Electrostatic Potential and Capacitance

ELECTROSTATIC POTENTIAL AND CAPACITANCE

2.1 INTRODUCTION

**Coulomb force** between stationary charges is a **conservative force**—work done depends only on initial and final positions, not on path taken. Like gravitational potential energy, we can define **electrostatic potential energy** of a charge in an electric field.

**Key Concept**: When an external force brings a test charge **q** from point R to point P against the electric force **F_E**, the external force **F_ext = –F_E** (for quasi-static motion with no acceleration). Work done by the external force equals the change in potential energy.

**Work-Potential Energy Relationship**:

Work done by external force from R to P: **W_RP = ΔU = U_P – U_R** (Equation 2.1)

This work is stored as potential energy

**Two Critical Points About Potential Energy**:

1. **Path Independence**: Work done by electrostatic field depends only on initial and final positions. This is the fundamental characteristic of a conservative force.

2. **Arbitrary Reference Point**: Potential energy is undetermined to within an additive constant. Only potential energy **differences** are physically meaningful. We choose the convenient reference: **potential energy is zero at infinity (U_∞ = 0)**.

With this choice, if R is at infinity:

**U_P = W_∞→P** (Equation 2.3)

**Physical Meaning**: Electrostatic potential energy of charge **q** at a point is the work done by an external force in bringing the charge from infinity to that point (quasi-statically).

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2.2 ELECTROSTATIC POTENTIAL

**Definition**: Since work done on a charge q is proportional to q itself (force is qE), we define **electrostatic potential V** as **work done per unit test charge**.

**Electrostatic Potential at a Point**: The work done by an external force in bringing a unit positive charge (without acceleration) from infinity to that point.

**Mathematical Relation**:

From Equation (2.1), dividing both sides by q:

**V_P – V_R = W_RP/q = (U_P – U_R)/q** (Equation 2.4)

where V_P and V_R are potentials at points P and R respectively.

**SI Unit**: **Volt (V)** or **J/C** (joule per coulomb)

**Relation Between E and V**:

The electric field is related to potential by:

**E = –dV/dr** (in radial direction)

Potential difference and work done: **W = q(V_P – V_R)** or **W = qΔV**

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2.3 POTENTIAL DUE TO A POINT CHARGE

**Derivation**: Consider a point charge **Q** at origin. We calculate work done in bringing a unit positive test charge from infinity to point P at distance r.

**Path chosen**: Radial direction (most convenient)

At intermediate point P' at distance r', electrostatic force on unit charge:

**F = (Q/4πε₀r'²) r̂'** (Equation 2.5)

Work done against this force from r' to r' + Δr':

**ΔW = –(Q/4πε₀r'²)Δr'** (Equation 2.6)

Integrating from r' = ∞ to r' = r:

W = ∫[∞→r] (–Q/4πε₀r'²)dr' = Q/4πε₀r (Equation 2.7)

**Potential Due to Point Charge**:

**V(r) = (1/4πε₀) × (Q/r)** (Equation 2.8)

**Key Properties**:

For **Q > 0**: V > 0 at all points (repulsive field)

For **Q < 0**: V < 0 at all points (attractive field)

V is **independent of test charge**—characteristic of the charge configuration

V → 0 as r → ∞ (consistent with zero potential at infinity)

Equipotential surfaces are **concentric spheres** centred at Q

**Worked Example 2.1**:

(a) Calculate potential at point P due to charge Q = 4 × 10⁻⁷ C at distance r = 9 cm = 0.09 m

V = (1/4πε₀) × (Q/r) = (9 × 10⁹ N·m²/C²) × (4 × 10⁻⁷ C)/(0.09 m)

= 9 × 10⁹ × 4 × 10⁻⁷/0.09 = 4 × 10⁴ V

(b) Work done in bringing charge q = 2 × 10⁻⁹ C from infinity to P:

W = qV = 2 × 10⁻⁹ × 4 × 10⁴ = 8 × 10⁻⁵ J

Work is **path-independent** because displacement perpendicular to radial direction contributes zero work.

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2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE

**Electric Dipole**: Two charges +q and –q separated by small distance 2a. Dipole moment **p = q × 2a**, pointing from –q to +q.

**Derivation**: Using superposition principle (potential obeys it because it's derived from conservative force):

V = (1/4πε₀)[q/r₁ – q/r₂] (Equation 2.9)

where r₁ and r₂ are distances of point P from +q and –q respectively.

From geometry (r >> a, first-order approximation in a/r):

r₁² = r² + a² – 2ar cosθ ≈ r²(1 – 2(a/r)cosθ)

r₂² = r² + a² + 2ar cosθ ≈ r²(1 + 2(a/r)cosθ)

Using binomial expansion (1 + x)^(-1/2) ≈ 1 – x/2 for small x:

1/r₁ ≈ (1/r)[1 + (a/r)cosθ]

1/r₂ ≈ (1/r)[1 – (a/r)cosθ]

**Dipole Potential Formula**:

**V(r,θ) = (1/4πε₀) × (p cosθ/r²)** (Equation 2.15)

Or in vector form: **V = (1/4πε₀) × (p·r̂/r²)**

Valid for r >> a (dipole approximation)

**Important Contrasts with Point Charge Potential**:

1. **Angle Dependence**: Dipole potential depends on both r and angle θ between position vector and dipole axis (point charge depends only on r)

2. **Distance Dependence**: Dipole potential ∝ 1/r² (point charge ∝ 1/r)

3. **Equipotential Locations**:

On dipole axis (θ = 0): **V = ±(1/4πε₀)(p/r²)**

In equatorial plane (θ = π/2): **V = 0**

4. **Axial Symmetry**: Potential has cylindrical symmetry about dipole axis

**Physical Insight**: Since dipole has zero net charge, potential falls off faster at large distances.

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2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES

**Superposition Principle for Potential**: Total potential at any point is the algebraic sum of potentials due to individual charges.

For discrete charges q₁, q₂, ..., qₙ at distances r₁ₚ, r₂ₚ, ..., rₙₚ from point P:

**V = Σᵢ(qᵢ/4πε₀rᵢₚ)** (Equation 2.18)

**For Continuous Charge Distribution**: With charge density ρ(r), we integrate:

**V = (1/4πε₀) ∫ ρ(r')/|r – r'| dv'**

The integral is over entire charge distribution.

**Worked Example 2.2**:

Two charges: q₁ = 3 × 10⁻⁸ C and q₂ = –2 × 10⁻⁸ C separated by 15 cm. Find point where V = 0.

Let x be distance from q₁ (positive charge at origin, negative charge at 15 cm).

For V = 0: (1/4πε₀)[q₁/x + q₂/(15–x)] = 0

3 × 10⁻⁸/x = 2 × 10⁻⁸/(15–x)

3(15–x) = 2x → 45 = 5x → **x = 9 cm**

Check if x lies between charges: 3/(15–9) = 3/6 = 1/2 ✓

Also: **x = 45 cm** (on extended line beyond negative charge)

At both points, potentials due to two charges cancel.

**Special Case — Uniformly Charged Spherical Shell**:

**Outside (r ≥ R)**: **V = (1/4πε₀)(q/r)** — same as point charge

**Inside (r < R)**: **V = (1/4πε₀)(q/R)** — constant (no work to move charge inside)

This follows from E = 0 inside the shell → dV/dr = 0 → V = constant.

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2.6 EQUIPOTENTIAL SURFACES

**Definition**: An **equipotential surface** is a surface at which the electric potential has the same value at all points.

**Key Property**: All equipotential surfaces are **perpendicular to field lines**.

**Proof**: On an equipotential surface, dV = 0 for any displacement. Since E = –dV/dr, only the component of E perpendicular to the surface contributes. This means E must be perpendicular to surface, and hence field lines are perpendicular to equipotentials.

**Equipotential Surfaces for Common Charge Distributions**:

1. **Single Point Charge**: V = (Q/4πε₀r) = constant when r = constant

Equipotentials: **Concentric spheres** centred at charge

2. **Uniform Electric Field**: V = –E·r = constant when E·r = constant

Equipotentials: **Planes perpendicular to field direction**

3. **Uniform Charged Infinite Plane**: V ∝ constant

Equipotentials: **Planes parallel to charged plane**

4. **Electric Dipole**: V = (p cosθ/4πε₀r²) = constant

Equipotentials: Complex surfaces (axially symmetric)

**Worked Example 2.3**:

Analyzing field lines of positive and negative point charges:

(a) **V_P – V_Q** for positive charge: Since closer to charge has higher potential (V ∝ 1/r), V_P > V_Q → **difference is positive**

(b) **V_B – V_A** for negative charge: V ∝ Q/r, with Q < 0, points further away have less negative potential. V_B > V_A → **difference is positive**

(c) **Work by field** moving positive charge from Q to P: Field does negative work (opposes radially outward motion) → **negative**

(d) **Work by external agency** moving negative charge from B to A: Against attractive force → **positive work**

(e) **Kinetic energy** of negative charge from B to A: Repulsive force from negative charge → velocity decreases → **KE decreases**

**Relation Between Field and Potential**:

In uniform field: E = |ΔV|/Δd (where Δd is perpendicular distance between equipotentials)

In general: **E = –∇V** or **E_r = –dV/dr**

No work done moving charge along equipotential: **W = q(V_f – V_i) = 0**

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2.7 ELECTROSTATIC POTENTIAL ENERGY OF A SYSTEM OF CHARGES

**Potential Energy of Two Charges**:

Work to bring charge q₂ from infinity in field of q₁ at distance r₁₂:

**U₁₂ = (1/4πε₀) × (q₁q₂/r₁₂)** (Equation not numbered in original)

**Important**: This is positive for like charges (repulsive) and negative for unlike charges (attractive).

**For System of Multiple Charges**:

Total potential energy is sum of all pair interactions:

**U = (1/4πε₀) Σᵢ<ⱼ (qᵢqⱼ/rᵢⱼ)**

Alternatively, using potential at each charge location:

**U = (1/2) Σᵢ qᵢVᵢ**

where V_i is potential at charge q_i due to all other charges. Factor 1/2 avoids double counting.

**Physical Meaning**:

**U > 0**: Configuration is unstable; energy required to assemble (repulsive dominates)

**U < 0**: Configuration is stable; energy released in assembly (attractive dominates)

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2.8 ELECTROSTATIC POTENTIAL ENERGY IN EXTERNAL FIELD

**Single Charge in External Field**:

If external potential is V_ext at location of charge q:

**U = qV_ext**

**Work done by external agent** to move charge:

**W = q(V_B – V_A) = ΔU**

**Dipole in External Field**:

Potential energy of dipole **p** in uniform field **E**:

**U = –p·E = –pE cosθ**

where θ is angle between dipole moment and field.

**Interpretation**:

Minimum energy: θ = 0 (dipole aligned with field) → **U_min = –pE**

Maximum energy: θ = π (anti-aligned) → **U_max = +pE**

Zero energy: θ = π/2 (perpendicular)

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2.9 CAPACITANCE

**Definition**: **Capacitance (C)** is the ability of a conductor to store electric charge. It is defined as the ratio of charge Q on conductor to the potential V of conductor:

**C = Q/V** (Equation 2.20)

**SI Unit**: **Farad (F) = Coulomb/Volt (C/V)**

**Physical Meaning**: Capacitance measures how much charge can be stored per unit potential difference. Higher capacitance → more charge stored at same voltage.

**Capacitance of Isolated Conductor**:

For single conductor, if Q is charge and V its potential (relative to infinity):

From V = (Q/4πε₀R) for spherical conductor of radius R:

**C = Q/V = 4πε₀R** (for isolated sphere)

For conductor of arbitrary shape: C depends on geometry and surrounding medium.

**Worked Example**:

Capacitance of isolated conducting sphere of radius R = 10 cm:

C = 4πε₀R = 4π × 8.85 × 10⁻¹² × 0.1 = 1.11 × 10⁻¹¹ F ≈ 11.1 pF (picofarads)

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2.10 PARALLEL PLATE CAPACITOR

**Configuration**: Two parallel conducting plates of area A separated by distance d. Plate 1 has charge +Q, plate 2 has charge –Q.

**Field Between Plates**:

Each infinite plate produces field E = σ/2ε₀ (where σ = Q/A is surface charge density).

Plate 1 (positive): Creates field pointing outward

Plate 2 (negative): Creates field pointing inward

Between plates, fields add: **E_total = Q/ε₀A**

Outside plates: Fields cancel = 0

**Potential Difference**:

Integrating field across gap of distance d:

**V = Ed = Qd/ε₀A** (Equation 2.21)

**Capacitance of Parallel Plate Capacitor**:

**C = Q/V = ε₀A/d** (Equation 2.22)

**Key Features**:

**C ∝ Area A**: Larger plates store more charge

**C ∝ 1/distance d**: Plates closer together → higher capacitance

**C is independent of Q and V**: Depends only on geometry and material

**C depends on dielectric between plates** (to be covered next)

**Worked Example**:

Parallel plate capacitor: A = 100 cm² = 0.01 m², d = 1 mm = 10⁻³ m

C = ε₀A/d = 8.85 × 10⁻¹² × 0.01/10⁻³ = 8.85 × 10⁻¹¹ F ≈ 88.5 pF

If V = 100 V applied:

Q = CV = 8.85 × 10⁻¹¹ × 100 = 8.85 × 10⁻⁹ C ≈ 8.85 nC

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2.11 EFFECT OF DIELECTRIC ON CAPACITANCE

**Dielectric Material**: An insulator material that becomes polarized in an electric field.

**Mechanism of Polarization**:

Non-polar molecules: E field induces dipoles (alignment of electron cloud and nucleus)

Polar molecules: Pre-existing dipoles align with field

**Dielectric Constant (Relative Permittivity) K or ε_r**:

When dielectric is introduced between capacitor plates:

**V' = V/K** (Potential reduces by factor K) (Equation 2.23)

**E' = E/K** (Field reduces by factor K) (Equation 2.24)

**Capacitance with Dielectric**:

Since C = Q/V and Q remains constant:

**C' = KC = Kε₀A/d** (Equation 2.25)

where K is dielectric constant (relative permittivity).

**Permittivity of Medium**:

**ε = Kε₀** (absolute permittivity)

**Typical Dielectric Constants**:

Vacuum: K = 1

Air: K ≈ 1.0006 ≈ 1

Paper: K ≈ 3.7

Glass: K ≈ 6–8

Mica: K ≈ 5–7

Water: K ≈ 80

**Physical Explanation**:

When dielectric inserted:

1. External field polarizes molecules

2. Polarized molecules create **induced dipole field** E_induced (opposite to E_external)

3. Net field E_net = E_external – E_induced = E_external/K

4. Reduced field → reduced potential difference → increased capacitance (for fixed Q)

**Worked Example**:

Parallel plate capacitor with air: C₀ = 100 pF, V₀ = 1000 V

If dielectric (K = 4) inserted:

C' = KC₀ = 4 × 100 = 400 pF

For same charge Q:

V' = V₀/K = 1000/4 = 250 V

For same voltage:

Q' = KQ = 4Q (capacitor can store 4 times more charge)

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2.12 ENERGY STORED IN A CAPACITOR

**Derivation**:

When capacitor is charged by bringing charge dq against potential V:

dW = V dq = (q/C) dq

Total energy stored:

U = ∫₀^Q (q/C) dq = (1/C) × (Q²/2) = **Q²/2C**

Since Q = CV:

**U = (1/2)CV²** (Equation 2.26)

Also:

**U = (1/2)QV**

**Key Points**:

Energy stored depends on both charge and voltage

Energy is stored in electric field between plates

For fixed C: U ∝ V² (doubling voltage stores 4× energy)

For fixed V: U ∝ C (larger capacitor stores more energy)

**Energy Density (Energy per unit volume)**:

Between plates: E = σ/ε₀ = Q/ε₀A

Volume = Ad

Energy density:

**u = U/(Ad) = (1/2)ε₀E²** (Equation 2.27)

**Worked Example**:

Capacitor: C = 10 μF, V = 100 V

Energy stored:

U = (1/2)CV² = 0.5 × 10 × 10⁻⁶ × (100)² = 0.5 × 10⁻⁶ × 10⁴ = 5 × 10⁻³ J = 5 mJ

Alternatively:

Q = CV = 10 × 10⁻⁶ × 100 = 10⁻³ C

U = Q²/2C = (10⁻³)²/(2 × 10 × 10⁻⁶) = 10⁻⁶/(2 × 10⁻⁵) = 0.05 J... [Check: 5 × 10⁻³ J = 5 mJ]

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2.13 COMBINATION OF CAPACITORS

Series Connection

**Configuration**: Capacitors C₁, C₂, ..., Cₙ connected end-to-end. Same charge Q flows through all.

**Voltage Distribution**:

V₁ = Q/C₁, V₂ = Q/C₂, ..., Vₙ = Q/Cₙ

Total voltage:

V = V₁ + V₂ + ... + Vₙ = Q(1/C₁ + 1/C₂ + ... + 1/Cₙ)

**Equivalent Capacitance**:

**1/C_eq = 1/C₁ + 1/C₂ + ... + 1/Cₙ** (Equation 2.28)

**Properties**:

C_eq < C_min (equivalent is less than smallest individual)

Voltage divides inversely as capacitance: smaller C gets larger V

**Worked Example**:

Three capacitors: C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF in series; V = 180 V

1/C_eq = 1/2 + 1/4 + 1/6 = 6/12 + 3/12 + 2/12 = 11/12

C_eq = 12/11 ≈ 1.09 μF

Charge: Q = C_eq × V = 1.09 × 10⁻⁶ × 180 ≈ 196 × 10⁻⁶ C ≈ 196 μC

Voltages:

V₁ = Q/C₁ = 196/2 = 98 V

V₂ = Q/C₂ = 196/4 = 49 V

V₃ = Q/C₃ = 196/6 ≈ 32.7 V

Check: 98 + 49 + 32.7 = 179.7 ≈ 180 V ✓

Parallel Connection

**Configuration**: All capacitors connected between same two points. Same voltage V across all.

**Charge Distribution**:

Q₁ = C₁V, Q₂ = C₂V, ..., Qₙ = CₙV

Total charge:

Q = Q₁ + Q₂ + ... + Qₙ = V(C₁ + C₂ + ... + Cₙ)

**Equivalent Capacitance**:

**C_eq = C₁ + C₂ + ... + Cₙ** (Equation 2.29)

**Properties**:

C_eq > C_max (equivalent is greater than largest individual)

Charge divides as capacitance: larger C draws more charge

Useful for increasing capacitance

**Worked Example**:

Three capacitors: C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF in parallel; V = 100 V

C_eq = C₁ + C₂ + C₃ = 2 + 4 + 6 = 12 μF

Total charge:

Q = C_eq × V = 12 × 10⁻⁶ × 100 = 12 × 10⁻⁴ C = 1.2 mC

Individual charges:

Q₁ = 2 × 10⁻⁶ × 100 = 200 μC

Q₂ = 4 × 10⁻⁶ × 100 = 400 μC

Q₃ = 6 × 10⁻⁶ × 100 = 600 μC

Total: 200 + 400 + 600 = 1200 μC = 1.2 mC ✓

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EXAM-IMPORTANT FORMULAS SUMMARY

**Electric Potential**:

Point charge: V = (Q/4πε₀r)

Dipole: V = (p cosθ/4πε₀r²)

System of charges: V = Σ(qᵢ/4πε₀rᵢ)

**Potential Energy**:

Two charges: U = q₁q₂/4πε₀r₁₂

Dipole in field: U = –p·E

**Capacitance**:

Definition: C = Q/V

Parallel plate (no dielectric): C = ε₀A/d

Parallel plate (with dielectric): C = Kε₀A/d

Series: 1/C_eq = Σ(1/Cᵢ)

Parallel: C_eq = ΣCᵢ

**Energy Storage**:

U = (1/2)CV² = (1/2)QV = Q²/2C

Energy density: u = (1/2)ε₀E²

**Dielectric Effects**:

Field reduction: E' = E/K

Voltage reduction: V' = V/K

Capacitance increase: C' = KC