From Coulomb’s law between two point charges, to the field and potential of a dipole, to continuous charge on a ring and a disk. Every picture is live: drag the charges and move the sliders.
The basics from the lecture notes, then a playground where you can try them.
Charge comes in whole multiples of the elementary charge, so it is quantized:
An electron carries −e, a proton +e, and a neutron 0. Charge is conserved: it can move, but it is never created or destroyed. The rate at which it flows past a point is the current, i = dq/dt.
Conductors let charge move freely, insulators do not, and semiconductors sit in between.
F is the force on q₂, and r̂ points from q₁ toward q₂. Like signs push apart (along r̂); opposite signs pull together (along −r̂). The force on q₁ is −F, by Newton’s third law. With several charges, the forces add as vectors: Fnet = F₁₂ + F₁₃ + ⋯.
ε₀ = 8.85 × 10⁻¹² C²/(N·m²) is the permittivity of free space.
Here r̂ points away from the other mass, so the minus sign makes gravity pull the two together. The 1/r² comes from geometry: whatever spreads out from a point is shared over a sphere of area 4πr².
Near Earth’s surface r ≈ RE, so Fg = G mEm/r² ≈ m(G mE/RE²) = mg, with g ≈ 9.8 m/s².
A uniformly charged spherical shell acts on a charge outside it as if all its charge sat at the center. On a charge inside the shell, the net force is zero.
These follow neatly from Gauss’s law, which comes later in the course.
How do two charges “know” about each other? Each one fills the space around it with a field, and the other charge feels that field. E is the force per unit test charge q₀, in N/C.
Faraday’s field lines have E as their tangent; they start on positive charges and end on negative ones. Fields add as vectors: E = E₁ + E₂ + ⋯.
Measured from V(∞) = 0. Potentials add as plain numbers, which is much easier than adding vectors, and the field follows as E = −∇V.
The potential energy of a pair is U = k q₁q₂/r: positive for like signs, negative for opposite signs. For more charges, add it up pair by pair.
Both forces are proportional to a product and fall off as 1/r². But charge has two signs, so the electric force can attract or repel, while mass is never negative, so gravity only attracts. And between particles, the electric force is overwhelmingly stronger.
Pair
Drag the charges and the test charge q₀. Arrows on each charge are the Coulomb forces from every other charge (thin, colored by the charge exerting them) and their vector sum (thick). The blue arrow at q₀ is E there. Turn on the potential to see equipotential curves cross the field lines at right angles.
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At the test charge q₀ = 1 nC
Two equal and opposite charges ±q a distance d apart. Molecules such as water behave like this: the oxygen side is slightly negative and the hydrogen side slightly positive.
On the axis, at distance z from the center, the two fields point opposite ways:
For z ≫ d the last factor is close to 1. With the dipole moment p = qd, pointing from − to +:
Seen from far away, the + and − almost cancel, so the field falls as 1/z³, faster than the 1/z² of a single charge. On the log–log plot that is a slope of −3 instead of −2. The potential falls as 1/r², one power slower, because E = −∂V/∂z takes one more power of z.
Add up the vectors dE, or add up the scalars dV and take the gradient. On the axis, symmetry makes the vector route easy. Move P off the axis and the sideways parts stop cancelling, while the potential is still a plain scalar sum.
Both curves follow the same straight line through P. The dot is P.
The tangent to V at P has slope −Es, the field component along the line. That is E = −∇V, one direction at a time.
Going back, V at P is the shaded area under Es from P out to infinity: V = −∫∞P E·ds, with V(∞) = 0.
A slice through the axis (the x–z plane). The shading and magenta curves are V; the blue arrows are E = −∇V, always pointing straight downhill across the equipotentials. Click to move P there, or switch to launching test charges and watch where the field takes them.
Look at the ring’s center: the equipotentials cross in an X. That is a saddle point. The field there is zero, yet V rises toward the wire in the plane and falls along the axis.
Computed from exact ring solutions with complete elliptic integrals (via the arithmetic–geometric mean). The disk is built from many such rings.
A test charge of mass m = 1 and charge ±1 starts at rest and moves under F = QE, using the exact field of the ring or disk. Nudge it off the center to test the stability there, or release it anywhere.
Launch point
You can also set it by clicking the map above with “Sets the launch point” selected.
Release a test charge
Following the lecture notes. Boxed lines are the results.
Coulomb’s law for each piece, then symmetry
Potential of each piece, then differentiate
Numbers in the demo use ke = 1/(4πε₀) = 1. For the ring, q = 1. For the disk, σ/(2ε₀) = 1, so the total charge q = σπR² grows with R. Disk formulas are written for z ≥ 0 as in the notes; below the disk V is even in z and Ez is odd. Off the axis, the disk is summed numerically from 400 exact rings.