Electric Charges and Fields

Certain materials, when rubbed against each other (for example, a glass rod rubbed with silk, or a plastic comb run through dry hair), acquire the property of attracting small bits of paper. This property is called electric charge. Charge is a scalar property of matter that exists in two kinds, called positive and negative. Like charges repel each other; unlike charges attract. Materials that allow charge to move freely through them are called conductors (metals, the human body, earth); materials that do not are called insulators or dielectrics (glass, rubber, plastic). A charged body can charge a nearby conductor without touching it, through induction: bringing a charged rod near an uncharged conductor separates charge on the conductor, with the near end acquiring the opposite charge.

Three basic properties of electric charge matter for this chapter:

• Additivity — charges add up algebraically like real numbers. A body with charges +q₁, +q₂ and −q₃ has total charge q₁ + q₂ − q₃.
• Conservation — the total charge of an isolated system stays constant; charge is only transferred, never created or destroyed. When glass is rubbed with silk, electrons move from glass to silk, so glass becomes positive and silk becomes negative by an equal amount.
• Quantisation — charge on any body is always an integral multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C, so q = ne, where n is an integer. This is because charge is transferred in units of the electron/proton charge. Quantisation is ignored for everyday (macroscopic) charges because e is so tiny compared to typical charges (~10⁻⁶ to 10⁻⁹ C) that charge appears to vary continuously.

Coulomb's law gives the force between two stationary point charges q₁ and q₂ separated by distance r in vacuum: the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, and acts along the line joining them.

In SI units, the constant of proportionality is written as 1/4πε₀, where ε₀ is the permittivity of free space (ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²). The combination 1/4πε₀ = k ≈ 9 × 10⁹ N m² C⁻². In vector form, the force on charge q₂ due to q₁ is directed along the unit vector from q₁ to q₂; by Newton's third law the force on q₁ due to q₂ is equal and opposite. If the charges are placed in a medium of relative permittivity (dielectric constant) K, the force reduces by a factor of K compared to vacuum.

Worked example: Two small charged spheres carry charges 2 × 10⁻⁷ C and 3 × 10⁻⁷ C and are placed 30 cm apart in air. Find the force between them.

r = 30 cm = 0.3 m. Using F = k q₁q₂/r²:

F = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)² = (9 × 10⁹ × 6 × 10⁻¹⁴) / 0.09 = 5.4 × 10⁻⁴ / 0.09 = 6 × 10⁻³ N.

Since both charges are taken as positive, the 6 × 10⁻³ N force is repulsive, directed along the line joining the spheres.

Rather than speaking of the force one charge exerts on another directly, it is useful to say that a charge q modifies the space around it, creating an electric field. Any other charge placed in this field experiences a force. The electric field at a point is defined as the force per unit positive test charge placed at that point (in the limit that the test charge is small enough not to disturb the source charges): E = F/q₀. Electric field is a vector quantity with SI unit newton per coulomb (N/C), also expressed as volt per metre (V/m). For a point charge q, the field at distance r is E = kq/r², directed radially outward from q if q is positive, and radially inward (toward q) if q is negative.

When several charges q₁, q₂, …, q_n are present, the resultant field at any point is the vector sum of the fields each charge would produce individually, as if the others were absent. This is the principle of superposition, and it follows directly from the fact that Coulomb forces from different charges add vectorially.

Electric field lines are a way to visualise the field: a field line is drawn so that the tangent to it at any point gives the direction of E at that point. Key properties: field lines start on positive charges (or at infinity) and end on negative charges (or at infinity); two field lines can never cross each other, because the field would then have two directions at the crossing point, which is impossible; the density of field lines (number of lines per unit area, perpendicular to them) indicates the field strength — closely spaced lines mean a strong field; field lines of a static field never form closed loops; and, for a uniform field, field lines are straight, parallel and equally spaced.

Worked example: Two point charges q_A = +3 μC and q_B = −3 μC are located 20 cm apart in vacuum. Find the electric field at the midpoint of the line AB.

The midpoint is 10 cm = 0.1 m from each charge. Field due to q_A at the midpoint (magnitude) = kq_A/r² = (9 × 10⁹ × 3 × 10⁻⁶)/(0.1)² = 2.7 × 10⁶ N/C, directed away from A (i.e. from A toward B, since q_A is positive). Field due to q_B at the midpoint has the same magnitude, 2.7 × 10⁶ N/C, but points toward B (since q_B is negative, the field points toward it) — which is also the direction from A to B. Since both fields point in the same direction, they add: E_net = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C, directed from A to B.

An electric dipole is a pair of equal and opposite point charges, +q and −q, separated by a small distance 2a. Its electric dipole moment is a vector p of magnitude p = q × 2a, directed from the negative charge to the positive charge, with SI unit coulomb-metre (C m).

Using superposition, the field of a dipole can be found at two special points for r >> a (r much larger than the charge separation):

• On the axial line (the line through both charges, extended), the field is directed along p, with magnitude approximately E_axial ≈ 2kp/r³.
• On the equatorial line (the perpendicular bisector of the line joining the charges), the field is directed opposite to p, with magnitude approximately E_eq ≈ kp/r³.

So for the same distance r (with r >> a), the axial field is exactly twice the equatorial field in magnitude, and the two fields point in opposite senses relative to p. Both fall off as 1/r³, faster than the 1/r² fall-off of a single point charge, because the dipole's net charge is zero and the fields of +q and −q nearly cancel at large distances.

When a dipole is placed in a uniform external electric field E, making angle θ with E, the two equal and opposite forces qE on the two charges have the same magnitude but act at different points, so the net force on the dipole is zero, but they form a couple that produces a net torque: τ = p × E, with magnitude τ = pE sinθ. The torque tends to rotate the dipole to align p with E. Torque is zero when θ = 0° (p parallel to E, stable equilibrium) or θ = 180° (p antiparallel to E, unstable equilibrium), and is maximum, equal to pE, when θ = 90°. The work needed to rotate the dipole from angle θ₁ to θ₂ against this torque gives the dipole its potential energy in the field, U(θ) = −pE cosθ (taking U = 0 at θ = 90°), which is least (most negative, most stable) when p is aligned with E.

Worked example: A dipole of moment 2 × 10⁻⁶ C m is placed at 60° to a uniform field of magnitude 2 × 10⁵ N/C. Find the torque on it.

τ = pE sinθ = (2 × 10⁻⁶)(2 × 10⁵)(sin 60°) = (0.4)(0.866) ≈ 0.346 N m.

Macroscopic charged objects (a charged rod, ring, disc, sphere, or shell) contain an enormously large number of elementary charges, so it is convenient to treat the charge as if it were spread continuously over the object rather than as discrete point charges. Three kinds of charge density are used depending on the geometry of the object:

• Linear charge density λ = charge per unit length, for charge distributed along a line or curve (unit C/m).
• Surface charge density σ = charge per unit area, for charge distributed over a surface (unit C/m²).
• Volume charge density ρ = charge per unit volume, for charge distributed throughout a volume (unit C/m³).

To find the field of a continuous distribution in principle, the object is imagined to be divided into a very large number of small elements, each carrying an infinitesimal charge Δq that can be treated as a point charge. Each element produces a field ΔE = kΔq/r² at the point of interest, and the total field is the vector sum (in the limit, an integral) of all these contributions — this is simply the superposition principle applied to infinitesimal charges. Direct summation this way is often mathematically involved; for distributions with high symmetry (a long straight wire, an infinite plane sheet, a uniformly charged spherical shell), the next sections show how Gauss's law gives the same result far more quickly, without detailed integration.

Worked example: A thin rod of length 40 cm carries a uniform linear charge density of 2 × 10⁻⁶ C/m. Find the total charge on the rod.

Since λ is uniform, Q = λ × L = (2 × 10⁻⁶ C/m)(0.4 m) = 8 × 10⁻⁷ C = 0.8 μC.

The electric flux through a small area element ΔS in an electric field E is defined as ΔΦ = E · ΔS = EΔS cosθ, where θ is the angle between E and the outward normal to the area. Flux is a scalar quantity; loosely, it measures the number of field lines passing through the surface. Its SI unit is N m²/C (equivalently, V m). For a large surface, the total flux is obtained by summing (integrating) E · ΔS over the whole surface: Φ = ∮E · dS for a closed surface.

Gauss's law states that the total electric flux through any closed surface (called a Gaussian surface) equals 1/ε₀ times the net charge enclosed by that surface: Φ = q_enc/ε₀. Remarkably, this result is independent of the size or shape of the closed surface and of exactly where the charges are located inside it; charges outside the closed surface contribute zero net flux through it (though they may affect the field at individual points on the surface, their net contribution to the total flux cancels out). Gauss's law can be derived from Coulomb's law and the superposition principle, but it is more general and, for symmetric charge distributions, gives the electric field far more directly than summing Coulomb's law contributions.

Worked example: A point charge of +3 μC is placed at the centre of a cube of side 10 cm. Find (a) the total electric flux through the cube, and (b) the flux through one face.

(a) By Gauss's law, total flux depends only on the enclosed charge, not on the cube's size: Φ = q/ε₀ = (3 × 10⁻⁶)/(8.854 × 10⁻¹²) ≈ 3.39 × 10⁵ N m²/C.

(b) Since the charge is at the centre, by symmetry the flux is shared equally among the 6 faces: Φ_face = 3.39 × 10⁵/6 ≈ 5.65 × 10⁴ N m²/C.

Infinite line charge: for a thin, infinitely long straight wire with uniform linear charge density λ, symmetry requires the field to point radially outward (for λ > 0) and to have the same magnitude at every point at a given perpendicular distance r from the wire. Choosing a cylindrical Gaussian surface of radius r and length l, coaxial with the wire, the flux through the two flat end-caps is zero (E is perpendicular to their normals), and the flux through the curved surface is E × 2πrl. The enclosed charge is λl, so Gauss's law gives E(2πrl) = λl/ε₀, or E = λ/(2πε₀r). The field falls off as 1/r, more slowly than a point charge's 1/r².

Infinite plane sheet of charge: for a thin, infinite plane sheet with uniform surface charge density σ, symmetry requires the field to be perpendicular to the sheet and equal in magnitude at equal distances on either side. Using a cylindrical ("pillbox") Gaussian surface straddling the sheet, with each flat face of area A parallel to the sheet, flux passes only through the two flat faces (none through the curved side): total flux = 2EA. The enclosed charge is σA, so 2EA = σA/ε₀, giving E = σ/(2ε₀), independent of the distance from the sheet. For a charged conductor's surface (where the field just inside the conductor is zero), the same pillbox argument gives a field just outside the surface of E = σ/ε₀ — twice the isolated-sheet value, because all the flux emerges through the outer face only.

Uniformly charged thin spherical shell of radius R and total charge q: by spherical symmetry, E must be radial and depend only on the distance r from the centre. Applying Gauss's law with a concentric spherical Gaussian surface of radius r: for r < R (inside the shell), the Gaussian surface encloses no charge, so E = 0 everywhere inside; for r ≥ R (on or outside the shell), the Gaussian surface encloses the entire charge q, so E(4πr²) = q/ε₀, giving E = kq/r² — identical to the field of a point charge q placed at the centre. Thus the field is zero inside, jumps discontinuously to σ/ε₀ at the surface, and falls off as 1/r² outside, exactly as if all the charge were concentrated at the centre.

Worked example: A thin spherical shell of radius 10 cm carries a total charge of 2 μC uniformly spread over it. Find the electric field at (a) 5 cm from the centre, (b) 10 cm (on the surface), and (c) 20 cm from the centre.

(a) r = 5 cm < R = 10 cm, inside the shell, so E = 0.

(b) r = R = 0.1 m: E = kq/R² = (9 × 10⁹ × 2 × 10⁻⁶)/(0.1)² = 1.8 × 10⁴/0.01 = 1.8 × 10⁶ N/C.

(c) r = 0.2 m > R: E = kq/r² = (9 × 10⁹ × 2 × 10⁻⁶)/(0.2)² = 1.8 × 10⁴/0.04 = 4.5 × 10⁵ N/C.