Conductors, Capacitors & Dielectrics: Master Notes

Chapter Overview

Capacitors store electrostatic potential energy in electric fields. In IIT-JEE (Advanced), the most tested problems involve dielectric slab insertion forces, spatially variable permittivity ($\kappa (x)$), charge redistribution heat dissipation, and RC transient charging/discharging curves.

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1. Capacitance & Standard Geometries

$$C=\frac{Q}{V},U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}=\frac{1}{2}QV$$

Geometry	Capacitance Formula $C$	Notes
Parallel Plate Capacitor	$C=\frac{{\epsilon }_{0}A}{d}$	With dielectric: $C=\frac{\kappa {\epsilon }_{0}A}{d}$
Spherical Capacitor (Radii $a,b$)	$C=4\pi {\epsilon }_0\left(\frac{ab}{b-a}\right)$	Isolated sphere ($b\to \mathrm{\infty }$): $C=4\pi {\epsilon }_{0}a$
Cylindrical Capacitor (Radii $a,b$)	$C=\frac{2\pi {\epsilon }_{0}L}{\ln (b/a)}$	Per unit length: $\frac{C}{L}=\frac{2\pi {\epsilon }_0}{\ln (b/a)}$

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2. Dielectric Slabs & Variable Permittivity

Multi-Mode DiagramCapacitors: Inhomogeneous Dielectric Insertion

Option 1: Publication-Grade Scientific Vector SVG

Parallel plate capacitor with dielectric slab insertion force and capacitance C = kappa * C_0.

Capacitance: $C = \frac{\varepsilon_0 A}{d - t + t/\kappa}$

Insertion Force: $F = \frac{1}{2} V^2 \frac{dC}{dx}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate electrostatic pulling force on dielectric slab.

Dielectric Constant κ: 5

Capacitance Multiplier ($C / C_0$):5x

Option 3: Interactive 3D Spatial Viewport

3D capacitor plates with dielectric insertion.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Stored Energy:$U = \frac{1}{2} C V^2$

Option 5: Reactive Parametric Physics with Live Calculus

Capacitance calculator with variable dielectric slab thickness.

Dielectric κ: 4

: 45

Capacitance ($C$):$50.0 pF$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

2.1 Slicing Strategy for Inhomogeneous Media

• If $\kappa =\kappa (y)$ varies perpendicular to plates: Slice into infinitesimal series capacitors of thickness $dy$:$$\frac{1}{C}={\int }_0^d\frac{dy}{\kappa (y){\epsilon }_{0}A}$$
• If $\kappa =\kappa (x)$ varies along the length of plates: Slice into infinitesimal parallel capacitors of area $dA=Wdx$:$$C={\int }_0^L\frac{\kappa (x){\epsilon }_{0}Wdx}{d}$$

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3. Force on a Dielectric Slab Entering a Capacitor

Consider a dielectric slab of width $w$ and dielectric constant $\kappa$ inserted by length $x$ into a parallel plate capacitor:

$$C(x)=\frac{{\epsilon }_{0}w}{d}[L+(\kappa -1)x]⟹\frac{dC}{dx}=\frac{(\kappa -1){\epsilon }_{0}w}{d}$$

• Case 1: Battery Connected ($V=\text{const}$):

  $$\boxed{{\displaystyle F=+\frac{1}{2}{V}^2\frac{dC}{dx}=\frac{(\kappa -1){\epsilon }_{0}w{V}^2}{2d}}}$$

  (The slab is pulled inside into the capacitor!)

• Case 2: Battery Disconnected ($Q=\text{const}$):

  $$\boxed{{\displaystyle F=-\frac{1}{2}\frac{Q^2}{C(x{)}^2}\frac{dC}{dx}}}$$

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4. Charge Redistribution & Heat Loss Formula

When two capacitors $(C_1,V_1)$ and $(C_2,V_2)$ are connected in parallel:

$$\text{Common Potential: }{V}_{\text{common}}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$$$$\boxed{{\displaystyle \text{Heat Dissipated }\mathrm{\Delta }H=U_i-U_f=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(V_1-V_2{)}^2}}$$

(Notice the exact mathematical identity to kinetic energy loss in completely inelastic collisions!)

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5. RC Circuit Transients

• Charging:$$q(t)=Q_0(1-e^{-t/\tau }),i(t)=I_0{e}^{-t/\tau }(\tau =RC)$$
• Discharging:$$q(t)=Q_0{e}^{-t/\tau },i(t)=-I_0{e}^{-t/\tau }$$

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6. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 — Linearly Varying Dielectric Constant

Question:
A parallel plate capacitor has plate area $A$ and separation $d$. A dielectric slab is placed between the plates such that its dielectric constant varies linearly with distance $x$ from one plate ($x=0$) to the other ($x=d$) as:

$$\kappa (x)={\kappa }_1+\left(\frac{{\kappa }_2-{\kappa }_1}{d}\right)x$$

Find the total capacitance $C$ of the system.

Step-by-Step Solution:

1. The permittivity varies perpendicular to the plates, so we treat it as an infinite stack of capacitors in series.
2. For an infinitesimal slice of thickness $dx$ at position $x$:$$d\left(\frac{1}{C}\right)=\frac{dx}{\kappa (x){\epsilon }_{0}A}$$
3. Integrating across the gap $x\in [0,d]$:$$\frac{1}{C}=\frac{1}{{\epsilon }_{0}A}{\int }_0^d\frac{dx}{{\kappa }_1+\left(\frac{{\kappa }_2-{\kappa }_1}{d}\right)x}$$Let $u={\kappa }_1+\left(\frac{{\kappa }_2-{\kappa }_1}{d}\right)x⟹du=\left(\frac{{\kappa }_2-{\kappa }_1}{d}\right)dx$:$$\frac{1}{C}=\frac{1}{{\epsilon }_{0}A}\left(\frac{d}{{\kappa }_2-{\kappa }_1}\right)[\ln u{]}_{{\kappa }_1}^{{\kappa }_2}=\frac{d}{({\kappa }_2-{\kappa }_1){\epsilon }_{0}A}\ln \left(\frac{{\kappa }_2}{{\kappa }_1}\right)$$
4. Inverting:$$\boxed{{\displaystyle C=\frac{({\kappa }_2-{\kappa }_1){\epsilon }_{0}A}{d\ln ({\kappa }_2/{\kappa }_1)}}}$$

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PYQ 2: JEE Advanced 2019 — Force & Work Done on Dielectric

Question:
A parallel plate capacitor of area $A$ and gap $d$ is connected to a battery of voltage $V$. A dielectric slab ($\kappa$) of thickness $d$ is slowly pulled completely out of the capacitor. Find the mechanical work $W_{\text{agent}}$ done by the external agent.

Step-by-Step Solution:

1. Initial and final capacitance:$$C_i=\frac{\kappa {\epsilon }_{0}A}{d},C_f=\frac{{\epsilon }_{0}A}{d}$$
2. Change in stored energy:$$\mathrm{\Delta }U=\frac{1}{2}(C_f-C_i)V^2=-\frac{1}{2}(\kappa -1)\frac{{\epsilon }_{0}A}{d}{V}^2$$
3. Work done by the battery ($\mathrm{\Delta }Q=(C_f-C_i)V$):$$W_{\text{battery}}=\mathrm{\Delta }Q\cdot V=(C_f-C_i)V^2=-(\kappa -1)\frac{{\epsilon }_{0}A}{d}{V}^2$$
4. First Law of Thermodynamics:$$W_{\text{agent}}+W_{\text{battery}}=\mathrm{\Delta }U⟹W_{\text{agent}}=\mathrm{\Delta }U-W_{\text{battery}}$$$$W_{\text{agent}}=-\frac{1}{2}(\kappa -1)\frac{{\epsilon }_{0}A}{d}{V}^2-[-(\kappa -1)\frac{{\epsilon }_{0}A}{d}{V}^2]=\boxed{{\displaystyle +\frac{1}{2}(\kappa -1)\frac{{\epsilon }_{0}A}{d}{V}^2}}$$

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7. High-Yield Formula Sheet

Concept	Formula	Notes
Energy Density	$u=\frac{1}{2}\kappa {\epsilon }_0{E}^2$	Stored in electric field
Heat Loss in Connection	$\mathrm{\Delta }H=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(V_1-V_2{)}^2$	Independent of resistance
Force on Slab (V const)	$F=\frac{1}{2}{V}^2\frac{dC}{dx}$	Attractive inward force
RC Time Constant	$\tau =RC$	$63.2\mathrm{\%}$ charged at $t=\tau$