Unit 01: Electrostatics ā¢ Chapter 02
Dielectrics, Polarization & Parallel Plate Capacitor
Microscopic dipole alignment, bound surface charge density, dielectric constant $K$, and parallel plate capacitance with slabs.
Socho ek spring-loaded door damper ko: jab aap door ko force se push karte hain, toh damper ek opposing resistance create karta hai, jisse net motion ease ho jati hai. Dielectric material bilkul wahi kaam capacitor plates ke beech karta hai. Iske atoms ke electron clouds external electric field ke opposite stretch ho jaate hain (polarization). Yeh polarized surface charges ek opposing internal field $\vec{E}_p$ create karte hain, jisse plates ke beech ka net electric field drop ho jata hai[cite: 13]. Aur jab potential difference $V$ ghat jata hai, toh same voltage par aur zyada charge hold karne ki ability yani Capacitance ($C = Q/V$) direct multiply ho jati hai[cite: 13]!
Polarization & Reduced Field
Net electric field inside a dielectric medium of dielectric constant $K$[cite: 13]:
Here $\vec{P}$ is the polarization vector (dipole moment per unit volume) and $\chi_e$ is electric susceptibility ($K = 1 + \chi_e$).
Parallel Plate Capacitance Formulae
Vacuum vs Partial Dielectric Slab insertion of thickness $t < d$[cite: 13]:
When completely filled ($t = d$): $C = K C_0$[cite: 13]. For a conducting slab of thickness $t$: $K = \infty \implies C = \frac{\varepsilon_0 A}{d - t}$.
Capacitance of a Parallel Plate Capacitor in Vacuum ($C_0 = \frac{\varepsilon_0 A}{d}$)[cite: 13]
1. Consider two large parallel conducting plates of area $A$ separated by distance $d$, carrying charges $+Q$ and $-Q$[cite: 13]. Surface charge density $\sigma = \frac{Q}{A}$[cite: 13].
2. In the outer regions (above plate 1 and below plate 2), electric fields cancel out: $E_{\text{outer}} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$[cite: 13].
3. In the inner gap, both fields point from positive to negative plate and add up constructively[cite: 13]:
4. Potential difference $V_0$ between the plates ($V_0 = E_0 d$):
5. By definition of capacitance $C_0 = \frac{Q}{V_0}$[cite: 13]:
Capacitance with a Dielectric Slab of Thickness $t < d$[cite: 13]
1. Introduce a dielectric slab of thickness $t$ and constant $K$ between the plates ($t < d$)[cite: 13].
2. The remaining vacuum gap of thickness $(d - t)$ maintains field $E_0$. Inside the slab, the field is reduced to $E = \frac{E_0}{K}$[cite: 13].
3. Total potential difference $V$ across the plates is the sum of potentials across both regions[cite: 13]:
4. Substitute $E_0 = \frac{Q}{\varepsilon_0 A}$:
5. Effective capacitance $C = \frac{Q}{V}$[cite: 13]:
Trap 1: Induced Surface Charge $q_p$ is NEVER equal to $q$
Students often set $q_p = q$. For a dielectric, induced bound charge is always strictly less: $q_p = q\left(1 - \frac{1}{K}\right)$[cite: 13]. It equals $q$ only if the inserted slab is a perfect metallic conductor ($K = \infty$).
Trap 2: Metallic Slab vs Dielectric Slab Distinction
If a copper/aluminium sheet of thickness $t$ is inserted inside a capacitor, electric field inside it becomes exactly zero ($\vec{E} = 0$)[cite: 13]. Do not write $K$; plug $K = \infty$ directly into the formula to get $C = \frac{\varepsilon_0 A}{d - t}$.
Trap 3: "Capacitance depends on Q or V" Fallacy
Just like resistance $R$ does not depend on current $I$, Capacitance $C = Q/V$ does NOT depend on charge $Q$ or potential difference $V$[cite: 13]. It is purely a geometric constant determined by plate shape ($A$), gap ($d$), and medium ($K$)[cite: 13].
Trap 4: Dielectric Breakdown Voltage
Every dielectric has a maximum dielectric strength (for air $\approx 3 \times 10^6\text{ V/m}$). Increasing $Q$ excessively will ionize the medium, causing a spark discharge between plates and destroying capacitor action.
Problem: A parallel plate capacitor has plate area $A = 100\text{ cm}^2$ and plate separation $d = 1\text{ mm}$. A dielectric slab of thickness $t = 0.6\text{ mm}$ and dielectric constant $K = 5$ is introduced between the plates. Calculate its capacitance before and after inserting the slab[cite: 13].
Solution:
1. Area $A = 100 \times 10^{-4}\text{ m}^2 = 10^{-2}\text{ m}^2$, $d = 10^{-3}\text{ m}$.
2. Initial capacitance in vacuum ($C_0$)[cite: 13]:
3. Capacitance with partial slab ($t = 0.6\text{ mm} = 0.6 \times 10^{-3}\text{ m}$)[cite: 13]:
Problem: A dielectric slab of constant $K$ fills half the space of a parallel plate capacitor. What is the equivalent capacitance if (i) slab divides the plate area $A$ into two halves ($A/2$), (ii) slab divides the plate separation $d$ into two halves ($d/2$)?
Solution:
Case (i) Area division (Parallel combination): Two capacitors in parallel, each of area $A/2$ and gap $d$[cite: 13]:
Case (ii) Thickness division (Series combination): Two capacitors in series, each of area $A$ and gap $d/2$[cite: 13]:
Capacitance in Vac
$C_0 = \frac{\varepsilon_0 A}{d}$[cite: 13]
Filled Dielectric
$C = K C_0$[cite: 13]
Partial Slab ($t < d$)
$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$[cite: 13]
Bound Charge
$q_p = q(1 - 1/K)$[cite: 13]