \vec{D}\cdot d\vec{S} = Q_{\text{free}}$.

Answer

Exam tip

What do you mean by polarization of a dielectric medium? Define the polarization vector.

Recall

Steps

1. Meaning: an external field stretches or aligns the molecules of the dielectric into tiny dipoles. The medium gains bound surface charge on its faces, but stays neutral overall (equal $+q$ and $-q$).
2. Define $\vec{P} = \sum\vec{p}/\Delta V$, the dipole moment per unit volume. Direction: from $-$ to $+$ inside the dielectric.
3. Unit: C/m² (C·m ÷ m³). For linear media: $\vec{P} = \varepsilon_0\chi_e\vec{E}$.

Answer

Exam tip

Show that the magnitude of polarization equals the surface density of induced (bound) charge.

Recall

Steps

1. Take a slab of area $A$ and thickness $d$, containing $N$ aligned molecular dipoles of charge $q$ and tip-to-tip separation $x$. Total dipole moment of the slab: $N q x$ (reason: moments of identical dipoles add).
2. By definition: $P = \dfrac{N q x}{A d}$ (reason: divide the total moment by the volume $A d$).
3. Bound charge on one face: each dipole deposits $+q$ on one face and $-q$ on the other. Total $N q$ spreads over area $A$, giving $\sigma_p = \dfrac{N q}{A}$ — but to relate it to $P$, note that shifting $N q$ across the slab thickness $d$ deposits charge $N q$ on the face of area $A$: $\sigma_p = \dfrac{N q x}{A d}$ (reason: charge × shift per volume = charge per area).
4. Compare with step 2: $\sigma_p = P$. ✓

Answer

Exam tip

Distinguish between polar and non-polar dielectrics. Give examples.

Recall

Steps

1. Polar dielectrics: molecules have a permanent $\vec{p} \neq 0$ even at zero field (reason: centres of positive and negative charge do not coincide, e.g. bent H₂O, HCl, NH₃). Without field, thermal motion randomises the orientations → net $P = 0$. A field partly aligns them against thermal disorder, so the effect falls as temperature rises.
2. Non-polar dielectrics: molecules have symmetric charge centres, so $\vec{p} = 0$ at $E = 0$ (reason: positive and negative centres coincide, e.g. H₂, O₂, N₂, CO₂). An applied field stretches them: induced $\vec{p} = \alpha\vec{E}$, almost independent of temperature.
3. Tabulate: permanent $\vec{p}$ vs zero $\vec{p}$; alignment vs stretching; temperature dependent vs not; examples on each side.

Answer

Exam tip

Show that the electric field decreases when a dielectric slab is introduced between the plates of a capacitor.

Recall

Steps

1. Before the slab: air-filled, $E_0 = \sigma/\varepsilon_0$ (reason: field of a single charged sheet, by Gauss's law on a pillbox).
2. After inserting the slab: bound charges $\pm\sigma_p$ appear on the two faces, with $\sigma_p$ opposite in sign to the adjacent free charge $\sigma$ (reason: dipoles align against the field).
3. Net field at any point in the gap: $E = (\sigma - \sigma_p)/\varepsilon_0 < E_0$ (reason: $\sigma_p > 0$ subtracts).
4. With dielectric constant $K$: $E = E_0/K = \sigma/(K\varepsilon_0)$. Limit: $K = 1$ (air) gives $E = E_0$. ✓

Answer

Exam tip