Electrostatic Potential and Capacitance (Class 12 Physics) describes the energy and potential set up by static charges and how capacitors store that energy. Electric potential V = W/q is the work done per unit charge, and for a point charge V = kq/r. The potential difference between two points drives no current in electrostatics but stores energy; a capacitor of capacitance C = Q/V holds charge for a given voltage. For a parallel-plate capacitor C = epsilon0 A/d, energy stored U = (1/2) C V^2, and capacitors combine in series and parallel, the basis of energy storage in circuits.
Key Concepts
1. Electrostatic Potential
The electric potential at a point is the work done per unit positive charge in bringing a test charge from infinity to that point (against the electric field).
V = W/q₀ = kQ/r (due to a point charge Q)
Unit: Volt (V) = Joule/Coulomb
Potential is a scalar quantity - it has magnitude but no direction. This makes it easier to work with than electric field.
Potential Due to Various Configurations
| Configuration | Potential |
|---|---|
| Point charge Q at distance r | V = kQ/r |
| System of charges | V = k(q₁/r₁ + q₂/r₂ + … ) - algebraic sum |
| Dipole (axial point) | V = kp cos θ/r² |
| Dipole (equatorial point) | V = 0 |
| Uniformly charged sphere (outside) | V = kQ/r |
| Uniformly charged sphere (surface) | V = kQ/R |
Potential Difference
V_A − V_B = work done per unit charge in moving a test charge from B to A.
Relation with electric field: E = −dV/dr (field points in the direction of decreasing potential)
2. Equipotential Surfaces
An equipotential surface is a surface where every point has the same electric potential. No work is done in moving a charge along an equipotential surface.
- Electric field lines are always perpendicular to equipotential surfaces
- For a point charge: equipotential surfaces are concentric spheres
- For a uniform field: equipotential surfaces are parallel planes perpendicular to the field
- Two equipotential surfaces never intersect
- Closer equipotential surfaces = stronger electric field
3. Electrostatic Potential Energy
Potential energy of a system of charges is the work done in assembling the charges from infinity.
For two charges: U = kq₁q₂/r
For three charges: U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)
4. Conductors and Capacitors
Properties of Conductors in Electrostatics
- Electric field inside a conductor is zero
- Any excess charge resides on the surface
- Electric field at the surface is perpendicular to the surface
- The entire conductor is at the same potential (equipotential body)
Capacitor
A capacitor is a device that stores electric charge and energy. It consists of two conductors separated by an insulator (dielectric).
Capacitance: C = Q/V
Unit: Farad (F). 1 F = 1 C/V (very large - usually μF, nF, or pF are used)
Parallel Plate Capacitor
C = ε₀A/d
- A = area of each plate
- d = separation between plates
- With dielectric (constant κ): C = κε₀A/d - capacitance increases κ times
Combination of Capacitors
| Feature | Series | Parallel |
|---|---|---|
| Charge | Same on each (Q = Q₁ = Q₂) | Divides (Q = Q₁ + Q₂) |
| Voltage | Divides (V = V₁ + V₂) | Same across each (V = V₁ = V₂) |
| Equivalent | 1/C = 1/C₁ + 1/C₂ + … | C = C₁ + C₂ + … |
| Effect | Total C decreases | Total C increases |
Energy Stored in a Capacitor
U = ½CV² = ½QV = Q²/(2C)
Energy density (energy per unit volume) in electric field: u = ½ε₀E²
Important Definitions
| Term | Definition |
|---|---|
| Electric potential | Work done per unit charge in bringing a test charge from infinity to a point |
| Equipotential surface | Surface where every point has the same potential |
| Capacitance | Ratio of charge stored to potential difference: C = Q/V |
| Dielectric | Insulating material placed between capacitor plates that increases capacitance |
| Dielectric constant (κ) | Factor by which capacitance increases when dielectric is inserted |
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Solved Examples
Example 1
Two capacitors of 6 μF and 3 μF are connected in series across a 12 V battery. Find the equivalent capacitance and charge on each.
Answer: 1/C = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. So C = 2 μF.
Charge: Q = CV = 2 × 10⁻⁶ × 12 = 24 μC (same on both in series).
Example 2
A parallel plate capacitor has plates of area 100 cm² separated by 2 mm. Find the capacitance. If a dielectric of κ = 5 is inserted, find the new capacitance.
Answer: C = ε₀A/d = (8.854 × 10⁻¹²)(100 × 10⁻⁴)/(2 × 10⁻³) = 44.27 pF.
With dielectric: C’ = κC = 5 × 44.27 = 221.35 pF.
Example 3
Find the potential at a point 9 cm from a charge of 4 × 10⁻⁷ C.
Answer: V = kQ/r = (9 × 10⁹)(4 × 10⁻⁷)/0.09 = 3600/0.09 = 4 × 10⁴ V = 40 kV
Example 4
A capacitor of 10 μF is charged to 100 V. Find the energy stored.
Answer: U = ½CV² = ½ × 10 × 10⁻⁶ × (100)² = ½ × 10⁻⁴ × 10⁴ = 0.05 J = 50 mJ
Important Questions for Board Exams
1-Mark Questions
- What is an equipotential surface?
- How does the capacitance of a parallel plate capacitor change when a dielectric is inserted?
- What is the SI unit of capacitance?
3-Mark Questions
- Derive the expression for potential due to an electric dipole at a general point.
- Derive the formula for the equivalent capacitance of capacitors in series and parallel.
- Show that the electric field is always perpendicular to an equipotential surface.
5-Mark Questions
- Derive the expression for capacitance of a parallel plate capacitor with and without a dielectric medium.
- What is electrostatic potential energy? Derive the expression for energy stored in a capacitor. Also find the energy density.
Quick Revision Points
- Potential V = kQ/r (scalar); Field E = kQ/r² (vector); E = −dV/dr
- Equipotential surfaces ⊥ field lines; no work to move charge along them
- PE of two charges: U = kq₁q₂/r
- Capacitance C = Q/V; Parallel plate: C = ε₀A/d; with dielectric: C = κε₀A/d
- Series: 1/C = 1/C₁ + 1/C₂ (same charge, voltage divides)
- Parallel: C = C₁ + C₂ (same voltage, charge divides)
- Energy: U = ½CV² = ½QV = Q²/(2C); Energy density: u = ½ε₀E²
- Inside conductor: E = 0, V = constant, charge on surface only
Previous Chapter: Chapter 1 - Electric Charges and Fields
Next Chapter: Chapter 3 - Current Electricity
Class 12 Physics · Chapter 2 – swipe through all 10 cards to understand the whole chapter.
Electric Potential
Work done per unit charge to bring a test charge from infinity to a point.
Unit: volt (V) · scalar · k = 1/4πε0
- No direction (scalar)
- Reference: V = 0 at infinity
- 1 V = 1 J/C
Potential of a Point Charge
Potential at distance r from a point charge.
Positive Q → +V, negative Q → −V
- Falls off as 1/r
- Field E falls as 1/r2
- Total V = algebraic sum
Relation Between E and V
Field points from high to low potential and is the negative gradient of V.
E in V/m · uniform field: V = E d
- E along steepest V drop
- V constant ⇒ E = 0 along it
- Work W = q(V_A − V_B)
Equipotential Surfaces
Surfaces on which the potential is the same everywhere.
Always perpendicular to E
- No work to move charge on it
- Closer surfaces ⇒ stronger E
- Conductor surface is equipotential
Potential Energy of Charges
Energy stored in assembling a system of charges.
Like charges: U > 0 · unlike: U < 0
- Sum over all pairs
- In a field: U = qV
- Dipole: U = − p·E
Capacitance
Charge stored per unit potential difference across a conductor or capacitor.
Unit: farad (F) · 1 F = 1 C/V
- Depends on geometry + medium
- Isolated sphere: C = 4πε0R
- Bigger C ⇒ more charge per volt
Parallel Plate Capacitor
Two plates of area A separated by distance d.
K = dielectric constant (≥ 1)
- C rises with area A
- C falls as gap d grows
- Dielectric multiplies C by K
Capacitors in Series & Parallel
How capacitance combines when capacitors are joined.
Opposite of resistors
- Series: same charge Q
- Parallel: same voltage V
- Series C < smallest
Energy Stored in a Capacitor
Work done to charge a capacitor is stored in its field.
Energy density u = ½ ε0 E2
- Energy lives in the field
- Halves if C doubles at fixed Q
- Lost as heat when sharing charge
Effect of a Dielectric
An insulating slab inserted between the plates increases capacitance.
Polarisation opposes the applied field
- V drops if charge is fixed
- Stores more energy at fixed V
- K = 1 for vacuum
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Related Chapters in Class 12 Physics
- Electric Charges and Fields Class 12 Notes
- Current Electricity Class 12 Notes
- Moving Charges and Magnetism Class 12 Notes
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Frequently Asked Questions
Electric potential at a point is the work done per unit positive charge in bringing it from infinity to that point, V = W/q, measured in volts. Potential difference is the difference in potential between two points, which is what drives charge to move and equals the work done per unit charge between them.
The electric potential due to a point charge q at a distance r is V = kq/r, where k = 1/(4 pi epsilon0) is about 9 times 10^9 N m^2 per C^2. Potential is a scalar, so the total potential from several charges is the simple algebraic sum of their potentials.
Capacitance C = Q/V is the charge stored per unit voltage, measured in farads. For a parallel-plate capacitor C = epsilon0 A/d, so it depends only on the plate area A, the separation d and the dielectric between the plates, not on the charge or voltage applied.
The energy stored in a capacitor is U = (1/2) C V^2, which can also be written as U = (1/2) Q V or U = Q^2 / (2C). This energy is stored in the electric field between the plates and is released when the capacitor discharges.
In parallel the capacitances add: C = C1 + C2 + … In series the reciprocals add: 1/C = 1/C1 + 1/C2 + …, giving a smaller total capacitance. Parallel combinations store more charge at the same voltage, which is why they are used to increase capacitance.