Coulomb force and electric field

Electric charge q is signed and measured in coulombs (C). In a vacuum, the SI permittivity ε₀ relates charge to the electrostatic field; write k=1/(4π ε₀). In engineering estimates, k≈8.99×10⁹ N·m²/C². Coulomb's force law models the interaction of stationary point charges separated by a nonzero distance, not a self-force at the source location.

The field produced by prescribed static sources is not altered in this ideal model by an infinitesimal test charge. A finite probe can perturb a source distribution; real current-carrying or moving systems require a broader electromagnetic description.

Place a source charge q₁ at position r₁ and a second stationary charge q₂ at r₂. The displacement from the source to the observation point is R=r₂−r₁, with R=|R| and unit vector R̂=R/R. The force ON charge 2 BY charge 1 is

If q₁q₂ is positive, force on 2 points from source 1 toward observation 2 (repulsion); if negative it points toward the source (attraction). Interchanging source and observation gives R→−R and F₁←₂=−F₂←₁. The vector formula stores the sign; do not attach a second ad hoc sign to R̂.

Choose a test charge qₜ at observation point r. Let Fₜ,ext be the electrostatic force exerted on it by OTHER charges. Formally the field at r is the small-test-charge limit of force divided by charge, with prescribed sources held fixed:

The limit is operational: it removes probe-induced disturbance. A negative qₜ experiences force opposite E; the field direction itself is defined by the force on a positive test charge. At the location of an ideal point source its own field formula diverges, so never evaluate that source's self-field by substituting r=r₁.

Insert the vector force of §01.2 into the definition of E and cancel qₜ. For a point source qᵢ located at rᵢ, and observation r distinct from every source position,

Superposition is vector addition. In Cartesian coordinates calculate Ex, Ey and Ez separately; adding field magnitudes generally fails. The formula assumes a fixed linear vacuum model. In a material, polarization can change the total source distribution and must be modeled explicitly.

Let q=+3.00 nC be at the origin and observe at r=(0.200 m,0,0). Here R=0.200 m and R̂=+x̂, so

For a −2.00 nC test charge at the same point, F=qₜE≈−1.35×10⁻⁶ x̂ N. The negative force sign follows solely from qₜ; the field did not reverse. The result scales as 1/R² and is undefined at R=0 in the point-source idealization.

Partition a finite distribution into small elements dq. A fixed observation r sees each source element at r′. Taking the continuum limit replaces the sum by an integral. Do not substitute a one-dimensional density into a volume element:

λ has C/m, σ has C/m², and ρ has C/m³; the primed coordinates label the SOURCE and the unprimed coordinates label the fixed observation point. The integral excludes or treats carefully singular source–observation coincidences; an ideal zero-thickness charged surface can have different one-sided limiting fields.

For a uniformly charged infinite line along the z axis, inspect the point (s,0,0), s>0. Source elements at +z′ and −z′ have equal x components and opposite z components. Hence Ez cancels and E points radially away from the line if λ>0 (toward it if λ<0). This cancellation follows by pairing integrand values; it is valid before integrating.

This method fails if the charge density is not symmetric: if λ(z′) differs from λ(−z′), the paired z components do not cancel. The full integral must then be retained.

After the odd component has vanished, perform the remaining convergent integral. For s>0,

The 1/s falloff differs from the point source's 1/R² because an infinite line contains arbitrarily remote contributing charges. The idealization fails near the finite wire's ends and at distances comparable to its finite radius; at s=0 the ideal infinitely thin line is singular.

Take +q at z=+a and −q at z=−a (q>0). On the positive z axis at z>a, the positive charge pushes toward +z while the negative charge pulls toward −z. Therefore

The exact expression is valid only on this axis segment and excludes z=a; the 1/z³ expression is the leading far-field term. Approaching a charge with the dipole approximation is invalid. The dipole vector p points from −q to +q; the full angular field will be derived later from the potential.

Field lines are tangent to E where E is nonzero; their arrow direction is the force direction for a POSITIVE test charge. Their drawn density is a visualization convention, not an additional equation or charge trajectory. A particle trajectory depends on inertia, initial velocity and any magnetic field. Never assume field lines can replace vector components or boundary conditions.

1. Prove F₁←₂=−F₂←₁ by exchanging r₁ and r₂ in the vector formula.
2. Compute E at r=(0.200 m,0,0) for q=−3.00 nC at the origin. What is the force on a positive +2.00 nC probe?
3. Two equal +1.00 nC charges sit at (−0.030 m,0) and (+0.030 m,0). Find the field at (0,+0.040 m); justify the vanished component.
4. For λ=+1.00 nC/m and s=0.020 m, compute the ideal-line field and check its units.
5. For +q at +a and −q at −a, derive the exact axial dipole expression by combining denominators and identify the condition for the far-field formula.
6. Give two reasons an electrostatic field line is generally not the trajectory of a released charged particle.

Next: Chapter 02 builds electric potential from the work done by this static field, and fixes the sign of the potential difference before introducing Gauss's law.