Class 12 Physics Notes: Electrostatic Potential and Capacitance (Chapter 2)
Master Chapter 2 of Class 12 Physics. Understand electric potential vs potential energy, equipotential surfaces, capacitor combinations, dielectrics, and Van de Graaff generator with key formulas and PYQs.
Students consistently conflate these two quantities. Here is the exact distinction:
- Electric Potential (V): A property of a point in space. Work done per unit positive charge to bring a test charge from infinity to that point. Unit: Volt (V = J/C).
- Electric Potential Energy (U): Energy stored when a specific charge q is placed at that point. U = qV.
Potential is the characteristic of the location. Potential energy belongs to the charge placed there.
⚠️ Watch Out! — Zero Field ≠ Zero Potential
Inside a conductor, E = 0, but potential is constant and not necessarily zero. The entire conductor is at the same potential — that potential can be non-zero depending on charge distribution.
Students who write "potential inside a conductor is zero" lose marks on this guaranteed exam topic.
2. Equipotential Surfaces
An equipotential surface is one on which every point is at the same potential.
- Electric field lines are always perpendicular to equipotential surfaces.
- No work is done moving a charge along an equipotential surface (ΔV = 0 → W = qΔV = 0).
- For a point charge: equipotential surfaces are concentric spheres.
- For a uniform field: equipotential surfaces are parallel planes perpendicular to E.
3. Capacitors: Combinations
Formula: Capacitor Combinations
Parallel: Ctotal = C₁ + C₂ + C₃ (capacitances add directly — same as resistors in series)
Series: 1/Ctotal = 1/C₁ + 1/C₂ + 1/C₃ (reciprocals add — same as resistors in parallel)
Memory trick: Capacitors behave opposite to resistors. In parallel, capacitors add. In series, resistors add. Never mix these up.
4. Energy Stored in a Capacitor — Three Equivalent Forms
| Form | Formula | Use when given... |
|---|---|---|
| Form 1 | U = ½CV² | C and V are known |
| Form 2 | U = Q²/2C | Q and C are known |
| Form 3 | U = ½QV | Q and V are known |
5. Effect of a Dielectric
Inserting a dielectric between capacitor plates increases capacitance by the dielectric constant K:
C = KC₀ = Kε₀A/d
Why? Dielectric molecules polarise and oppose the applied field. This reduces E between plates, which reduces V (since E = V/d). Since C = Q/V, a smaller V with the same Q gives a larger C.
Summary: Formula Sheet
| Concept | Formula |
|---|---|
| Electric Potential (point charge) | V = kQ/r |
| Potential Energy | U = qV = kq₁q₂/r |
| Capacitance | C = Q/V |
| Parallel plate capacitor | C = ε₀A/d |
| With dielectric | C = Kε₀A/d |
| Energy stored | U = ½CV² = Q²/2C = ½QV |
| Capacitors in parallel | C = C₁ + C₂ |
| Capacitors in series | 1/C = 1/C₁ + 1/C₂ |
Practice Questions (PYQs)
- Define electric potential at a point. How is it different from electric potential energy?
- Two capacitors of capacitance 4 μF and 6 μF are connected (i) in series and (ii) in parallel. Find the equivalent capacitance in each case.
- A parallel plate capacitor has plate area 100 cm² and plate separation 2 mm. Calculate its capacitance with and without a dielectric of K = 5. (ε₀ = 8.85 × 10⁻¹² C²/N·m²)
- Why are equipotential surfaces always perpendicular to electric field lines? What would happen if they were not?
- Explain the working principle of a Van de Graaff generator. Why does charge accumulate on the outer shell without flowing back?