Electric charge and its conservation
Charge is a property of matter that makes it push or pull other charged things. There are two kinds: positive (+) and negative (−). Like charges repel. Unlike charges attract.
Three basic facts
- Additivity: total charge is the plain sum with signs. +3 µC and −5 µC together make −2 µC.
- Conservation: in an isolated system the total charge never changes. Rubbing, touching or even nuclear reactions only move or pair charges; they never create net charge.
- Quantisation: every charge is a whole number of electron charges: q = ne, with e = 1.6 × 10⁻¹⁹ C and n an integer.
Charging methods
By friction (rubbing moves electrons), by conduction (touching shares charge) and by induction (bring a charged rod near, earth the far side, remove the earth, then the rod: the object keeps the opposite charge). Conductors let charge move freely; insulators do not.
Try it: rub a plastic ruler in dry hair and hold it near a thin stream of tap water. The stream bends toward the ruler – induced charge being pulled.
Coulomb's law
The force between two point charges q₁ and q₂ a distance r apart in vacuum is
F = k q₁q₂ / r², with k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
- The force acts along the line joining the charges.
- It is an inverse-square law, just like gravity – but it can repel as well as attract and it is about 10³⁶ times stronger between two protons.
- In a medium of dielectric constant K, the force becomes F/K.
Vector form: F₁₂ = k q₁q₂ r̂₂₁ / r², where r̂₂₁ points from q₂ to q₁. Newton's third law holds: F₁₂ = −F₂₁.
1 coulomb is the charge that, placed 1 m from an equal charge in vacuum, repels it with 9 × 10⁹ N – a huge force, so real lab charges are in µC or nC.
Forces between many charges: the superposition principle
When there are many charges, each pair acts as if the others were not there. The net force on one charge is the vector sum of the forces from all the others:
F₁ = F₁₂ + F₁₃ + F₁₄ + …
How to solve
- Draw each force as an arrow (repel = away, attract = toward).
- Find each size with Coulomb's law.
- Split into x and y parts, add each part, then combine: F = √(Fx² + Fy²).
Symmetry saves time: equal charges at the corners of a square or an equilateral triangle give zero net force at the centre.
Continuous charge distribution
On a charged rod, sheet or ball the charge is spread out, so we use densities:
- Linear λ = charge per length (C/m) – a wire.
- Surface σ = charge per area (C/m²) – a sheet or shell.
- Volume ρ = charge per volume (C/m³) – a solid ball.
We cut the body into tiny bits dq = λdl, σdA or ρdV, treat each as a point charge, and add (integrate) their fields: E = ∫ k dq r̂ / r². Gauss's law (below) avoids this integral when the shape is very symmetric.
Electric field of a point charge
The electric field at a point is the force on a tiny positive test charge q₀ placed there, per unit charge: E = F / q₀. Unit: N/C (same as V/m). It is a vector.
For a point charge q: E = kq / r², pointing away from + and toward −. Once you know E, the force on any charge q is simply F = qE.
The test charge must be tiny so it does not push the source charges out of place. The field of many charges again adds as vectors.
Electric field lines
A field line is a curve whose tangent at every point shows the direction of E.
- They start on + charges and end on − charges (or at infinity).
- They never cross – at a crossing E would have two directions.
- Closer lines = stronger field.
- They do not form closed loops in electrostatics.
- They meet a conductor's surface at 90°.
Pictures: single + (lines straight out), single − (straight in), equal + and − (lines bend from + to −), two equal + charges (lines push apart with an empty neutral point in the middle), and a uniform field (parallel, evenly spaced lines).
Electric dipole and its field
An electric dipole is a pair +q and −q separated by a small distance 2a. Its dipole moment is p = q × 2a, a vector pointing from − to +. Unit: C m. Water molecules are natural dipoles.
Field on the axis (end-on)
At distance r from the centre along the axis: E = 2kpr / (r² − a²)², which for r ≫ a becomes E_axial = 2kp / r³, along p.
Field on the equator (broad-on)
On the perpendicular bisector: E = kp / (r² + a²)^{3/2}, which for r ≫ a becomes E_eq = kp / r³, opposite to p.
So far from a dipole E falls as 1/r³, faster than a single charge (1/r²), because the + and − nearly cancel. And E_axial = 2 × E_eq at the same distance.
Derivation idea (axial)
The + charge is at distance r − a and the − charge at r + a. E = kq/(r − a)² − kq/(r + a)² = kq·4ar/(r² − a²)² = 2kpr/(r² − a²)².
Torque on a dipole in a uniform field
In a uniform field E, the + end feels qE along E and the − end feels qE opposite. Net force = 0, so the dipole does not move off. But the two forces are on different lines, so they form a couple:
τ = (qE)(2a sinθ) = pE sinθ, or in vector form τ = p × E.
- θ = 0°: τ = 0, stable (p along E).
- θ = 90°: τ = pE, the maximum.
- θ = 180°: τ = 0 but unstable.
In a non-uniform field the two forces are unequal, so there is also a net force – that is why a charged comb pulls neutral paper (the paper's induced dipoles sit in a non-uniform field).
Electric flux
Electric flux through a surface measures how many field lines pass through it:
Φ = E · A = EA cosθ, where θ is the angle between E and the outward normal of the area. Unit: N m² C⁻¹ (or V m). Scalar.
- E perpendicular to the surface (θ = 0): Φ = EA, maximum.
- E along the surface (θ = 90°): Φ = 0.
- Lines coming out count +, lines going in count −.
Gauss's law and its uses: wire, plane sheet, thin spherical shell
Gauss's law: the total flux out of any closed surface equals the charge enclosed divided by ε₀: ∮E·dA = q_in / ε₀. Charges outside the surface add zero net flux (every line that enters also leaves).
Recipe: pick a closed imaginary surface (a Gaussian surface) on which E is the same size and is either along or perpendicular to the surface.
1. Infinitely long straight wire (charge λ per metre)
Cylinder of radius r, length l. Flux only through the curved side: E(2πrl) = λl/ε₀ → E = λ / (2πε₀r), pointing radially out. E ∝ 1/r.
2. Infinite plane sheet (σ per m²)
Small cylinder through the sheet with end area A. Flux through both ends: 2EA = σA/ε₀ → E = σ / (2ε₀). Same at all distances.
3. Thin spherical shell (charge q, radius R)
Sphere of radius r. Outside (r > R): E(4πr²) = q/ε₀ → E = kq/r², as if all charge were at the centre. Inside (r < R): no charge enclosed → E = 0. At the surface: E = kq/R² = σ/ε₀.
Coulomb's law itself follows from Gauss's law applied to a sphere around a point charge.
Board exam focus
Units I and II together carry about 16 marks. Common asks: state and derive (dipole axial/equatorial field, torque on a dipole, E of wire/sheet/shell using Gauss's law), draw field-line patterns, a 2–3 mark numerical on Coulomb force with superposition, and a flux question. Always draw the Gaussian surface in derivations.
Key formulas and definitions
- q = ne, e = 1.6 × 10⁻¹⁹ C
- F = k q₁q₂ / r², k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻²
- F_net = F₁ + F₂ + … (vector sum)
- E = F / q₀; point charge E = kq / r²
- p = q × 2a; E_axial = 2kp/r³; E_equatorial = kp/r³
- τ = p × E, τ = pE sinθ
- Φ = EA cosθ; ∮E·dA = q_in/ε₀
- Wire E = λ/(2πε₀r); sheet E = σ/(2ε₀); shell E = kq/r² (out), 0 (in)
Worked examples
1. How many electrons must be removed from a neutral body to give it +3.2 µC?
n = q/e = 3.2 × 10⁻⁶ / 1.6 × 10⁻¹⁹ = 2 × 10¹³ electrons.
2. Two charges +2 µC and −3 µC are 30 cm apart in air. Find the force.
F = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶ / (0.3)² = 0.054/0.09 = 0.6 N, attractive.
3. The force between two charges is 16 N. The distance is doubled and each charge is halved. New force?
F ∝ q₁q₂/r². New = 16 × (½ × ½) / 2² = 16 × ¼ / 4 = 1 N.
4. Charges +1 µC at A and +4 µC at B are 3 m apart. Where between them is the net force on a third charge zero?
Let the point be x from A. k(1)/x² = k(4)/(3 − x)² → (3 − x) = 2x → x = 1 m from A (the smaller charge).
5. Three charges of +2 µC each sit at the corners of an equilateral triangle of side 10 cm. Find the force on one of them.
Each force = 9 × 10⁹ × (2 × 10⁻⁶)² / 0.01 = 3.6 N. Angle between the two forces = 60°. F = √(3.6² + 3.6² + 2 × 3.6² cos60°) = 3.6√3 ≈ 6.24 N, along the outward bisector.
6. A dipole has charges ±5 nC, 2 mm apart. Find p and the field 10 cm away on its axis.
p = 5 × 10⁻⁹ × 2 × 10⁻³ = 10⁻¹¹ C m. E_axial = 2kp/r³ = 2 × 9 × 10⁹ × 10⁻¹¹ / 10⁻³ = 180 N/C, along p.
7. A dipole of moment 4 × 10⁻⁹ C m is at 30° to a uniform field of 5 × 10⁴ N/C. Find the torque.
τ = pE sinθ = 4 × 10⁻⁹ × 5 × 10⁴ × 0.5 = 1 × 10⁻⁴ N m.
8. A square of side 10 cm lies in a uniform field of 200 N/C. The field makes 60° with the normal. Find the flux.
Φ = EA cosθ = 200 × 0.01 × 0.5 = 1 N m² C⁻¹.
9. A long wire carries λ = 2 µC/m. Find E at 20 cm.
E = 2kλ/r = 2 × 9 × 10⁹ × 2 × 10⁻⁶ / 0.2 = 1.8 × 10⁵ N/C, radially outward.
10. A shell of radius 10 cm has +8 nC. Find E at 5 cm and at 20 cm from the centre.
At 5 cm (inside): E = 0. At 20 cm: E = 9 × 10⁹ × 8 × 10⁻⁹ / 0.04 = 1800 N/C.
Common mistakes
- Forgetting to convert µC to C (× 10⁻⁶) and cm to m before using Coulomb's law.
- Adding the sizes of forces from many charges instead of adding them as vectors.
- Thinking the flux through a closed surface depends on charges outside it – outside charges give zero net flux (but they do change E at points on the surface).
- Mixing up dipole formulas: axial field is 2kp/r³, equatorial is kp/r³, and the dipole direction is from − to +.