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Electrostatics

Electrostatics is the regime of electromagnetism in which charges are at rest and fields do not change in time. The governing laws are Gauss’s law ∇⋅E=ρ/ε0\nabla \cdot \mathbf{E} = \rho/\varepsilon_0 and ∇×E=0\nabla \times \mathbf{E} = 0 (the field is conservative, derived from a scalar potential).

When it is relevant. Electrostatics is the slow, static limit of Electrodynamics. It applies whenever the charges of interest are at rest or move slowly enough that the fields they produce are essentially constant on the timescale of the problem. The criterion is that the system’s size LL is small compared with the distance light travels in the relevant time, L≪c τL \ll c\,\tau (equivalently, the signal wavelength is much larger than the system). Under that condition the time-derivative terms in Maxwell’s equations vanish, the electric field becomes conservative, and the whole problem collapses to a single scalar potential obeying Poisson’s equation. This is the regime behind capacitors, electrostatic shielding (Faraday cages), electrostatic precipitators and paint spraying, contact and triboelectric charging, the forces between charged objects, and the non-radiative electric structure of atoms and molecules.

This page treats vacuum electrostatics only: the permittivity is the vacuum value ε0\varepsilon_0 throughout. How a material modifies the field — dielectrics, polarization, and the constitutive relation D=εE\mathbf{D}=\varepsilon\mathbf{E} — is the subject of Electromagnetic Media.

Authoritative sources

Part of Electromagnetism; the static limit of Electrodynamics.

1. Coulomb’s law

Definition. The electrostatic force between two point charges q1q_1 and q2q_2 separated by a distance rr is

F12=14πε0 q1q2r2 r^12\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^2}\,\hat{\mathbf{r}}_{12}

where r^12\hat{\mathbf{r}}_{12} points from charge 1 toward charge 2 and k≡1/(4πε0)≈8.99×109 N\cdotpm2/C2k \equiv 1/(4\pi\varepsilon_0) \approx 8.99\times10^9\ \text{N·m}^2/\text{C}^2. The force is central (along the line joining the charges), repulsive for like signs and attractive for opposite signs, and it obeys Newton’s third law (F12=−F21\mathbf{F}_{12} = -\mathbf{F}_{21}).

Details

The idea, step by step.

  1. An inverse-square, central force. Coulomb measured the force between two charged spheres and found it scales as 1/r21/r^2 and is proportional to the product of the charges. The constant kk is fixed by experiment; in SI units it is 1/(4πε0)1/(4\pi\varepsilon_0) with ε0≈8.85×10−12 F/m\varepsilon_0 \approx 8.85\times10^{-12}\ \text{F/m}. Because the force points along the line of centres, it is a central force, and the 1/r21/r^2 form is what makes the field of a point charge spherically symmetric.

  2. Superposition. With more than two charges, the force on any one charge is the vector sum of the pairwise Coulomb forces from all the others:

Fi=∑j≠i14πε0 qiqjrij2 r^ij\mathbf{F}_i = \sum_{j\neq i} \frac{1}{4\pi\varepsilon_0}\,\frac{q_i q_j}{r_{ij}^2}\,\hat{\mathbf{r}}_{ij}

There is no “three-body” force: every pair interacts independently. This linearity is why the field (Section 2) and the potential add so simply.

  1. Field lines of a point charge. The field of a single charge is radial: straight lines pointing away from a positive charge and toward a negative one, with density falling as 1/r21/r^2.
+ r E ∝ 1/r², radial; lines point outward for +q
  1. Authoritative source. Coulomb’s law is the starting point of electrostatics; see Griffiths, Introduction to Electrodynamics, Ch. 2, and Ellingson, Electrostatics.
Example problem

Two point charges, q1=+3.0 μCq_1 = +3.0\,\mu\text{C} and q2=−5.0 μCq_2 = -5.0\,\mu\text{C}, are 0.40 m0.40\,\text{m} apart in vacuum. Find the magnitude and direction of the force on q1q_1.

Solution F=14πε0 ∣q1q2∣r2=(8.99×109) (3.0×10−6)(5.0×10−6)(0.40)2=(8.99×109) 1.5×10−110.16≈0.84 NF = \frac{1}{4\pi\varepsilon_0}\,\frac{|q_1 q_2|}{r^2} = (8.99\times10^9)\,\frac{(3.0\times10^{-6})(5.0\times10^{-6})}{(0.40)^2} = (8.99\times10^9)\,\frac{1.5\times10^{-11}}{0.16} \approx 0.84\ \text{N}

Because the charges have opposite signs the force is attractive: the force on q1q_1 points toward q2q_2.

2. Electric field and electric potential

Definition. The electric field is the force per unit test charge:

E=Fq0\mathbf{E} = \frac{\mathbf{F}}{q_0}

For a point charge qq at the origin, E=14πε0qr2r^\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}. Because ∇×E=0\nabla\times\mathbf{E}=0, the electrostatic field is conservative and derives from a scalar potential VV:

E=−∇V,V(r)=−∫refrE⋅dℓ\mathbf{E} = -\nabla V,\qquad V(\mathbf{r}) = -\int_{\text{ref}}^{\mathbf{r}} \mathbf{E}\cdot d\boldsymbol{\ell}

The potential difference between two points is the work per unit charge, Va−Vb=−∫baE⋅dℓV_a - V_b = -\int_b^a \mathbf{E}\cdot d\boldsymbol{\ell}. For a point charge (with V→0V\to 0 at infinity), V=14πε0qrV = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}.

Details

The idea, step by step.

  1. Field = force per charge. Instead of computing the force between every pair of charges, we let the source charges create a field E\mathbf{E} at every point, and a test charge q0q_0 placed there simply feels F=q0E\mathbf{F} = q_0 \mathbf{E}. The field decouples the source from the test charge and, by superposition, is the vector sum of the fields of all the sources.

  2. A conservative field has a potential. In electrostatics ∇×E=0\nabla\times\mathbf{E}=0, so the line integral of E\mathbf{E} between two points is path-independent. That lets us define a single number at each point — the potential VV — whose gradient (with a minus sign) is the field: E=−∇V\mathbf{E} = -\nabla V. The minus sign means the field points “downhill,” in the direction of decreasing potential, just as a ball rolls down a slope.

  3. Equipotential surfaces. Points at the same potential form equipotential surfaces. Because E=−∇V\mathbf{E} = -\nabla V is perpendicular to surfaces of constant VV, field lines always cross equipotentials at right angles. Near a point charge the equipotentials are concentric spheres (V∝1/rV \propto 1/r); for a dipole they are closed loops around each charge joined by a zero-potential plane between them.

+ V = const E E ⟂ equipotentials; V ∝ 1/r

  1. The potential is a bookkeeping shortcut. For many charges the potential is just a sum of scalars, V(r)=14πε0∑iqi/riV(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_i q_i/r_i, which is far easier to add than vectors. Once you have VV, the field is its gradient. This “sum scalars, then differentiate” strategy is the workhorse of electrostatics.

  2. Authoritative source. See Griffiths, Introduction to Electrodynamics, Ch. 2, and OpenStax, University Physics Vol. 2, Ch. 7 (Electric Potential).

Example problem

A point charge q=+2.0 nCq = +2.0\,\text{nC} sits at the origin. Find the potential at r=0.10 mr = 0.10\,\text{m} and the electric field there (magnitude and direction).

Solution V=14πε0qr=(8.99×109) 2.0×10−90.10≈180 VV = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = (8.99\times10^9)\,\frac{2.0\times10^{-9}}{0.10} \approx 180\ \text{V} E=14πε0qr2=(8.99×109) 2.0×10−9(0.10)2≈180 V/mE = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} = (8.99\times10^9)\,\frac{2.0\times10^{-9}}{(0.10)^2} \approx 180\ \text{V/m}

The field points radially outward (positive charge). Notice E=V/rE = V/r here, which holds for a point charge because V∝1/rV \propto 1/r.

Interactive: drag the charges

Drag the charges and switch between the electric field (arrows, coloured by strength) and the potential (colour map with equipotential contours). Watch how the field is always perpendicular to the equipotentials, and how a positive and a negative charge together make a dipole pattern.

3. Gauss’s law

Definition. The total electric flux through any closed surface equals the enclosed charge divided by ε0\varepsilon_0:

∮SE⋅dA=Qencε0,∇⋅E=ρε0\oint_S \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0},\qquad \nabla\cdot\mathbf{E} = \frac{\rho}{\varepsilon_0}

Gauss’s law is one of Maxwell’s equations — the integral form of the divergence of the field. It is exact for any closed surface, but it is a calculation tool only when the charge distribution has enough symmetry.

Details

The idea, step by step.

  1. Flux counts the net field leaving a surface. The flux ΦE=∮E⋅dA\Phi_E = \oint \mathbf{E}\cdot d\mathbf{A} sums the outward component of the field over a closed surface. Field lines entering contribute negative flux, lines leaving contribute positive flux. Gauss’s law says the net flux depends only on how much charge is inside.

  2. External charge contributes zero net flux. A charge outside the surface sends field lines in and out that cancel exactly, so it adds nothing to the net flux. Only the enclosed charge QencQ_{\text{enc}} matters. This is the deep content of the law: the net flux is a “charge counter.”

  3. Symmetry turns it into a calculator. If the field has a known direction and constant magnitude over a cleverly chosen Gaussian surface, the integral reduces to E×(area)E \times (\text{area}):

    • Spherical symmetry (point charge, charged sphere): a concentric sphere gives E(4πr2)=Qenc/ε0E(4\pi r^2) = Q_{\text{enc}}/\varepsilon_0, recovering Coulomb’s law.
    • Cylindrical symmetry (infinite line charge): a coaxial cylinder gives E(2πrℓ)=λℓ/ε0E(2\pi r \ell) = \lambda \ell/\varepsilon_0, so E=λ/(2πε0r)E = \lambda/(2\pi\varepsilon_0 r).
    • Planar symmetry (infinite sheet): a “pillbox” straddling the sheet gives E(2A)=σA/ε0E(2A) = \sigma A/\varepsilon_0, so E=σ/(2ε0)E = \sigma/(2\varepsilon_0), independent of distance.
Q Gaussian surface S ∮ E·dA = Q_enc/ε₀ (all lines leave; net flux ∝ enclosed charge)
  1. Authoritative source. Gauss’s law is the first of Maxwell’s equations; see Maxwell (1865) and Griffiths, Introduction to Electrodynamics, Ch. 1–2.
Example problem

A long straight wire carries a uniform linear charge density λ=5.0 μC/m\lambda = 5.0\,\mu\text{C/m}. Use Gauss’s law to find the electric field at a distance r=0.05 mr = 0.05\,\text{m} from the wire.

Solution

Choose a coaxial cylindrical Gaussian surface of radius rr and length ℓ\ell. By symmetry E\mathbf{E} is radial and constant on the curved wall, and there is no flux through the end caps. The enclosed charge is Qenc=λℓQ_{\text{enc}} = \lambda \ell:

E(2πrℓ)=λℓε0⇒E=λ2πε0r=5.0×10−62π(8.85×10−12)(0.05)≈1.8×104 V/mE(2\pi r \ell) = \frac{\lambda \ell}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{5.0\times10^{-6}}{2\pi(8.85\times10^{-12})(0.05)} \approx 1.8\times10^{4}\ \text{V/m}

The field points radially outward (positive line charge) and falls as 1/r1/r.

4. Conductors in electrostatic equilibrium and shielding

Definition. In a conductor in electrostatic equilibrium: (i) the electric field inside the conducting material is zero; (ii) any net charge resides on the surface; (iii) the conductor is an equipotential; (iv) the field just outside the surface is perpendicular to it, with magnitude E=σs/ε0E = \sigma_s/\varepsilon_0 (where σs\sigma_s is the local surface charge density). A closed conducting shell shields its interior from external static fields — the Faraday cage.

Details

The idea, step by step.

  1. Free charges move until the field vanishes. A metal contains mobile electrons. If any field existed inside the metal, those electrons would feel a force and drift. They keep moving until the field they create exactly cancels the applied one, so the net field inside the conducting material is zero. Equilibrium is reached in a fraction of a nanosecond.

  2. Charge is pushed to the surface. Because the interior field is zero, Gauss’s law applied to a surface just inside the boundary gives zero enclosed charge: all excess charge must sit on the surface. The surface charge arranges itself so that the field it produces, added to any external field, is zero inside and perpendicular to the surface outside.

  3. Shielding (the Faraday cage). Place a closed conductor in an external static field. The external field polarizes the conductor: charge accumulates on the two faces, and the field from that induced charge cancels the external field everywhere inside the cavity. The interior is field-free regardless of the external field’s strength or shape. This is why a car body protects its occupants from a lightning strike, why a metal box protects sensitive electronics from radio interference, and why the inside of a charged hollow conductor is at a constant potential.

− − − + + + E = 0 external field is cancelled inside the closed conductor (Faraday cage)
  1. Authoritative source. See Griffiths, Introduction to Electrodynamics, Ch. 2 (conductors), and OpenStax, University Physics Vol. 2, Ch. 6 (Conductors in Electrostatic Equilibrium).
Example problem

A hollow conducting sphere of radius R=0.10 mR = 0.10\,\text{m} carries a net charge Q=+6.0 μCQ = +6.0\,\mu\text{C}. Find the electric field (a) at the centre, (b) inside the metal, (c) just outside the surface, and (d) at r=0.30 mr = 0.30\,\text{m}.

Solution

(a) At the centre and (b) anywhere inside the metal, the field is zero — the conductor is in electrostatic equilibrium.

(c) Just outside the surface the charge is spread uniformly over the sphere, so the field is that of a point charge at r=Rr = R:

E=14πε0QR2=(8.99×109) 6.0×10−6(0.10)2≈5.4×105 V/mE = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2} = (8.99\times10^9)\,\frac{6.0\times10^{-6}}{(0.10)^2} \approx 5.4\times10^{5}\ \text{V/m}

(d) Outside, the sphere behaves as a point charge at the centre:

E=14πε0Qr2=(8.99×109) 6.0×10−6(0.30)2≈6.0×104 V/mE = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} = (8.99\times10^9)\,\frac{6.0\times10^{-6}}{(0.30)^2} \approx 6.0\times10^{4}\ \text{V/m}

The field is zero inside, jumps to its maximum at the surface, and then falls as 1/r21/r^2.

5. Capacitance and capacitors

Definition. The capacitance of a two-conductor system is the charge stored per unit potential difference:

C=QVC = \frac{Q}{V}

For a parallel-plate capacitor in vacuum (plate area AA, separation dd, ignoring edge effects),

C=ε0AdC = \frac{\varepsilon_0 A}{d}

The energy stored is U=12CV2=12QV=Q2/(2C)U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}QV = Q^2/(2C), and it resides in the electric field between the plates.

Details

The idea, step by step.

  1. Capacitance is a geometric property. A capacitor is two conductors holding equal and opposite charge ±Q\pm Q at a potential difference VV. The ratio C=Q/VC = Q/V depends only on the geometry (and the vacuum permittivity here), not on how much charge you put on it. Bigger plates and smaller gaps store more charge per volt.

  2. Deriving the parallel-plate value. By Gauss’s law (Section 3), a plate with surface charge density σ=Q/A\sigma = Q/A produces a field E=σ/(2ε0)E = \sigma/(2\varepsilon_0); the two plates’ fields add between them and cancel outside, giving a uniform field E=σ/ε0=Q/(ε0A)E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A) between the plates. The potential difference is V=Ed=Qd/(ε0A)V = E d = Qd/(\varepsilon_0 A), so

C=QV=ε0AdC = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}
  1. Energy lives in the field. Charging a capacitor stores energy U=12CV2U = \tfrac{1}{2}CV^2. This energy is not “in the plates” — it is in the electric field between them, with density u=12ε0E2u = \tfrac{1}{2}\varepsilon_0 E^2 (the electric half of the energy density in Poynting’s theorem). Integrating uu over the volume between the plates reproduces U=12CV2U = \tfrac{1}{2}CV^2.
++++ −−−− d E = V/d (uniform)
  1. Authoritative source. See Griffiths, Introduction to Electrodynamics, Ch. 4 (capacitance), and OpenStax, University Physics Vol. 2, Ch. 8 (Capacitors and Capacitance).
Example problem

A parallel-plate capacitor has square plates of side a=5.0 cma = 5.0\,\text{cm} separated by d=1.0 mmd = 1.0\,\text{mm} in vacuum. Find its capacitance, the charge on each plate when it is connected to a 12 V12\,\text{V} source, and the energy stored.

Solution

The plate area is A=a2=(0.05)2=2.5×10−3 m2A = a^2 = (0.05)^2 = 2.5\times10^{-3}\ \text{m}^2. The capacitance is

C=ε0Ad=(8.85×10−12)(2.5×10−3)1.0×10−3≈2.2×10−11 F≈22 pFC = \frac{\varepsilon_0 A}{d} = \frac{(8.85\times10^{-12})(2.5\times10^{-3})}{1.0\times10^{-3}} \approx 2.2\times10^{-11}\ \text{F} \approx 22\ \text{pF}

The charge is Q=CV=(2.2×10−11)(12)≈2.7×10−10 C≈0.27 nCQ = CV = (2.2\times10^{-11})(12) \approx 2.7\times10^{-10}\ \text{C} \approx 0.27\ \text{nC}. The stored energy is

U=12CV2=12(2.2×10−11)(12)2≈1.6×10−9 JU = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(2.2\times10^{-11})(12)^2 \approx 1.6\times10^{-9}\ \text{J}

A small gap and large area give a usable capacitance; inserting a dielectric would multiply CC by εr\varepsilon_r (see Electromagnetic Media).

Part of Electromagnetism.

Quizz

Test yourself

Problem. Two identical conducting spheres, each of radius R=0.05 mR = 0.05\,\text{m}, carry charges Q1=+4.0 μCQ_1 = +4.0\,\mu\text{C} and Q2=−4.0 μCQ_2 = -4.0\,\mu\text{C} and are held d=0.50 md = 0.50\,\text{m} apart (centre to centre) in vacuum.

  1. What is the magnitude of the electrostatic force between the spheres?
  2. What is the electric potential at the midpoint between the two centres (take V→0V \to 0 at infinity)?
  3. If the two spheres are now connected by a thin conducting wire, what charge does each carry afterwards?
Solution
  1. Because d≫Rd \gg R, each sphere’s charge is well approximated as a point charge at its centre. Coulomb’s law gives
F=14πε0 ∣Q1Q2∣d2=(8.99×109) (4.0×10−6)2(0.50)2≈0.58 NF = \frac{1}{4\pi\varepsilon_0}\,\frac{|Q_1 Q_2|}{d^2} = (8.99\times10^9)\,\frac{(4.0\times10^{-6})^2}{(0.50)^2} \approx 0.58\ \text{N}

The force is attractive (opposite signs).

  1. The potential is a scalar sum. At the midpoint each sphere is a distance d/2=0.25 md/2 = 0.25\,\text{m} away, so
V=14πε0(Q1d/2+Q2d/2)=14πε0 Q1+Q2d/2=0V = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_1}{d/2} + \frac{Q_2}{d/2}\right) = \frac{1}{4\pi\varepsilon_0}\,\frac{Q_1 + Q_2}{d/2} = 0

The equal and opposite charges cancel exactly — the midpoint lies on the zero-potential plane of the dipole-like pair.

  1. Connecting the spheres with a wire lets charge flow until both are at the same potential. Identical spheres at the same potential carry the same charge, so the total charge is shared equally:
Qeach=Q1+Q22=+4.0−4.02=0Q_{\text{each}} = \frac{Q_1 + Q_2}{2} = \frac{+4.0 - 4.0}{2} = 0

Both spheres end up neutral. (For unequal spheres the final charges would not be equal — they would be proportional to the radii, since V=Q/(4πε0R)V = Q/(4\pi\varepsilon_0 R).)