Electric potential and potential difference
The electric force is conservative: the work done moving a charge from A to B does not depend on the path. So we can give every point a 'height' called potential.
Potential V at a point = work done by an outside agent to bring a unit positive charge from infinity to that point slowly (no speeding up). V = W/q. Unit: volt (1 V = 1 J/C). It is a scalar.
Potential difference V_B − V_A = W_AB / q, the work per unit charge to move from A to B.
Link with field: E = −dV/dr. The field points from high to low potential, along the steepest drop. In a uniform field, V = Ed.
Potential due to a point charge, a dipole and a system of charges
Point charge
Bringing unit charge from ∞ to r against the push of q: W = ∫ kq/r² dr from ∞ to r → V = kq/r. Positive near +q, negative near −q, zero at infinity.
Dipole
At a far point at distance r and angle θ from the axis: V = kp cosθ / r².
- On the axis (θ = 0): V = kp/r² (max).
- On the equator (θ = 90°): V = 0, even though E is not zero there.
Dipole potential falls as 1/r², a single charge's as 1/r.
System of charges
Potential is a scalar, so just add with signs: V = k(q₁/r₁ + q₂/r₂ + …). No arrows, no angles – easier than adding fields.
Charged spherical shell
Outside: V = kq/r. Inside and on the surface: V = kq/R, the same everywhere (because E = 0 inside).
Equipotential surfaces
An equipotential surface is a surface on which every point has the same potential.
- No work is done moving a charge along it (ΔV = 0).
- E is always perpendicular to it (if E had a part along the surface, moving along it would need work).
- Two equipotentials never cross.
- Where they are close together, the field is strong (E = −ΔV/Δr).
Shapes: point charge → concentric spheres; uniform field → parallel planes perpendicular to E; dipole → curved surfaces, with the equator plane at V = 0. The surface of a conductor in equilibrium is an equipotential.
Potential energy of a system of charges and of a dipole in a field
Two charges
Work to bring q₂ from ∞ to distance r from q₁ is stored as U = kq₁q₂ / r. U > 0 for like charges (they want to fly apart), U < 0 for unlike (bound). For three charges, add the energy of each pair: U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃).
Charge in an outside field
U = qV(r), where V is the potential of the outside field at that point.
Dipole in a uniform field
Work to turn a dipole from θ₁ to θ₂ is W = pE(cosθ₁ − cosθ₂). Taking U = 0 at θ = 90°: U = −pE cosθ = −p·E.
- θ = 0°: U = −pE, minimum → stable.
- θ = 180°: U = +pE, maximum → unstable.
- Work to flip from 0° to 180° = 2pE.
Conductors and insulators in an electric field
Conductors (metals) have many free electrons. Insulators hold their electrons tightly.
Electrostatics of conductors
- E = 0 inside the material of a conductor.
- Just outside, E is perpendicular to the surface.
- Net charge lives only on the outer surface.
- The whole conductor is at one potential.
- Surface field E = σ/ε₀; charge crowds at sharp points (σ is large there).
- Inside a hollow conductor with no charge in the cavity, E = 0 – electrostatic shielding. That is why a car or a metal cage protects you from lightning.
Free and bound charges
Free charges can travel through the whole material – the outer electrons in a metal. They carry current and rearrange until E inside becomes zero.
Bound charges are tied to their atom or molecule. In a field they can only shift a tiny bit or turn. They cannot flow, but their small shift still creates surface charge on a dielectric.
Dielectrics and polarisation
A dielectric is an insulator that becomes polarised in a field (glass, mica, paper, water).
- Non-polar molecules (O₂, N₂, CO₂): centres of + and − coincide. A field pulls them apart a little → induced dipoles.
- Polar molecules (H₂O, HCl): already dipoles, randomly pointed. A field lines them up partly.
Polarisation P = dipole moment per unit volume. The lined-up dipoles leave − charge on one face and + on the other. These bound surface charges make a field opposite to E₀, so the field inside drops to E = E₀/K. K (dielectric constant, relative permittivity) = ε/ε₀ is always > 1 (K ≈ 80 for water, ∞ for a metal).
Capacitors and capacitance
A capacitor is two conductors separated by an insulator. Given +Q and −Q, a potential difference V appears, with Q ∝ V. C = Q/V is the capacitance. Unit: farad (1 F = 1 C/V), a huge unit – real ones are µF, nF, pF. C depends only on size, shape, gap and the material between, not on Q or V.
Parallel plate capacitor with and without a dielectric
Two plates of area A, gap d. Between them the field is uniform: E = σ/ε₀ = Q/(ε₀A). V = Ed = Qd/(ε₀A). So
C₀ = ε₀A/d (vacuum or air).
Filled with a dielectric
The field drops to E₀/K, so V drops K times for the same Q: C = Kε₀A/d = KC₀.
Slab of thickness t (t < d)
V = E₀(d − t) + E₀t/K → C = ε₀A / (d − t + t/K). For a metal slab (K → ∞) C = ε₀A/(d − t).
Battery connected or not?
- Battery stays connected (V fixed): inserting a dielectric raises C, Q and U by K.
- Battery removed first (Q fixed): C rises K times, V and U fall K times.
Combination of capacitors: series and parallel
Series (one after another)
Every capacitor carries the same charge Q. Voltages add: V = V₁ + V₂ + … = Q/C₁ + Q/C₂ + …. So 1/C = 1/C₁ + 1/C₂ + 1/C₃ + …. C_eq is less than the smallest one. Two in series: C = C₁C₂/(C₁ + C₂).
Parallel (side by side)
Every capacitor has the same V. Charges add: Q = C₁V + C₂V + …. So C = C₁ + C₂ + C₃ + ….
Note: this is the reverse of resistors.
Energy stored in a capacitor
Charging a capacitor means pushing charge against the growing voltage. The work done is stored as electric potential energy in the field between the plates:
U = ½CV² = Q²/(2C) = ½QV.
Energy density (energy per volume) in the field: u = ½ε₀E². The syllabus asks for the formula only – use whichever form matches the known quantities.
Try it at home
Wrap your phone in two layers of kitchen aluminium foil (a closed box, no gaps) and call it. The call usually fails: the foil is a conducting shell and shields the inside – electrostatic shielding. Unwrap it and the call comes through.
In the 3D free play, set d = 2 mm, then 4 mm: C halves. Then set K = 5: C is five times bigger.
Key formulas and definitions
- V = W/q; V_B − V_A = W_AB/q; E = −dV/dr
- Point charge V = kq/r; dipole V = kp cosθ / r²
- System: V = k Σ qᵢ/rᵢ
- U (two charges) = kq₁q₂/r; U (dipole) = −pE cosθ
- C = Q/V; parallel plate C = Kε₀A/d
- Slab: C = ε₀A/(d − t + t/K)
- Series 1/C = 1/C₁ + 1/C₂ + …; parallel C = C₁ + C₂ + …
- U = ½CV² = Q²/2C = ½QV
Worked examples
1. Find the potential 9 cm from a +4 nC charge.
V = kq/r = 9 × 10⁹ × 4 × 10⁻⁹ / 0.09 = 400 V.
2. How much work moves a 2 µC charge from a point at 10 V to a point at 60 V?
W = q(V_B − V_A) = 2 × 10⁻⁶ × 50 = 1 × 10⁻⁴ J.
3. Charges +3 nC and −3 nC are at the ends of a 6 cm line. Find V at the midpoint and say if E is zero there.
V = k(3 × 10⁻⁹)/0.03 − k(3 × 10⁻⁹)/0.03 = 0. But E is not zero: both fields point toward the −3 nC charge, so they add. Zero V does not mean zero E.
4. Find the potential energy of +2 µC and −4 µC placed 20 cm apart.
U = kq₁q₂/r = 9 × 10⁹ × 2 × 10⁻⁶ × (−4 × 10⁻⁶)/0.2 = −0.36 J (negative: they attract, the system is bound).
5. A dipole p = 2 × 10⁻⁸ C m lies along a field of 5 × 10⁴ N/C. How much work turns it through 180°?
W = pE(cos0° − cos180°) = 2pE = 2 × 2 × 10⁻⁸ × 5 × 10⁴ = 2 × 10⁻³ J.
6. A parallel plate capacitor has plates of 0.02 m² and gap 1 mm in air. Find C. What if mica (K = 6) fills the gap?
C₀ = ε₀A/d = 8.85 × 10⁻¹² × 0.02 / 10⁻³ = 1.77 × 10⁻¹⁰ F = 177 pF. With mica C = 6 × 177 = 1062 pF ≈ 1.06 nF.
7. 2 µF, 3 µF and 6 µF are joined in series across 12 V. Find C_eq, Q and the voltage on each.
1/C = 1/2 + 1/3 + 1/6 = 1 → C = 1 µF. Q = CV = 12 µC (same on each). V₁ = 12/2 = 6 V, V₂ = 12/3 = 4 V, V₃ = 12/6 = 2 V; total 12 V.
8. The same three are now joined in parallel across 12 V. Find C_eq, the charges and the energy stored.
C = 2 + 3 + 6 = 11 µF. Q₁ = 24 µC, Q₂ = 36 µC, Q₃ = 72 µC (total 132 µC). U = ½CV² = ½ × 11 × 10⁻⁶ × 144 = 7.92 × 10⁻⁴ J.
9. A 10 µF capacitor is charged to 100 V and the battery is removed. A slab of K = 5 now fills the gap. Find the new C, V and energy.
Q stays 1 mC. C = 50 µF, V = Q/C = 20 V. U before = ½ × 10⁻⁵ × 10⁴ = 0.05 J; after = ½ × 5 × 10⁻⁵ × 400 = 0.01 J (energy falls K times; the slab is pulled in).
10. A capacitor with plates of area A and gap d has a slab of K = 4 and thickness d/2 inserted. Compare the new C with C₀.
C = ε₀A/(d − d/2 + d/8) = ε₀A/(5d/8) = 8C₀/5 = 1.6 C₀.
Common mistakes
- Adding potentials as vectors – potential is a scalar; add with signs only.
- Thinking V = 0 means E = 0 (the dipole's equator has V = 0 but E ≠ 0), or E = 0 means V = 0 (inside a shell V is constant, not zero).
- Using the resistor rules for capacitors – series capacitors add as 1/C, parallel ones add directly.
- Forgetting whether the battery is still connected when a dielectric is inserted: connected keeps V fixed, disconnected keeps Q fixed.