Chapter 02 · Detailed Study Hub

Electrostatic Potential & Capacitance Full Notes.

Chapter 2 exhaustive notes, theory derivations, equipotential geometries, dielectric shifts, and board numericals by Jatin Sandilya.

"Potential is stored work waiting for the right moment to act." — Jatin Sandilya
Engineered by Jatin Sandilya · High-Yield Board Portal
High-voltage capacitor discharge and energy circuit
energy confined within field lines
energy is stored work,
build your potential! ⚡
100% CompleteZero Skipped Proofs
Equipotential SymmetriesConcentric Contours
Dielectric Modifications$K \cdot C_0$ Boundary Conditions
Zero Paywalls100% Free Study Material
every equation runs in reality

Physics Meets Daily Life.
Hover or tap to flip for technological uses and profound life parallels.

Real-World Tech 01

Defibrillators & Energy Release

Batteries deliver energy steadily and slowly. A capacitor accumulates charge over seconds and discharges thousands of volts in a millisecond to shock a stopped heart back to rhythm.

Flip for life wisdom → $U = \frac{1}{2}CV^2$
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Life Connection: Patience & Burst
"Accumulate skills slowly in silence; release your impact when opportunity knocks." — Jatin Sandilya

Preparation is like charging a capacitor. The hours of unseen, disciplined study quietly build internal potential. When exam day arrives, that stored energy is released effortlessly.

Potential Energy in Action
Real-World Tech 02

Touchscreens & Equipotential Planes

Projected capacitive touchscreens maintain an equipotential electrostatic grid. Touching the glass alters the field gradient and pinpoints coordinate touches.

Flip for life wisdom → $E = -\frac{dV}{dr}$
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Life Connection: Zero Friction Living
"On an equipotential surface, moving a charge takes zero work." — Jatin Sandilya

When your mind is balanced and free of internal conflict, navigating daily challenges requires minimum emotional friction. Move without wasting mental energy on doubt.

Balance Yields Power
Real-World Tech 03

Dielectric Slabs & Capacity Boost

Air breaks down under high electric fields ($3 \times 10^6\text{ V/m}$). Inserting an insulating dielectric slab dampens the internal field, allowing the capacitor to hold $K$-times more charge safely.

Flip for life wisdom → Permittivity ($C = K C_0$)
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Life Connection: Inner Resilience
"A dielectric absorbs external electric stress to amplify capacity." — Jatin Sandilya

Life will test your breakdown threshold with pressure and stress. Developing emotional resilience acts like a dielectric insulator—it dampens external negativity and expands your capacity to handle challenges.

Resilience Prevents Breakdown
Real-World Tech 04

Lightning Protection & Aircraft Cages

Airplanes are struck by lightning regularly during storms, yet passengers are untouched because charge stays on the outer metallic skin while the interior field is strictly zero ($E = 0$).

Flip for life wisdom → Electrostatic Shielding
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Life Connection: Protecting What Matters
"Guard your inner calm like a cavity inside a conductor." — Jatin Sandilya

External storms of distractions, criticism, and chaos will strike daily. Maintain a solid mental shield around your goals so external turbulence never penetrates your core focus.

Invulnerability of the Focused Mind
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📌 Core NCERT Definitions

1 Mark Essentials
1. Conservative Force

A force field where total work done in transporting a charge between two points is completely independent of the path taken.

2. Electric Field Intensity

The electrostatic force experienced per unit infinitesimal positive test charge placed at that spatial location.

3. Electrostatic Potential ($V$)

External mechanical work required to bring a unit positive test charge from infinity up to a designated point without acceleration.

4. Work Done ($W$)

Energy transferred to or from a system via vector force acting across displacement ($W = \vec{F} \cdot \vec{d}$).

5. Potential Energy ($U$)

Energy stored within a collective system of charges governed by their relative spatial geometric configuration.

6. Test Charge ($q_0$)

An infinitesimal charge magnitude chosen explicitly so it does not distort the source electric field being mapped.

7. Point Charge Potential

The localized scalar potential generated at distance $r$ from an isolated point charge ($V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$).

8. Electric Dipole & Moment ($\vec{p}$)

Pair of equal and opposite charges ($\pm q$) separated by $2a$. Vector $\vec{p} = q(2\vec{a})$ points conventionally from $-q$ to $+q$.

9. Multi-Charge Superposition

Scalar superposition property: total potential equals direct algebraic sum of potentials produced by all individual charges.

💡 High-Yield Conceptual Pearls

Board Highlights
01

Work done moving a charge in an electrostatic field is independent of path, proving the conservative nature of electrostatic forces.

02

By standard universal reference convention, both electrostatic potential and potential energy are set to zero at infinity ($\infty$).

03

Point charge potential varies inversely with distance: $V \propto 1/r$, while electric field varies as $E \propto 1/r^2$.

04

Electrostatic potential along the entire perpendicular bisector (equatorial line) of a dipole is always exactly zero ($V = 0$), though electric field $\vec{E} \neq 0$.

05

In static equilibrium, electric potential inside any hollow or solid conductor is completely uniform and equal to its surface potential.

06

Potential is positive for positive source charges ($+q$) and negative for negative charges ($-q$). It is strictly a scalar quantity.

📐 Essential Formula Reference Bank

Formula Summary
1. External Work & Line Integral

Work required to move charge $q$ against field $\vec{E}$:

$$W_{RP} = \int_{R}^{P} \vec{F}_{\text{ext}} \cdot d\vec{r} = -\int_{R}^{P} \vec{F}_{\text{elec}} \cdot d\vec{r}$$
2. Point Charge Potential

Scalar potential at distance $r$ from charge $Q$:

$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$
3. Electric Dipole Potential ($r \gg a$)

Axial ($V = \pm\frac{kp}{r^2}$) | Equatorial ($V = 0$):

$$V_{\text{general}} = \frac{1}{4\pi\varepsilon_0}\frac{\vec{p}\cdot\hat{r}}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}$$
4. Multi-Charge System Potential

Superposition sum across $n$ discrete charges:

$$V = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i}$$

📊 Comparison: Point Charge vs Dipole Potential

Symmetry Matrix
Characteristic Isolated Point Charge Electric Dipole ($\vec{p}$)
Distance Scaling Inversely proportional to distance: $V \propto \frac{1}{r}$ Falls off faster: $V \propto \frac{1}{r^2}$
Angular Dependence Spherically symmetric (independent of angle $\theta$) Depends explicitly on polar angle ($\propto \cos\theta$)
Equatorial Potential Non-zero at all finite points Strictly zero at all equatorial bisector points ($V=0$)
Spatial Symmetry Spherical symmetry Axial / Cylindrical symmetry

⚔️ Contrast: Electric Field vs Electric Potential

2–3 Mark Questions
Parameter Electric Field ($\vec{E}$) Electrostatic Potential ($V$)
Physical Nature Vector quantity (magnitude + spatial direction) Scalar quantity (magnitude + algebraic sign)
Physical Role Electrostatic force experienced per unit test charge External work required per unit test charge from $\infty$
SI Unit $\text{N/C}$ or $\text{V/m}$ $\text{Volt (V)}$ or $\text{J/C}$
Gradient Formula $\vec{E} = -\frac{dV}{dr}\hat{r}$ (negative spatial gradient) $V = -\int_\infty^r \vec{E} \cdot d\vec{r}$
Field Direction Points toward decreasing potential Has no spatial direction vector attribute

⚡ Derivation: Electric Potential Due to a Point Charge

Core Proof
∞ +q (O) P q₀ (at x) B dx F r (Point P) x (Position of A)

Fig 2.2: Coordinate layout for deriving electrostatic potential at point $P$ from charge $+q$.

Step 1 Repulsive Coulomb force acting on test charge $q_0$ at intermediate distance $x$:
$$F = \frac{1}{4\pi\varepsilon_0}\frac{q q_0}{x^2}$$
Step 2 Incremental work done against repulsive force across displacement $dx$ ($\theta = 180^\circ$):
$$dW = \vec{F} \cdot d\vec{x} = F dx \cos(180^\circ) = -F dx$$
Step 3 Total work required to bring charge $q_0$ from infinity ($\infty$) to radial distance $r$:
$$W = -\int_{\infty}^{r} \left( \frac{1}{4\pi\varepsilon_0}\frac{q q_0}{x^2} \right) dx = -\frac{q q_0}{4\pi\varepsilon_0} \left[ -\frac{1}{x} \right]_{\infty}^{r} = \frac{q q_0}{4\pi\varepsilon_0 r}$$
Step 4 By definition, potential $V = W / q_0$:
$$V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$$

📈 Graphical Decay: Potential ($V$) vs Field ($E$)

Asymptotic Curves
r → V, E ↑ E = kq / r² (Steeper) V = kq / r (Smoother)

Fig 2.3: Comparison showing steeper inverse-square drop ($E \propto 1/r^2$) vs smoother decay ($V \propto 1/r$).

🌐 Equipotential Geometries & Energy Assemblies

Visual Topology

Equipotential Contours (Point Charge)

V₁ V₂

Concentric spherical surfaces perpendicular to radial field lines ($V_1 > V_2 > V_3$).

Three-Charge Energy Assembly

q₃ q₁ q₂ r₁₃ r₂₃ r₁₂

Total Stored Energy: $U = \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1q_2}{r_{12}} + \frac{q_2q_3}{r_{23}} + \frac{q_3q_1}{r_{31}}\right]$.

⚡ Mathematical Framework & Quantitative Relations

Analytical Overview

Spatial gradient relationships linking fields and scalar potentials, multi-charge interaction energies within external fields, and dipole rotational dynamics.

Gradient Connection

Spatial Field-Potential Link

The magnitude of the local electric field intensity equals the maximum spatial rate of potential decrease.

$$|E| = -\frac{\delta V}{\delta l}$$
External Field

Charge Energy in External Potential

Electrostatic potential energy acquired by positioning a charge within an existing spatial potential distribution.

$$U = q \cdot V(r)$$
Orientation State

Dipole Alignment Energy

Potential energy stored in a dipole oriented at angle $\theta$ relative to a uniform electric field.

$$U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$$
Torque & Work

Rotational Work on Dipole

Net external mechanical work required to pivot an electric dipole between two angular orientations.

$$W = pE(\cos\theta_1 - \cos\theta_2)$$
Multi-Body System

Potential Energy of Two Charges in External Field

Includes external field interactions of each individual charge along with their mutual Coulombic interaction energy:

$$U = q_1 V(r_1) + q_2 V(r_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$
x y O q₁ q₂ r⃗₁ r⃗₂ r₁₂

Spatial position vectors $\vec{r}_1$, $\vec{r}_2$, and mutual separation $r_{12}$.

🗺️ Spatial Field Mapping & Equipotential Topologies

Geometry Specs
Source

Isolated Point Charge

Contour Geometry

Concentric spherical shells radiating outward symmetrically from charge center.

Source

Uniform Electric Field

Contour Geometry

Equidistant parallel planar sheets perpendicular to electric field vectors.

Source

Electric Dipole Pairing

Contour Geometry

Non-uniform distorted envelopes closing around opposing charges; flat plane at center ($V=0$).

Source

Two Positive Charges (+q, +q)

Contour Geometry

Repelling surfaces close to charges, merging into a single dumbbell envelope further away.

Source

Infinite Line Charge

Contour Geometry

Coaxial cylindrical surfaces concentric with the charged wire axis.

Proof: Electric Field as Negative Potential Gradient ($E = -dV/dr$)

Core Derivation
E⃗ B (V) A (V + dV) dr⃗

Displacement $d\vec{r}$ connecting adjacent equipotential levels within field $\vec{E}$.

1. External work done to move unit test charge $q_0$ from surface $A$ to $B$ against electric field $\vec{E}$:

$$dW = \vec{F}_{\text{ext}} \cdot d\vec{r} = -q_0 \vec{E} \cdot d\vec{r}$$

2. By potential difference definition: $\frac{dW}{q_0} = V_B - V_A = V - (V + dV) = -dV$.

3. Equating both expressions:

$$-\vec{E} \cdot d\vec{r} = -dV \implies E = -\frac{dV}{dr}$$

Physical Takeaway: The negative sign dictates that electric field lines point in the direction of steepest potential decline.

Unit Equivalency: $\text{V/m} \equiv \text{N/C}$
$$\frac{\text{Volt}}{\text{metre}} = \frac{\text{Joule / Coulomb}}{\text{metre}} = \frac{(\text{Newton} \cdot \text{metre}) / \text{Coulomb}}{\text{metre}} = \frac{\text{Newton}}{\text{Coulomb}} \implies \mathbf{1\text{ V/m} = 1\text{ N/C}}$$

Potential Energy Assembly: Two Point Charges ($U = \frac{kq_1q_2}{r_{12}}$)

Core Derivation
Step 1 Bringing charge $q_1$ from infinity to point $P_1$ requires zero work since no external field is initially present ($W_1 = 0$).
Step 2 Potential established by $q_1$ at location $P_2$ at distance $r_{12}$:
$$V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_{12}}$$
Step 3 Work performed bringing charge $q_2$ from infinity to point $P_2$:
$$W_2 = q_2 V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$
$$U = W_1 + W_2 = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$

⚡ Mathematical Framework & Quantitative Relations

Analytical Overview

Spatial gradient relationships linking fields and potentials, multi-charge interaction energies within external fields, and dipole rotational dynamics.

Gradient Connection

Spatial Field-Potential Link

The magnitude of local electric field intensity equals the maximum spatial rate of potential decrease.

$$|E| = -\frac{\delta V}{\delta l}$$
External Field

Charge Energy in External Potential

Electrostatic potential energy acquired by positioning a charge within an existing spatial potential distribution.

$$U = q \cdot V(r)$$
Orientation State

Dipole Alignment Energy

Potential energy stored in a dipole oriented at angle $\theta$ relative to a uniform electric field.

$$U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$$
Torque & Work

Rotational Work on Dipole

Net external mechanical work required to pivot an electric dipole between two angular orientations.

$$W = pE(\cos\theta_1 - \cos\theta_2)$$
Multi-Body System

Potential Energy of Two Charges in External Field

Includes external field interactions of each charge along with their mutual Coulombic interaction energy:

$$U = q_1 V(r_1) + q_2 V(r_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$
x y O q₁ q₂ r⃗₁ r⃗₂ r₁₂

Spatial position vectors $\vec{r}_1$, $\vec{r}_2$, and mutual separation $r_{12}$.

🗺️ Spatial Field Mapping & Equipotential Topologies

Geometry Specs
Source

Isolated Point Charge

Contour Geometry

Concentric spherical shells radiating outward symmetrically from charge center.

Source

Uniform Electric Field

Contour Geometry

Equidistant parallel planar sheets perpendicular to electric field vectors.

Source

Electric Dipole Pairing

Contour Geometry

Distorted envelopes closing around opposing charges; flat plane at center ($V=0$).

Source

Two Positive Charges (+q, +q)

Contour Geometry

Repelling surfaces close to charges, merging into a dumbbell envelope further away.

Source

Infinite Line Charge

Contour Geometry

Coaxial cylindrical surfaces concentric with the charged wire axis.

Analytical Coupling: Field as Potential Gradient ($E = -dV/dr$)

Core Derivation
E⃗ B (V) A (V + dV) dr⃗

Displacement $d\vec{r}$ between adjacent equipotential levels within field $\vec{E}$.

1. External work done to move unit test charge $q_0$ from surface $A$ to $B$ against field $\vec{E}$:

$$dW = \vec{F}_{\text{ext}} \cdot d\vec{r} = -q_0 \vec{E} \cdot d\vec{r}$$

2. By potential difference definition: $\frac{dW}{q_0} = V_B - V_A = V - (V + dV) = -dV$.

3. Equating both expressions:

$$-\vec{E} \cdot d\vec{r} = -dV \implies E = -\frac{dV}{dr}$$

Physical Takeaway: Electric field vectors point in the direction of steepest potential decline.

4. Potential from known field via integration ($\vec{r}_1 \to \infty, V_1 = 0$):

$$V(\vec{r}) = -\int_{\infty}^{\vec{r}} \vec{E} \cdot d\vec{r}$$
Unit Equivalency: $\text{V/m} \equiv \text{N/C}$
$$\frac{\text{Volt}}{\text{metre}} = \frac{\text{Joule / Coulomb}}{\text{metre}} = \frac{(\text{Newton} \cdot \text{metre}) / \text{Coulomb}}{\text{metre}} = \frac{\text{Newton}}{\text{Coulomb}} \implies \mathbf{1\text{ V/m} = 1\text{ N/C}}$$

Potential Energy Assembly: Two Point Charges ($U = \frac{kq_1q_2}{r_{12}}$)

Core Derivation
q₁ (at P₁) q₂ (at P₂) r₁₂

Positioning charge $q_1$ followed by the arrival of charge $q_2$ from infinity.

Step 1 Bringing charge $q_1$ from infinity to point $P_1$ requires zero work since no external field opposes it ($W_1 = 0$).
Step 2 Potential established by $q_1$ at location $P_2$ at distance $r_{12}$:
$$V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_{12}}$$
Step 3 Work performed bringing charge $q_2$ from infinity to point $P_2$:
$$W_2 = q_2 V_1 = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$
$$U = W_1 + W_2 = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$

Electrostatic Potential of a Charged Spherical Shell (Radius $R$, Charge $Q$)

Radial Zones
O R P (r > R) r

Spherical shell of radius $R$ and observation point $P$.

r V R V ∝ 1/r V = Constant

Radial potential decay: constant inside, $1/r$ outside.

1. External Point ($r > R$)

Acts as if entire charge $Q$ is concentrated at center:

$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$$
2. Surface Point ($r = R$)

Potential at the outer conductive boundary:

$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$$
3. Internal Point ($r < R$)

Because internal field $E = 0$, potential remains constant:

$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R} \quad (\text{Constant})$$

Conductors in Equilibrium & Electrostatic Shielding

Physical Laws
Property I: Zero Internal Field

Free electrons redistribute instantly until induced internal fields cancel external fields completely ($\vec{E} = 0$).

Property II: Surface Normal Field

Electric field lines are strictly perpendicular to the outer surface. Any tangential component would trigger continuous current.

Property III: Absence of Interior Charge

Applying Gauss's Law inside the metal reveals zero net volume charge; all excess charge resides strictly on the outer boundary.

Property IV: Equipotential Volume

Because $\vec{E} = -\frac{dV}{dr} = 0$, the potential gradient vanishes, maintaining identical potential everywhere inside and on the surface.

Surface Field Derivation: Gaussian Pillbox ($E = \sigma / \varepsilon_0$)

Constructing a tiny cylindrical pillbox across the surface of a charged conductor with cross-sectional area $\Delta S$:

$$\Phi_E = E \cdot \Delta S = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\sigma \Delta S}{\varepsilon_0} \implies \mathbf{E = \frac{\sigma}{\varepsilon_0}}$$

Faraday Cage Principle: Field inside a cavity within a conductor is zero regardless of external fields. Protects passengers in cars/airplanes during lightning strikes.

Electric Dipole: Torque, Potential Energy & Equilibrium States

Torque & Energy
E⃗ -q +q θ

Rotational couple $\vec{\tau} = \vec{p} \times \vec{E}$ acting on dipole in uniform field.

1. Restoring torque magnitude: $\tau = qE(2a\sin\theta) = pE\sin\theta \implies \vec{\tau} = \vec{p} \times \vec{E}$.

2. Integrating work done during rotation from $\theta_1$ to $\theta_2$:

$$W = \int_{\theta_1}^{\theta_2} pE\sin\theta \, d\theta = pE(\cos\theta_1 - \cos\theta_2)$$

3. Potential energy definition ($U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$):

Stable Equilibrium ($\theta = 0^\circ$) $\tau = 0$, $U_{\min} = -pE$
Zero Energy ($\theta = 90^\circ$) $\tau = pE$, $U = 0$
Unstable Equilibrium ($\theta = 180^\circ$) $\tau = 0$, $U_{\max} = +pE$

Capacitance Standards, Metric Units & Isolated Spherical Conductor

The Farad Unit

Fundamental Definition: $Q \propto V \implies Q = CV \implies C = \frac{Q}{V}$. The SI unit is the Farad ($\text{F}$), where $1\text{ F} = 1\text{ C/V}$. Dimensional formula: $[M^{-1}L^{-2}T^4A^2]$.

Capacitance of an Isolated Spherical Conductor ($R$)

Surface potential due to uniform charge $Q$: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$.

$$C = \frac{Q}{V} = \mathbf{4\pi\varepsilon_0 R} \quad (C \propto R)$$

Planetary Context: For an isolated sphere to have $C = 1\text{ F}$, radius $R = 9 \times 10^9\text{ m} = 9 \times 10^6\text{ km}$, roughly 1,500 times larger than Earth's radius ($R_E \approx 6400\text{ km}$). This proves a single isolated farad is impossibly huge; practical units use $\mu\text{F}, \text{nF}, \text{pF}$.

Solved Exemplar: Coalescence of 8 Identical Charged Droplets

Numerical Analysis

Problem: Eight identical spherical droplets, each containing charge $q = 1\text{ nC}$ and at potential $V = 900\text{ V}$, merge without loss to form a single large drop. Compute the resultant potential $V'$.

1. Volume Conservation: $\frac{4}{3}\pi R^3 = 8\left(\frac{4}{3}\pi r^3\right) \implies R = 2r$.

2. Charge Conservation: Total charge $Q' = 8q$.

3. Capacitance Scaling: $C' = 4\pi\varepsilon_0 R = 4\pi\varepsilon_0(2r) = 2C$.

4. Resultant Potential:

$$V' = \frac{Q'}{C'} = \frac{8q}{2C} = 4 \left(\frac{q}{C}\right) = 4 \times 900\text{ V} = \mathbf{3600\text{ V}}$$

Redistribution of Charges & Proof of Energy Loss ($\Delta U$)

Sharing Theorem
A C₁, V₁ B C₂, V₂ Conducting Wire

Conductors linked by a wire equalize at common potential $V$.

1. Common Potential Formula:

$$V = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$

2. Charge Sharing Ratio: After sharing, $Q_1' = C_1 V$ and $Q_2' = C_2 V$:

$$\frac{Q_1'}{Q_2'} = \frac{C_1}{C_2} \quad \text{(Charges divide in direct ratio of capacities)}$$

3. Mathematical Proof of Energy Dissipation ($\Delta U = U_i - U_f$):

$$\Delta U = \frac{1}{2}\frac{C_1 C_2 (V_1 - V_2)^2}{C_1 + C_2}$$

Because $(V_1 - V_2)^2 > 0$, energy loss is strictly positive ($\Delta U > 0$). This energy dissipates as heat in the connecting wire and spark radiation.

Multi-Dielectric Composite Slabs: Series vs Parallel Formulations

Composite Geometry
Tandem Series Slabs ($d = t_1 + t_2 + \dots$)

Slabs stacked sequentially across thickness $d$. System behaves as capacitors in series:

$$C = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2} + \dots + \frac{t_n}{K_n}}$$
Side-by-Side Parallel Slabs ($A = A_1 + A_2 + \dots$)

Slabs placed side-by-side spanning separation $d$. System behaves as capacitors in parallel:

$$C = \frac{\varepsilon_0}{d}(K_1 A_1 + K_2 A_2 + \dots + K_n A_n)$$

Concentric Spheres: Potential Difference & Unidirectional Charge Transfer

Core Principle
Outer Shell (R, Q) r, q

Concentric sphere ($r, q$) inside hollow shell ($R, Q$).

1. Potential of inner sphere: $V_r = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{r} + \frac{Q}{R}\right]$.

2. Potential of outer shell: $V_R = \frac{1}{4\pi\varepsilon_0}\left[\frac{Q + q}{R}\right]$.

3. Potential Difference:

$$V_r - V_R = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r} - \frac{1}{R}\right)$$

Because $r < R$, the term $\left(\frac{1}{r} - \frac{1}{R}\right) > 0$. Hence, if $q > 0$, $V_r$ is always strictly greater than $V_R$. When connected by a wire, charge $q$ flows entirely to the outer shell regardless of how large charge $Q$ is.

🛡️️ Conductors in Electrostatic Equilibrium & Shielding

Fundamental Laws
E⃗_ext → E⃗_ext → − − − − −σ_ind + + + + +σ_ind ← E⃗_ind E⃗_net = E⃗_ext + E⃗_ind = 0

Fig 2.6: Free electrons migrate until induced internal field completely cancels applied external field ($\vec{E} = 0$).

I. Interior Field is Zero

Induced boundary charges create counter-fields neutralizing any applied electric field inside the metallic bulk.

II. Surface Normal Field

Field lines meet conductor surfaces strictly at $90^\circ$. Any tangential component would induce continuous surface current.

III. Zero Interior Net Charge

Gauss's law on any closed loop inside where $\vec{E}=0$ yields $q_{\text{enclosed}} = 0$. All excess charge sits on the outer boundary.

IV. Constant Potential Volume

Because $E = -dV/dr = 0$, potential gradient vanishes. Potential throughout the volume and surface is completely uniform.

Surface Electric Field Proof: Gaussian Pillbox ($E = \sigma / \varepsilon_0$)

Gauss Application
Conductor Bulk (E = 0 inside) + + + + Charge: σΔS Area ΔS E⃗

Fig 2.7: Cylindrical Gaussian pillbox spanning the conductor boundary.

1. Inside the conductor, field is zero; bottom cap contributes zero flux.

2. Field is normal to the surface; curved cylindrical walls are parallel to $\vec{E}$, contributing zero flux.

3. Flux emerges solely through the upper cap of area $\Delta S$:

$$\Phi_E = E \cdot \Delta S$$

4. Total charge enclosed with surface density $\sigma$ is $q = \sigma \Delta S$. Applying Gauss's law:

$$E \cdot \Delta S = \frac{\sigma \Delta S}{\varepsilon_0} \implies \mathbf{E = \frac{\sigma}{\varepsilon_0} \hat{n}}$$

Rotational Dynamics & Potential Energy of an Electric Dipole

Torque & Work
E⃗ −q −qE⃗ +q +qE⃗ θ 2a

Fig 2.8: Equal and opposite electrostatic forces creating couple torque $\vec{\tau} = \vec{p} \times \vec{E}$.

1. Net Translatory Force: $\vec{F}_{\text{net}} = (+q\vec{E}) + (-q\vec{E}) = 0$. Dipole experiences no net linear acceleration.

2. Torque Magnitude: Force $\times$ perpendicular arm: $\tau = qE(2a\sin\theta) = pE\sin\theta \implies \vec{\tau} = \vec{p} \times \vec{E}$.

3. Angular Work: Work in rotating dipole from $\theta_1$ to $\theta_2$ against restoring torque:

$$W = \int_{\theta_1}^{\theta_2} pE\sin\theta \, d\theta = pE(\cos\theta_1 - \cos\theta_2)$$

4. Potential Energy: $U(\theta) = -\vec{p}\cdot\vec{E} = -pE\cos\theta$.

Stable Equilibrium ($\theta = 0^\circ$) $\tau = 0$, $U_{\min} = -pE$ (aligned)
Zero Energy State ($\theta = 90^\circ$) $\tau = pE$ (maximum), $U = 0$
Unstable Equilibrium ($\theta = 180^\circ$) $\tau = 0$, $U_{\max} = +pE$ (opposed)

Working Principle of a Capacitor: Electrostatic Induction & Grounding

Capacity Boost
A (+) B − − − + + + Step 1: Induction lowers V_A slightly A (+) B (−) − − − Ground Step 2: V_A drops sharply → C jumps up!

Fig 2.9: Grounding outer positive charge leaves bound negative charge, dramatically lowering potential and amplifying capacitance ($C = Q/V$).

1. When uncharged plate $B$ approaches charged plate $A$, induced negative charge on $B$'s inner face lowers $A$'s potential, while induced positive charge on $B$'s outer face raises it.

2. Because negative charges are nearer to $A$, the net potential of $A$ decreases slightly.

3. Earthing Effect: When plate $B$ is grounded, free positive charges are neutralized by electrons from earth. Bound negative charges remain, driving potential $V_A$ down substantially.

4. Since $C = Q / V$, lowering potential for the same charge expands capacitance exponentially.

Parallel-Plate Capacitor: Field Superposition & Capacitance Derivation

Core Derivation
Upper Plate: Charge +Q, Density +σ (Area A) Lower Plate: Charge −Q, Density −σ (Area A) E = σ / ε₀ Fringing d Outer Zone I: E = (σ/2ε₀) − (σ/2ε₀) = 0 Outer Zone II: E = (σ/2ε₀) − (σ/2ε₀) = 0

Fig 2.10: Complete regional field distribution showing outer cancellation ($E=0$) and inner reinforcement ($E = \sigma/\varepsilon_0$).

1. Outer Regions: Opposing fields from plates cancel: $E_{\text{outer}} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.

2. Inner Region: Fields add up constructively:

$$E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$

3. Potential Difference: Uniform field across distance $d$ gives $V = E \cdot d = \frac{Q d}{\varepsilon_0 A}$.

4. Capacitance Expression: Substituting into $C = Q / V$:

$$C_0 = \frac{\varepsilon_0 A}{d}$$
Governing Structural Factors: • Area ($C \propto A$): Larger plates store more charge.
• Separation ($C \propto 1/d$): Bringing plates closer boosts attraction.
• Medium ($C \propto K$): Dielectrics amplify capacity by factor $K$.

📐 Parallel-Plate Capacitor: Field Superposition & Capacitance Derivation

Core Derivation

Two expansive planar conducting armatures of active area $A$, separated across distance $d$, carrying uniform surface charge densities $\pm\sigma = \pm Q/A$.

Upper Plate: +Q, Density +σ (Area A) Lower Plate: −Q, Density −σ (Area A) E = σ / ε₀ Fringing d Outer Zone I: E = (σ/2ε₀) − (σ/2ε₀) = 0 Outer Zone II: E = (σ/2ε₀) − (σ/2ε₀) = 0

Fig 2.10: Regional field distribution showing cancellation in outer regions ($E=0$) and inner constructive reinforcement ($E = \sigma/\varepsilon_0$).

1. Outer Cancellation: Opposing fields from plates cancel: $E_{\text{outer}} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.

2. Inner Reinforcement: Internal fields add constructively: $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$.

3. Terminal Potential Drop: Uniform field across distance $d$ yields: $V = E \cdot d = \frac{\sigma d}{\varepsilon_0} = \frac{Q d}{\varepsilon_0 A}$.

4. Capacitance Expression ($C = Q / V$):

$$C_0 = \frac{\varepsilon_0 A}{d}$$
1. Plate Surface Area ($C \propto A$) Expansive plate areas accommodate larger total charge accumulations.
2. Inter-Plate Separation ($C \propto 1/d$) Decreasing plate distance strengthens electrostatic attraction, raising capacity.
3. Medium Permittivity ($C \propto \varepsilon$) Inserting dielectric media ($K$) dampens internal field and enhances capacity ($C = K C_0$).

🌐 Concentric Spherical Storage Architectures & Capacitance Derivation

Spherical Shells

Dual concentric spherical armatures: inner shell of radius $a$ carrying charge $-Q$, and outer shell of radius $b$ carrying charge $+Q$.

Outer Shell (b, +Q) Gaussian Surface (r) a, −Q O

Fig 2.11: Concentric spherical configuration with intermediate Gaussian surface $r$ ($a < r < b$).

1. Field in the Gap: Applying Gauss's law for $a < r < b$: $E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$.

2. Integrating Potential Difference ($V$):

$$V = \int_{a}^{b} E dr = \frac{Q}{4\pi\varepsilon_0}\int_{a}^{b} r^{-2} dr = \frac{Q}{4\pi\varepsilon_0}\left(\frac{b - a}{ab}\right)$$

3. Capacitance Formula ($C = Q / V$):

$$C = \frac{4\pi\varepsilon_0 ab}{b - a}$$

🔢 Solved Exemplars: Voltage-Charge Variations

Board Problem Sets
Problem 1: Voltage Reduction & Subsequent Rise

A capacitor holds $360\,\mu\text{C}$ at potential $V$. Dropping potential by $120\text{ V}$ reduces charge to $120\,\mu\text{C}$. Find $V$, $C$, and charge if voltage rises by $120\text{ V}$.

• $\frac{V}{V - 120} = \frac{360}{120} = 3 \implies V = 180\text{ V}$

• $C = \frac{Q}{V} = \frac{360\,\mu\text{C}}{180\text{ V}} = \mathbf{2\,\mu\text{F}}$

• If $V$ rises by $120\text{ V}$: $Q'' = C(V + 120) = 2\,\mu\text{F}(300\text{ V}) = \mathbf{600\,\mu\text{C}}$

Problem 2: Unknown Capacitance & Voltage Shifts

A capacitor holds $120\,\mu\text{C}$ at potential $V$. Lowering potential by $40\text{ V}$ drops charge to $40\,\mu\text{C}$. Find base potential $V$, capacity $C$, and charge if voltage increases by $40\text{ V}$.

• $\frac{120}{V} = \frac{40}{V - 40} \implies 3(V - 40) = V \implies V = 60\text{ V}$

• $C = \frac{120\,\mu\text{C}}{60\text{ V}} = \mathbf{2\,\mu\text{F}}$

• If $V$ rises by $40\text{ V}$: $Q_{\text{new}} = C(V + 40) = 2\,\mu\text{F}(100\text{ V}) = \mathbf{200\,\mu\text{C}}$

🔌 Network Architectures: Series vs. Parallel Circuits

Equivalent Capacitance
Tandem Series Circuit

Charge Constant ($Q$), Voltages Divide

C₁ V₁ C₂ V₂ C₃ V₃ +Q −Q
$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

• Total voltage: $V = V_1 + V_2 + V_3$.
• Equivalent capacity is smaller than the smallest component in the chain.

Parallel Bus Circuit

Voltage Constant ($V$), Charges Add

+V 0V C₁ (Q₁) C₂ (Q₂) C₃ (Q₃)
$$C_p = C_1 + C_2 + C_3$$

• Total charge: $Q = Q_1 + Q_2 + Q_3$.
• Equivalent capacity is larger than the largest component in the network.

⚡ Stored Electrostatic Energy & Field Energy Density ($u$)

Volumetric Density
Plate 1 (+Q') Plate 2 (−Q') Field E⃗ (Stored Energy) dQ' Work done $dW = V' dQ' = \frac{Q'}{C} dQ'$

Fig 2.12: Incremental charge transfer $dQ'$ establishing internal field energy density.

1. Integration of Work Done: $W = \int_{0}^{Q} \frac{Q'}{C} dQ' = \frac{1}{C}\left[\frac{Q'^2}{2}\right]_{0}^{Q} = \frac{1}{2}\frac{Q^2}{C}$.

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

2. Volumetric Energy Density Derivation ($u$):

$$u = \frac{U}{\text{Volume}} = \frac{\frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)(E d)^2}{A \cdot d} \implies \mathbf{u = \frac{1}{2}\varepsilon_0 E^2}$$

Universal Principle: This energy density formula holds true for electric fields of any geometry or origin (measured in $\text{J/m}^3$).

SHORVEER NOTES

⚡ Mathematical Framework & Quantitative Relations

Analytical Overview

Spatial gradient relationships linking fields and potentials, multi-charge interaction energies within external fields, and dipole rotational dynamics.

Gradient Connection

Spatial Field-Potential Link

The magnitude of local electric field intensity equals the maximum spatial rate of potential decrease.

$$|E| = -\frac{\delta V}{\delta l}$$
External Field

Charge Energy in External Potential

Electrostatic potential energy acquired by positioning a charge within an existing spatial potential distribution.

$$U = q \cdot V(r)$$
Orientation State

Dipole Alignment Energy

Potential energy stored in a dipole oriented at angle $\theta$ relative to a uniform electric field.

$$U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$$
Torque & Work

Rotational Work on Dipole

Net external mechanical work required to pivot an electric dipole between two angular orientations.

$$W = pE(\cos\theta_1 - \cos\theta_2)$$
Multi-Body System

Potential Energy of Two Charges in External Field

Includes external field interactions of each charge along with their mutual Coulombic interaction energy:

$$U = q_1 V(r_1) + q_2 V(r_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$$
x y O q₁ q₂ r⃗₁ r⃗₂ r₁₂

Spatial position vectors $\vec{r}_1$, $\vec{r}_2$, and mutual separation $r_{12}$.

🗺️️ Spatial Field Mapping & Equipotential Topologies

Geometry Specs
Source

Isolated Point Charge

Contour Geometry

Concentric spherical shells radiating outward symmetrically from charge center.

Source

Uniform Electric Field

Contour Geometry

Equidistant parallel planar sheets perpendicular to electric field vectors.

Source

Electric Dipole Pairing

Contour Geometry

Distorted envelopes closing around opposing charges; flat plane at center ($V=0$).

Parallel-Plate Capacitor: Field Superposition & Capacitance Derivation

Core Derivation

Two expansive planar conducting armatures of active area $A$, separated across distance $d$, carrying uniform surface charge densities $\pm\sigma = \pm Q/A$.

Upper Plate: +Q, Density +σ (Area A) Lower Plate: −Q, Density −σ (Area A) E = σ / ε₀ Fringing d Outer Zone I: E = (σ/2ε₀) − (σ/2ε₀) = 0 Outer Zone II: E = (σ/2ε₀) − (σ/2ε₀) = 0

Regional field distribution showing cancellation in outer regions ($E=0$) and inner reinforcement ($E = \sigma/\varepsilon_0$).

1. Outer Cancellation: Opposing fields cancel: $E_{\text{outer}} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.

2. Inner Reinforcement: $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$.

3. Potential Difference: $V = E \cdot d = \frac{Q d}{\varepsilon_0 A}$.

$$C_0 = \frac{\varepsilon_0 A}{d}$$
1. Plate Area ($C \propto A$) Larger plates store more charge.
2. Plate Separation ($C \propto 1/d$) Closer plates boost electrostatic attraction.

⚡ Stored Electrostatic Energy & Field Energy Density ($u$)

Volumetric Density

1. Work Done during Charging: $W = \int_{0}^{Q} \frac{Q'}{C} dQ' = \frac{1}{2}\frac{Q^2}{C}$.

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

2. Volumetric Energy Density ($u$):

$$u = \frac{U}{\text{Volume}} = \mathbf{\frac{1}{2}\varepsilon_0 E^2}$$

1. Parallel-Plate Capacitor: Internal Field & Geometric Capacitance

Upper Plate: +Q, Surface Density +σ (Area A) Lower Plate: −Q, Surface Density −σ (Area A) E = σ / ε₀ = Q / (Aε₀) d Outer Zone I: E = (σ/2ε₀) − (σ/2ε₀) = 0 Outer Zone II: E = (σ/2ε₀) − (σ/2ε₀) = 0

Fig 2.13: Uniform field formation between plates and cancellation in external regions.

1. Field Formulation: Electric field intensity between plates is given by $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$.

2. Terminal Potential Difference: $V = E \cdot d = \frac{Q d}{\varepsilon_0 A}$.

3. Capacitance Equation:

$$C_0 = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}$$
Plate Area ($C \propto A$) Directly proportional: larger surface areas allow greater charge accumulation.
Separation Distance ($C \propto 1/d$) Inversely proportional: decreasing distance elevates electrostatic coupling.
Medium Permittivity ($C \propto K$) Directly proportional: dielectric constant $K$ multiplies total storage capacity ($C = K C_0$).

2. Concentric Spherical Capacitor Derivation

Outer Shell: Radius b (+Q) Gaussian Surface r Radius a (−Q) O

Fig 2.14: Concentric spherical configuration with intermediate Gaussian surface $r$.

1. Field in the Gap ($a < r < b$): By Gauss's Law: $E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \implies E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$.

2. Potential Difference Line Integral:

$$V = \int_{a}^{b} E dr = \frac{Q}{4\pi\varepsilon_0} \int_{a}^{b} \frac{1}{r^2} dr = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{Q}{4\pi\varepsilon_0}\left(\frac{b - a}{ab}\right)$$

3. Resultant Capacitance ($C = Q / V$):

$$C = \frac{4\pi\varepsilon_0 ab}{b - a}$$

3. Solved Board Numericals: Voltage & Charge Variations

Exemplar 1: Voltage Drop & Escalation

A capacitor holds $360\,\mu\text{C}$ at potential $V$. Lowering potential by $120\text{ V}$ reduces charge to $120\,\mu\text{C}$. Calculate $V$, $C$, and charge if voltage rises by $120\text{ V}$.

• Ratio: $\frac{V}{V - 120} = \frac{360}{120} = 3 \implies V = 180\text{ V}$

• Capacity: $C = \frac{360\,\mu\text{C}}{180\text{ V}} = \mathbf{2\,\mu\text{F}}$

• Charge at $(V + 120\text{ V})$: $Q = 2\,\mu\text{F} \times (180 + 120) = \mathbf{600\,\mu\text{C}}$

Exemplar 2: Capacity Under Potential Variation

A capacitor holds $120\,\mu\text{C}$ at potential $V$. Lowering potential by $40\text{ V}$ drops charge to $40\,\mu\text{C}$. Find $V$, $C$, and charge if voltage rises by $40\text{ V}$.

• Ratio: $\frac{120}{V} = \frac{40}{V - 40} \implies 3(V - 40) = V \implies V = 60\text{ V}$

• Capacity: $C = \frac{120\,\mu\text{C}}{60\text{ V}} = \mathbf{2\,\mu\text{F}}$

• Charge at $(V + 40\text{ V})$: $Q = 2\,\mu\text{F} \times (60 + 40) = \mathbf{200\,\mu\text{C}}$

4. Capacitor Combinations: Series & Parallel Topologies

Series Combination ($Q = \text{Constant}$)

Charge is uniform across each plate; terminal voltages divide: $V = V_1 + V_2 + V_3$.

C₁ C₂ C₃ +Q −Q
$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$$
Parallel Combination ($V = \text{Constant}$)

Voltage is identical across all branches; charges accumulate: $Q = Q_1 + Q_2 + Q_3$.

+V 0V C₁ C₂ C₃
$$C_p = C_1 + C_2 + \dots + C_n$$

5. Stored Electrostatic Energy & Field Energy Density ($u$)

Plate 1 (+Q') Plate 2 (−Q') Stored Field E⃗ dQ' Work done: $dW = V' dQ' = \frac{Q'}{C} dQ'$

Fig 2.15: Incremental charge transfer $dQ'$ against potential $V'$, establishing stored potential energy.

1. Energy Integration: Total work done during charging: $W = \int_{0}^{Q} \frac{Q'}{C} dQ' = \frac{1}{2}\frac{Q^2}{C}$.

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

2. Volumetric Energy Density ($u$): Dividing total energy by enclosed plate volume ($A \cdot d$):

$$u = \frac{U}{A \cdot d} = \frac{\frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)(E d)^2}{A \cdot d} \implies \mathbf{u = \frac{1}{2}\varepsilon_0 E^2}$$

6. Charge Redistribution & Energy Dissipation ($\Delta U$)

A (C₁, V₁) B (C₂, V₂) Thin Conducting Wire

Fig 2.16: Charges flow until potential reaches common value $V$.

1. Common Potential: $V = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.

2. Charge Sharing Ratio: $Q_1' = C_1 V$ and $Q_2' = C_2 V \implies \mathbf{\frac{Q_1'}{Q_2'} = \frac{C_1}{C_2}}$ (Charges divide in direct ratio of capacities).

3. Energy Loss Equation ($\Delta U = U_i - U_f$):

$$\Delta U = \frac{1}{2}\frac{C_1 C_2 (V_1 - V_2)^2}{C_1 + C_2}$$

Because $(V_1 - V_2)^2 > 0$, energy loss is strictly positive ($\Delta U > 0$). This energy dissipates as heat in the wire and electromagnetic radiation.

7. Partial Dielectric & Conducting Slab Insertion

Upper Plate (+Q) Lower Plate (−Q) Slab (Thickness t, Constant K) t d

Fig 2.17: Field reduction across thickness $t$ inside a dielectric slab.

1. Reduced field inside dielectric slab: $E = \frac{E_0}{K}$.

2. Potential drop summation: $V = E_0(d - t) + \left(\frac{E_0}{K}\right)t = E_0\left(d - t + \frac{t}{K}\right)$.

3. Modified Capacitance Expression:

$$C = \frac{\varepsilon_0 A}{d - t\left(1 - \frac{1}{K}\right)}$$
Completely Filled ($t = d$) $C = K \frac{\varepsilon_0 A}{d} \implies \mathbf{C = K \cdot C_0}$
Conducting Slab ($K = \infty$) $C = \frac{\varepsilon_0 A}{d - t} \implies \mathbf{C = \left(\frac{d}{d-t}\right) C_0}$

8. Multi-Dielectric Slabs: Series vs Parallel Configurations

K₁, t₁ K₂, t₂

Series layers: division across thickness.

K₁, A₁ K₂, A₂

Parallel sections: division across area.

Series Stacking ($d = \sum t_i$):
$$C = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2} + \dots + \frac{t_n}{K_n}}$$
Parallel Partitioning ($A = \sum A_i$):
$$C = \frac{\varepsilon_0}{d}\left(K_1 A_1 + K_2 A_2 + \dots + K_n A_n\right)$$

9. Suspended Concentric Spherical Shells: Unidirectional Charge Transfer

Outer Shell: Radius R (+Q) r, +q Connecting Wire

Fig 2.18: Inner sphere $r$ connected to outer shell $R$.

1. Inner sphere potential: $V_r = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{r} + \frac{Q}{R}\right]$.

2. Outer shell potential: $V_R = \frac{1}{4\pi\varepsilon_0}\left[\frac{q + Q}{R}\right]$.

3. Potential difference:

$$V_r - V_R = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r} - \frac{1}{R}\right)$$

Because $r < R$, term $\left(\frac{1}{r} - \frac{1}{R}\right) > 0$. If $q > 0$, then $V_r > V_R$ always. When joined by a wire, charge $q$ flows entirely to the outer shell, regardless of the magnitude of $Q$.

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