Electric Charges, Fields, and Gauss's Law
Core concepts
Coulomb's law gives the force between point charges, (\vec F = kq_1q_2\hat r/r^2). The electric field is defined as force per unit positive test charge, (\vec E = \vec F/q_0), so a point charge produces (E = k|q|/r^2) pointing away from positive charge and toward negative charge. Superposition means you add fields as vectors, resolving into components before combining. Continuous distributions require integrating (d\vec E = k,dq,\hat r/r^2), exploiting symmetry to cancel components.
Gauss's law, (\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0), is always true but only useful for computing fields when symmetry is high: spherical, cylindrical (infinite line or cylinder), or planar (infinite sheet). For an infinite sheet the field is (\sigma/(2\varepsilon_0)); between two oppositely charged plates it is (\sigma/\varepsilon_0).
Traps
- Confusing the field of a conductor's surface with that of an infinite sheet: just outside a conductor, (E = \sigma/\varepsilon_0).
- Forgetting that (Q_{\text{enc}}) counts only charge inside the Gaussian surface; external charges contribute flux zero.
- Treating flux as a vector or giving it direction through a closed surface; net outward flux is a scalar signed by convention.
- Ignoring that field inside a spherical shell of charge is zero, but potential is not.
Worked example
A solid insulating sphere of radius (R) carries uniform charge density (\rho). For (r < R), choose a Gaussian sphere: (E(4\pi r^2) = \rho(\tfrac{4}{3}\pi r^3)/\varepsilon_0), giving (E = \rho r/(3\varepsilon_0)), linear in (r). Outside, (E = \rho R^3/(3\varepsilon_0 r^2)). Rule of thumb: pick a surface where (\vec E) is either parallel or perpendicular to (d\vec A) and constant in magnitude.