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Electric Charges and Fields — Notes

7 sections 7 min readBeta

1. Electric charge and its properties

Electric charge is a basic property of matter that makes it exert and feel electric forces. SI unit: coulomb (C).

  • Two kinds: positive and negative. Like charges repel; unlike charges attract. (Glass rubbed with silk becomes positive; the silk negative.)
  • Additivity: the total charge of a system is the algebraic sum of its charges.
  • Conservation: the net charge of an isolated system never changes. Charging by rubbing only transfers electrons.
  • Quantisation: q=neq = n e, where nn is an integer and e=1.6×10−19e = 1.6 \times 10^{-19} C. Everyday charges contain so many ee that charge looks continuous.

Conductors (metals, the human body, tap water) have free charge carriers; insulators (glass, plastic, dry wood) hold charge where it is placed.

Charging by induction: bring a charged rod near an insulated conductor; charges separate. Earth the far side, remove the earth connection, then remove the rod. The conductor keeps a charge opposite to the rod's. A charged body attracts a neutral light object (a paper bit) because it induces opposite charge on the near side.

Formulas

  • Quantisation
    q=ne,e=1.6×10−19 Cq = n e,\quad e = 1.6 \times 10^{-19}\ \text{C}

Key terms

Quantisation

Charge exists only in whole-number multiples of e=1.6×10−19e = 1.6 \times 10^{-19} C.

Conservation of charge

The total charge of an isolated system stays constant.

Induction

Separating charges in a conductor by bringing a charged body near it, without contact.

2. Coulomb’s law and superposition

For two point charges a distance rr apart in vacuum:

F=14πε0∣q1q2∣r2,14πε0=k=9×109 N m2 C−2,ε0=8.85×10−12 C2 N−1 m−2F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_2|}{r^2},\qquad \dfrac{1}{4\pi\varepsilon_0} = k = 9 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2},\qquad \varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}

Vector form: F⃗21=kq1q2r2r^12\vec F_{21} = k\dfrac{q_1 q_2}{r^2}\hat r_{12} (force on q2q_2; r^12\hat r_{12} points from q1q_1 to q2q_2), and F⃗12=−F⃗21\vec F_{12} = -\vec F_{21}.

In a medium of dielectric constant εr\varepsilon_r: F=FvacuumεrF = \dfrac{F_\text{vacuum}}{\varepsilon_r}.

Superposition: the force on a charge due to many charges is the vector sum of the individual Coulomb forces, each calculated as if the others were absent.

Method for multi-charge problems: (1) draw each force with its direction from the signs; (2) find magnitudes; (3) add by components or, for perpendicular forces, by Pythagoras; (4) state magnitude and direction.

Worked result (deck example 2): forces of 0.40 N and 0.30 N at right angles give 0.50 N.

Forces drawn to scale (300 pixels per newton); the resultant is the diagonal of the rectangle.

Formulas

  • Coulomb’s law
    F=k ∣q1q2∣r2F = k\,\dfrac{|q_1 q_2|}{r^2}
  • Superposition
    F⃗0=∑iF⃗0i\vec F_0 = \sum_i \vec F_{0i}

Key terms

Coulomb’s law

The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation.

Dielectric constant

The factor εr=ε/ε0\varepsilon_r = \varepsilon/\varepsilon_0 by which a medium reduces the force between charges.

3. Electric field and field lines

The electric field at a point is the force per unit small positive test charge: E⃗=F⃗/q0\vec E = \vec F / q_0 (unit N C−1^{-1}). A charge qq placed there feels F⃗=qE⃗\vec F = q\vec E (opposite to E⃗\vec E if q<0q < 0).

Point charge: E=k∣q∣r2E = k\dfrac{|q|}{r^2}, radially away from a positive charge and towards a negative one.

System of charges: E⃗=∑E⃗i\vec E = \sum \vec E_i (superposition).

Field lines — properties:

  • tangent gives the direction of E⃗\vec E; start on ++ charges, end on −- charges (or at infinity);
  • never cross; never form closed loops in electrostatics;
  • crowding shows strength; parallel equally spaced lines show a uniform field;
  • meet a conductor's surface at right angles.

Neutral point: where the resultant field is zero. Between two like charges it lies between them (nearer the smaller charge); for unlike charges it lies outside, on the side of the smaller charge.

Traced numerically from Coulomb’s law; arrows checked against the computed field.
Traced numerically; no line passes through the neutral point N.

Formulas

  • Electric field
    E⃗=F⃗q0,E=k ∣q∣r2\vec E = \dfrac{\vec F}{q_0},\qquad E = k\,\dfrac{|q|}{r^2}

Key terms

Electric field

Force per unit positive test charge at a point.

Field line

A curve whose tangent at every point is along the electric field there.

Neutral point

A point where the net electric field is zero.

4. Electric dipole

Dipole: charges +q+q and −q-q separated by 2a2a. Dipole moment p⃗=q(2a)\vec p = q(2a), from −q-q to +q+q, unit C m.

Point at distance rr from the centreExactShort dipole (r≫ar \gg a)Direction
Axis14πε02pr(r2−a2)2\dfrac{1}{4\pi\varepsilon_0}\dfrac{2pr}{(r^2-a^2)^2}2kpr3\dfrac{2kp}{r^3}along p⃗\vec p
Equatorial line14πε0p(r2+a2)3/2\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{(r^2+a^2)^{3/2}}kpr3\dfrac{kp}{r^3}opposite to p⃗\vec p

At the same large distance, Eaxis=2EeqE_\text{axis} = 2E_\text{eq}. A dipole field falls as 1/r31/r^3.

Dipole in a uniform field: net force zero; torque τ⃗=p⃗×E⃗\vec\tau = \vec p \times \vec E, τ=pEsin⁡θ\tau = pE\sin\theta. Stable equilibrium at θ=0∘\theta = 0^\circ, unstable at 180∘180^\circ, maximum torque pEpE at 90∘90^\circ. In a non-uniform field there is also a net force (towards stronger field when p⃗\vec p is along E⃗\vec E).

Formulas

  • Dipole moment
    p=q×2ap = q \times 2a
  • Short dipole
    Eaxis≈2kpr3,Eeq≈kpr3E_\text{axis} \approx \dfrac{2kp}{r^3},\quad E_\text{eq} \approx \dfrac{kp}{r^3}
  • Torque
    τ⃗=p⃗×E⃗\vec\tau = \vec p \times \vec E

Key terms

Dipole moment

Product of either charge and the separation, directed from the negative to the positive charge.

Polar molecule

A molecule with a permanent dipole moment, such as water.

5. Continuous charge distributions

Charge spread over an object is described by a density:

  • linear λ=dq/dl\lambda = dq/dl (C m−1^{-1}), e.g. a charged wire or ring;
  • surface σ=dq/dS\sigma = dq/dS (C m−2^{-2}), e.g. a charged plate or shell;
  • volume ρ=dq/dV\rho = dq/dV (C m−3^{-3}), e.g. a charged cloud.

Set-up: split into elements dqdq, use dE⃗=kdqr2r^d\vec E = k\dfrac{dq}{r^2}\hat r for each, and add vectors: E⃗=k∫dqr2r^\vec E = k\int \dfrac{dq}{r^2}\hat r. Use symmetry to see which components cancel.

Formulas

  • Continuous distribution
    E⃗=14πε0∫dqr2 r^\vec E = \dfrac{1}{4\pi\varepsilon_0}\int \dfrac{dq}{r^2}\,\hat r

Key terms

Linear charge density

Charge per unit length, λ\lambda.

Surface charge density

Charge per unit area, σ\sigma.

6. Electric flux and Gauss’s law

Flux through a small area: Δϕ=E⃗⋅ΔS⃗=E ΔScos⁡θ\Delta\phi = \vec E\cdot\Delta\vec S = E\,\Delta S\cos\theta, where ΔS⃗\Delta\vec S is normal to the area (outward for a closed surface). Unit N m² C−1^{-1}; a scalar.

Gauss’s law: ∮SE⃗⋅dS⃗=qencε0\displaystyle\oint_S \vec E\cdot d\vec S = \dfrac{q_\text{enc}}{\varepsilon_0}.

  • True for any closed surface of any shape.
  • E⃗\vec E in the integral is the total field (from charges inside and outside), but only enclosed charge contributes net flux.
  • Charge outside: net flux zero. Charge inside: flux independent of the surface's size and shape and of where the charge sits inside.
  • A charge qq at the centre of a cube sends q/6ε0q/6\varepsilon_0 through each face.

Formulas

  • Gauss’s law
    ϕ=∮SE⃗⋅dS⃗=qencε0\phi = \oint_S \vec E\cdot d\vec S = \dfrac{q_\text{enc}}{\varepsilon_0}

Key terms

Electric flux

The surface integral of the field, ∮E⃗⋅dS⃗\oint \vec E\cdot d\vec S; a measure of field lines crossing a surface.

Gaussian surface

A closed surface chosen to apply Gauss’s law.

7. Applications of Gauss’s law

Charged objectGaussian surfaceResult
Infinitely long straight wire, λ\lambdaCoaxial cylinder; flux only through curved sideE=λ2πε0rE = \dfrac{\lambda}{2\pi\varepsilon_0 r}, radial
Infinite plane sheet, σ\sigmaPillbox; flux only through the two endsE=σ2ε0E = \dfrac{\sigma}{2\varepsilon_0}, normal, independent of distance
Thin spherical shell, qq, radius RR, outside (r>Rr > R)Concentric sphereE=14πε0qr2E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}
Same shell, inside (r<Rr < R)Concentric sphere, qenc=0q_\text{enc} = 0E=0E = 0

Steps for a derivation: draw the Gaussian surface; state why E⃗\vec E is constant and normal on part of it and why other parts carry no flux; write flux == charge enclosed /ε0/\varepsilon_0; solve for EE; give the direction.

  • inside: E = 0
  • outside: E = kq/r²
Field of a uniformly charged thin spherical shell of radius R.
  • long wire: E ∝ 1/r
  • point charge: E ∝ 1/r²
  • plane sheet: E constant
Same value at r = 1 for comparison.

Formulas

  • Long wire
    Ewire=λ2πε0rE_\text{wire} = \dfrac{\lambda}{2\pi\varepsilon_0 r}
  • Plane sheet
    Esheet=σ2ε0E_\text{sheet} = \dfrac{\sigma}{2\varepsilon_0}
  • Thin shell
    Eshell=q4πε0r2 (r>R),0 (r<R)E_\text{shell} = \dfrac{q}{4\pi\varepsilon_0 r^2}\ (r > R),\quad 0\ (r < R)

Key terms

Shielding

The field inside a charged conducting or thin shell is zero, so the inside is screened from charges on the shell.

Exam tips

  • Units first: μ\muC =10−6= 10^{-6} C, nC =10−9= 10^{-9} C, cm =10−2= 10^{-2} m.
  • Direction always: a force or field answer without a direction loses a mark.
  • Derivations (dipole axial or equatorial field, wire, sheet, shell): figure, directions, the key step, the final result, the approximation used.
  • Gauss's law: name the surface and justify where the flux is zero.
  • Field lines: arrows on every line, no crossings, density shows strength.