Electromagnetism · Chapter 13

Electric Charges and Fields

Like charges repel, unlike charges attract — but the field concept turns this force-at-a-distance into something local and geometric.

PrerequisitesNewton′sLaws⋅Vectors⋅Basiccalculus(helpfulNewton's Laws \cdot Vectors \cdot Basic calculus (helpful
Learning Goals
  • Use Coulomb's law to compute electrostatic forces between point charges.
  • Explain why the electric field is a local vector description of force-at-a-distance.
  • Apply superposition to forces, fields, and potentials from multiple charges.
  • Distinguish electric potential energy from electric potential.
  • Use Gauss's law qualitatively and recognize when symmetry makes it powerful.

13.1 Electric Charge

Electric charge is a fundamental property of matter, carried by protons (+e+e) and electrons (−e-e), where e=1.602×10−19 Ce = 1.602 \times 10^{-19}\,\mathrm{C} is the elementary charge. Charge comes in two signs; like signs repel and unlike signs attract. Charge is quantized (always a multiple of ee) andconserved — the total charge of an isolated system never changes.

Definition 13.1 — Coulomb's Law
The electrostatic force between two point charges q1q_1 and q2q_2 separated by distance rr:F=k∣q1q2∣r2F = \frac{k|q_1q_2|}{r^2}     k=8.99×109 N m2/C2k = 8.99 \times 10^9\,\mathrm{N\,m^2/C^2}The force is along the line connecting the charges: repulsive if same sign, attractive if opposite. It obeys a 1/r21/r^2 inverse-square law — the same mathematical form as gravity, but enormously stronger (about 103610^{36} times for electrons vs. gravity).

Coulomb's constant k=1/(4πε0)k = 1/(4\pi\varepsilon_0), where ε0=8.85×10−12 C2/(N m2)\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{C^2/(N\,m^2)} is the permittivity of free space. For multiple charges, forces add as vectors (superposition principle).

Example 13.1 — Force Between Charges

Two charges, q1=+3 μCq_1 = +3\,\mu\mathrm{C} and q2=−2 μCq_2 = -2\,\mu\mathrm{C}, are 0.15 m0.15\,\mathrm{m} apart. Find the force between them.

Apply Coulomb's law:F=k∣q1q2∣r2=(8.99×109)(3×10−6)(2×10−6)(0.15)2F = \frac{k|q_1q_2|}{r^2} = \frac{(8.99\times10^9)(3\times10^{-6})(2\times10^{-6})}{(0.15)^2}
Calculate:F=(8.99×109)(6×10−12)0.0225=53.94×10−30.0225=2.40 NF = \frac{(8.99\times10^9)(6\times10^{-12})}{0.0225} = \frac{53.94\times10^{-3}}{0.0225} = 2.40\,\mathrm{N}
Direction:Attractive — unlike charges. The force pulls them toward each other.

13.2 The Electric Field

Rather than thinking about force-at-a-distance, Faraday1830s · The field idea starts as a pictureMichael Faraday had little formal mathematics, but his line-of-force diagrams were physically sharp. Maxwell later translated those pictures into equations. introduced the electric field: a charge creates a field everywhere in space, and other charges respond to that field locally. The field E\mathbf{E} at a point is the force per unit positive test charge placed there:

E=Fq0∣E∣=kqr2(point charge)\mathbf{E} = \frac{\mathbf{F}}{q_0} \qquad |\mathbf{E}| = \frac{kq}{r^2} \qquad \text{(point charge)}(13.1)

The electric field is a vector field — it has a direction (away from + charges, toward − charges) and a magnitude at every point in space. For multiple charges, fields add as vectors.

Definition 13.2 — Electric Field Lines
Field lines are a visual tool for representing electric fields:
  • Field lines originate on positive charges and terminate on negative charges.
  • The direction of the field at any point is tangent to the field line.
  • The magnitude is proportional to the density of field lines.
  • Field lines never cross (the field has a unique direction at each point).
Figure 13.1. Interactive electric field simulation. Start with one positive charge and one negative charge, then add a second positive charge to see superposition. Field lines show direction by their tangent and relative strength by their density; drag charges around and watch where lines crowd together or cancel.

13.3 Electric Potential Energy and Potential

Just as gravitational force has an associated potential energy U=mghU = mgh, the electric force is conservative and has a potential energy. For two point charges:

U=kq1q2rU = \frac{kq_1q_2}{r}(13.2)

The electric potential V (not to be confused with voltage) is potential energy per unit charge:

V=Uq0=kqr(point charge)ΔV=−∫E⋅dlV = \frac{U}{q_0} = \frac{kq}{r} \qquad \text{(point charge)} \qquad \Delta V = -\int \mathbf{E}\cdot d\mathbf{l}(13.3)

Potential is a scalar — it's easier to work with than the vector field. The field points from high to low potential: E=−∇V\mathbf{E} = -\nabla V. Equipotential surfaces (surfaces of constant VV) are always perpendicular to field lines.

Theorem 13.1 — Gauss's Law
ThetotalelectricfluxthroughanyclosedsurfaceequalstheenclosedchargedividedbyεThe total electric flux through any closed surface equals the enclosed charge divided by \varepsilon₀:ΦE=∮E⋅dA=Qencε0\Phi_E = \oint \mathbf{E}\cdot d\mathbf{A} = \frac{Q_\mathrm{enc}}{\varepsilon_0}This is equivalent to Coulomb's law for static charges but is far more powerful — it can determine the field from highly symmetric charge distributions (sphere, cylinder, plane) with a single integral. It is one of Maxwell's four equations.
Example 13.2 — Field from a Charged Sphere

A solid metal sphere of radius R=0.1 mR = 0.1\,\mathrm{m} carries charge Q=5 μCQ = 5\,\mu\mathrm{C}. Find EE at r=0.3 mr = 0.3\,\mathrm{m} from the center.

By Gauss's law:For r>Rr > R, the sphere looks like a point charge: E=kQ/r2E = kQ/r^2.
Calculate:E=(8.99×109)(5×10−6)(0.3)2=449500.09=4.99×105 N/C≈500 kN/CE = \frac{(8.99\times10^9)(5\times10^{-6})}{(0.3)^2} = \frac{44950}{0.09} = 4.99\times10^5\,\mathrm{N/C} \approx 500\,\mathrm{kN/C}
Inside:For r<Rr < R (inside a conductor): E=0E = 0. All charge resides on the surface.
Definition 13.3 — Common Traps
  • Force and field are not the same: Eisforceperunittestcharge;F=qEdependsonthechargeplacedinthefieldE is force per unit test charge; F = qE depends on the charge placed in the field
  • Potential is a scalar: add potentials algebraically, but add electric fields as vectors.
  • Signs matter for potential energy: opposite charges have negative U and are bound by attraction.
  • Field lines are a model: they visualize direction and density, not physical strings in space.
  • Gauss's law is always true: it is only easy to use when symmetry makes E constant on a chosen surface.
Exercises — 13.1–13.3 Electric Charges and Fields
1.
Twopointchargeseachof+1.0\muCare0.1mapart.FindthemagnitudeoftheelectrostaticTwo point charges each of +1.0 \muC are 0.1 m apart. Find the magnitude of the electrostatic force.
N
Straightforward
2.
Findtheelectricfield1.0mfroma+1.0\muCpointchargeFind the electric field 1.0 m from a +1.0 \muC point charge.
N/C
Straightforward
3.
Findtheelectricpotentialatadistanceof0.10mfroma+1.0\muCpointchargeFind the electric potential at a distance of 0.10 m from a +1.0 \muC point charge.
V
Intermediate
4.
Chargesq1=+3.0\muCandq2=−2.0\muCareseparatedby15cm.FindthemagnitudeoftheelCharges q_{1} = +3.0 \muC and q_{2} = -2.0 \muC are separated by 15 cm. Find the magnitude of the electrostatic force.
N
Intermediate
5.Two positive charges, q and 4q, are placed d apart. Find the location between them where the electric field is zero.
Challenging
Key Takeaways
  • Coulomb′slaw:F=kq1q2/r2—inversesquare,likegravity,but1036×strongerCoulomb's law: F = kq_{1}q_{2}/r^{2} — inverse square, like gravity, but 10^{36}\times stronger.
  • ElectricfieldE=F/q0:forceperunitpositivecharge,avectorfieldpointingawayfromElectric field E = F/q_{0}: force per unit positive charge, a vector field pointing away from +, toward−toward -
  • Superposition: fields and forces from multiple charges add vectorially.
  • ElectricpotentialV=kq/risascalar;E=−\nablaVElectric potential V = kq/r is a scalar; E = -\nablaV.
  • Gauss′slaw\ointE\cdotdA=Qenc/ε0isthemostpowerfultoolforsymmetricchargedistributionsGauss's law \ointE\cdotdA = Q_{enc}/\varepsilon_{0} is the most powerful tool for symmetric charge distributions.