One idea underneath everything: a charge fills the space around it with a field. Put a second charge there and it feels a push. Start with the force between two charges, then stop tracking charges and start tracking the field they build — drag charges and watch the lines and equipotentials snap into place. Finish with potential, the field's energy bookkeeping, and what happens inside a conductor.
§01
Charge & Coulomb's law
Two point charges attract or repel along the line joining them, with a force that dies off as the square of their separation. Like signs push apart, opposite signs pull together — and by Newton's third law the two forces are always equal and opposite, whatever the sizes of the charges.
COULOMB BENCH
separation r{{ crlab }} cm
|q₁| magnitude
|q₂| magnitude
F =k q₁ q₂r²={{ cForce }} N
{{ cVerdIcon }}{{ cVerdict }}
With k = 9×10⁹ N·m²/C² and charges in microcoulombs (µC). Halve r and the force quadruples; the two arrows are always the same length — Newton's third law holds even for a 1 µC and a 5 µC charge.
SUPERPOSITION · NET FORCE ON EACH CHARGE
charges
Drag any charge · tap it to flip its sign|each orange arrow is the vector sum of the Coulomb forces from all the others (number = length relative to the largest)
§02
The electric field & its lines
Instead of asking "what force on this charge?", ask what the region itself is doing: the electric field E is the force a unit positive test charge would feel at every point. Draw it as field lines — leaving +, entering −, closer together where the field is stronger — or as equipotentials, the contours of constant potential that always cross the lines at right angles. Drag the charges and watch the whole picture rebuild.
FIELD BENCH · DRAG THE CHARGES
at probe ◇
V = —
|E| = —
dir = —
presets
2 charges
key:● positive · E points away● negative · E points inheat map = potential (red +, blue −)drag ◇ to probe the field
§03
The electric dipole
A pair of equal, opposite charges a small distance 2a apart is a dipole, described by one vector — the dipole moment p = q·(2a), pointing from − to +. Drop it in a uniform field and the two forces cancel but leave a torque that twists it into line with the field.
TORQUE IN A UNIFORM FIELD
angle θ between p and E{{ dipDeg }}°
Right now
τ = pE sinθ ={{ dipTorque }}
U = −pE cosθ ={{ dipEnergy }}
Values in units of pE. Torque is largest at 90° and vanishes when p lines up (0°, stable) or points backwards (180°, unstable). Energy is lowest — most stable — when aligned.
Its own field falls off as 1/r³
on the axis
E =2kpr³
on the ⟂ bisector
E =kpr³
Twice as strong on the axis as on the bisector, at the same distance — and both drop off faster than a single charge because the + and − nearly cancel far away.
THE DIPOLE'S OWN FIELD · AXIAL vs EQUATORIAL
probe distance r (equal on both){{ dipProbeR }} px
on the axial line — E = 2kp/r³
E = {{ dipEaxLab }} · V = +{{ dipVaxLab }}
on the equatorial plane — E = kp/r³
E = {{ dipEeqLab }} · V = {{ dipVeqLab }}
ratio axial : equatorial = {{ dipRatioLab }} : 1
At equal distance the axial field is about twice the equatorial (→ exactly 2 : 1 for r ≫ a). Potential is positive on the axis but zero on the equatorial plane — every point there is equidistant from + and −.
Let go — what's the motion?hit ▶ release & watch above and set it swinging
Uniform field · pure rotation
The two forces are equal and opposite, so the net force is zero — the centre never moves. Released from rest, the dipole only rotates, swinging back and forth about the aligned position like a pendulum.
Small tilt · SHM
For a small angle, sinθ ≈ θ, so the restoring torque τ = −pEθ is Hooke-like — simple harmonic motion of period
T = 2πIpE
Big tilt → still oscillates, but slower (anharmonic). At 180° the balance is unstable — the tiniest nudge and it swings all the way round.
Non-uniform field · it drifts
Now the ends sit in different field strengths, so the forces no longer cancel: a net force survives and pulls the dipole toward the stronger field — it rotates and translates.
§04
Potential & the field as a slope
Potential V is the work done per unit charge to bring a test charge from infinity to a point — pure energy bookkeeping, a single number at every location. Its real power: the field is just how steeply V changes. E = −dV/dr — the field points downhill, toward lower potential, and its size is the slope. Drag the point and read the slope straight off the curve.
POTENTIAL ALONG A LINE
drag the point along xx = {{ potXlab }}
potential here
V = {{ potVlab }}
field = −slope
E = {{ potElab }}
{{ potNote }}
Vpoint =
kqr(per charge, summed)
RememberPotential is a scalar — just add the kq/r from each charge with sign, no angles. Along an equipotential (constant V) the field does zero work, so moving a charge there is free — which is exactly why equipotentials sit at right angles to the field lines.
§05
Conductors in a field
Drop a conductor into a field and its free electrons rush around until they've killed the field inside. The charge that piles up on the surface arranges itself so that, in equilibrium, E = 0 everywhere inside, the whole body is one equipotential, and every field line outside meets the surface dead perpendicular. Start with a neutral sphere and switch the field on — or hollow it out and drag a charge around a cavity.
{{ condTitle }}
{{ fieldOnHint }}
field strength E₀{{ condElab }}
cavities
Drag the charge around inside a cavity — the wall's induced charge crowds toward it (watch the dots) and the interior lines always strike the wall square-on, yet the outer surface stays uniform so the field outside never flinches. With two cavities, neither charge feels the other: that is electrostatic shielding.
1 · Field is zero inside
Free charges shift until the interior field cancels completely. No field means no potential difference — the whole conductor is one equipotential, surface included.
2 · Field ⟂ at the surface
Any sideways component would push surface charge along until it's gone, so just outside the surface E = σ/ε₀, pointing straight out.
3 · Shielding — the Faraday cage
A hollow inside a conductor is completely screened: no matter how strong the outside field, the cavity stays field-free. That's why you're safe in a car during lightning.
＋
Worked example — the null point
Charges +q and +4q are fixed a distance L apart. Where on the line joining them is the electric field zero?
Set it up
The null point sits between the two (both fields point outward, so they can only cancel in the middle region). Let it be a distance x from +q, hence L−x from +4q. Set the two field magnitudes equal.
Solve
kqx²=k(4q)(L−x)²
Take square roots: (L−x) = 2x, so x = L/3 — a third of the way from the smaller charge.
Sanity check
Closer to the weaker charge — as it must be, since you need to be nearer the small charge for its field to keep up with the big one. Note V is never zero here: two positive charges give V > 0 everywhere.
✎
Warm-up
quick checks · tap a card to reveal · {{ warmChev }} to fold