Electrostatic Potential and Capacitance
'Potential is the energy per unit charge — it tells you how much work it takes to bring a charge to a point in an electric field.'
1. Chapter Overview
This chapter introduces the concept of ELECTROSTATIC POTENTIAL (V) — the work done per unit charge in bringing a test charge from infinity to a point. The relationship between electric field and potential (E = −dV/dr) is explored. Equipotential surfaces (surfaces of constant potential) are discussed. The chapter then covers CAPACITANCE — the ability of a conductor to store charge — and introduces CAPACITORS, their combinations, and the effect of DIELECTRICS on capacitance.
2. Electrostatic Potential
Potential Due to a Point Charge
- V = kQ/r. 'Potential DECREASES as distance INCREASES (for positive Q).'
- Potential difference: V_B − V_A = W_AB/q₀.
Potential Due to a System of Charges
- V = Σ kqᵢ/rᵢ — scalar sum (EASIER than electric field, which requires vector sums).
Relationship Between E and V
- E = −dV/dr (in one dimension). 'The electric field is the NEGATIVE GRADIENT of the potential.'
- E points in the direction where V DECREASES most rapidly.
3. Equipotential Surfaces
- Any surface with the SAME potential at every point.
- Properties: (1) No work is done moving a charge along an equipotential surface. (2) E is ALWAYS perpendicular to equipotential surfaces. (3) For a point charge, equipotential surfaces are CONCENTRIC SPHERES.
4. Potential Energy of a System of Charges
- U = Σ (kqᵢqⱼ)/rᵢⱼ (sum over ALL pairs). 'The work required to assemble the charges from infinity.'
5. Capacitance
Definition
- C = Q/V. 'Capacity to store charge per unit potential.' Unit: Farad (F).
Capacitance of a Parallel Plate Capacitor
- C = ε₀A/d. Depends ONLY on geometry (area, separation) — NOT on charge.
Capacitance of Other Shapes
- Spherical capacitor: C = 4πε₀R (isolated sphere). C = 4πε₀(ab)/(b−a) (concentric spheres).
- Cylindrical capacitor: C = 2πε₀L/ln(b/a).
6. Combinations of Capacitors
| Combination | Equivalent Capacitance | Charge/Potential |
|---|---|---|
| SERIES | 1/C_eq = 1/C₁ + 1/C₂ + ... | Same charge on each, potential divides |
| PARALLEL | C_eq = C₁ + C₂ + ... | Same potential across each, charge divides |
Worked Example 1
Problem: Three capacitors of 2 μF, 3 μF, and 6 μF are connected in series. Find the equivalent capacitance. Solution: 1/C_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. C_eq = 1 μF. 'In series, the equivalent capacitance is LESS than the smallest individual capacitance.'
7. Energy Stored in a Capacitor
- U = ½ QV = ½ CV² = ½ Q²/C.
- 'A capacitor stores energy in the ELECTRIC FIELD between its plates.'
8. Dielectrics and Polarisation
Effect of Dielectric
- When a dielectric (insulator) is inserted between the plates: C increases by a factor K (dielectric constant).
- C' = KC₀ = Kε₀A/d.
- Polarisation: The alignment of dipole moments in the dielectric under an applied electric field.
Comparison Table: Conductor vs Dielectric
| Property | Conductor | Dielectric |
|---|---|---|
| Free charges | YES — charges move freely | NO — charges are bound |
| Inside E field | E = 0 (electrostatic equilibrium) | E reduces by factor K |
| Dielectric constant K | Effectively INFINITE | K ≥ 1 |
| Polarisation | Charges rearrange on surface | Molecules align/induce dipoles |
9. Common Mistakes
- Potential is scalar, field is vector: Adding potentials is arithmetic sum. Adding fields requires vector addition.
- E = −dV/dr: The negative sign means E points from HIGH to LOW potential — many students forget the sign.
- Series capacitor formula: 1/C_eq = sum of reciprocals, NOT C_eq = sum of capacitors (that is parallel).
- Dielectric inserted with battery connected vs disconnected: With battery connected, V constant, Q increases. With battery disconnected, Q constant, V decreases.
10. CBSE Exam Focus
- Potential due to a point charge and system of charges
- Equipotential surfaces — properties and examples
- Capacitance of parallel plate capacitor — formula and derivation
- Series and parallel combinations of capacitors — numerical problems
- Energy stored in a capacitor
- Effect of dielectric on capacitance — with battery ON vs OFF
11. Self-Test
Q1: Two charges 2 μC and −2 μC are 0.1 m apart. Find the potential at the midpoint. A1: V = k(2×10⁻⁶)/0.05 + k(−2×10⁻⁶)/0.05 = 0. 'Potential at the midpoint of a dipole is ZERO.'
Q2: A parallel plate capacitor has plates of area 0.1 m² separated by 1 mm. Find its capacitance. A2: C = ε₀A/d = (8.85×10⁻¹²)(0.1)/(10⁻³) = 8.85×10⁻¹⁰ = 885 pF.
Q3: Find the equivalent capacitance of 4 μF, 6 μF, and 12 μF in parallel. If connected to a 10 V battery, find total charge stored. A3: C_eq = 4+6+12 = 22 μF. Q = CV = 22×10⁻⁶×10 = 220 μC.
Q4: A 10 μF capacitor is charged to 100 V. Find the energy stored. If disconnected and then connected to an uncharged 10 μF capacitor, find the energy loss. A4: Initial energy = ½CV² = 0.5×10⁻⁵×10⁴ = 0.05 J. After connection: total Q = CV = 10⁻³ C. Combined C = 20 μF. V' = Q/C = 10⁻³/(20×10⁻⁶) = 50 V. Final energy = ½(20×10⁻⁶)(50)² = 0.025 J. Loss = 0.025 J. 'Energy is DISSIPATED when capacitors share charge.'
Q5: A dielectric slab of K=3 is inserted in a 5 μF capacitor charged to 20 V (battery disconnected). Find new capacitance, potential, and energy. A5: C' = KC = 3×5 = 15 μF. Q unchanged = 5×20 = 100 μC. V' = Q/C' = 100/15 = 6.67 V. U' = ½Q²/C' = ½(10⁻⁴)²/(15×10⁻⁶) = 3.33×10⁻⁴ J. Original U = 5×10⁻⁴ J. Energy DECREASED by 1.67×10⁻⁴ J.
12. Conclusion
Potential and capacitance are PRACTICAL electrostatics:
- POTENTIAL: 'The voltage — work per unit charge — a scalar quantity easier to work with than electric field.'
- CAPACITANCE: 'How much charge a system can store per volt — determined by geometry.'
- CAPACITORS: 'Energy storage devices — essential in electronics for filtering, timing, and power supply smoothing.'
- DIELECTRICS: 'Insulators that INCREASE capacitance by reducing the internal field.'
'Capacitance is to electrostatics what a battery is to circuits — the ability to store energy for later use.'
