Electric Potential
Three topics that replace vector bookkeeping with a single scalar. Potential energy of a configuration, signed and summed over distinct pairs, electric potential as energy per unit charge with the field recovered as its slope, and conservation of electric energy, which answers questions about speed and direction without ever writing an acceleration.
AP exam 10-20%3 topics
Topics
Key forms For every problem in this unit
Energy of a pair
U = k q₁ q₂ / r, one power of r, and the charge SIGNS are kept. Unlike charges give a negative U
Force vs energy
F = k|q₁||q₂| / r² strips signs and goes as 1/r². U keeps signs and goes as 1/r. Different rules, one rung apart
Potential
V = U / q₀, energy per unit charge. Volts ARE joules per coulomb. V describes a place
Energy of a charge at a place
U = qV. V is the place, U is the charge at the place. A potential is never an energy until it is multiplied by a charge
Point charge
V = kq / r, with V taken as zero infinitely far away. Signed: a negative charge makes V negative around it
Superposition of V
V = k q₁/r₁ + k q₂/r₂ + …, plain signed addition. Potential is a SCALAR: no components, no angles, no partial cancellation
Work
field: W = −ΔU = −qΔV. External agent moving the charge slowly: W = +ΔU = +qΔV. Two ledgers, opposite signs
Assembly energy
U = sum over DISTINCT PAIRS of k q(i) q(j) / r(ij). Three charges give exactly three terms. Summing q(i)V(i) over the charges counts every pair twice, so halve it
Electron volt
1 eV = 1.6 × 10⁻¹⁹ J, the energy an elementary charge picks up crossing 1 V
Field from potential
Ex = −dV/dx, and in general E = −grad V. The FIELD is the SLOPE of V, so it reports how V changes, not what V is
Potential from field
ΔV = −∫ E · dl, path independent. Pick the easiest route between the endpoints; every other route gives the same number
Uniform field ONLY
E = V / d, the parallel-plate result. Do not export it to a point charge or to any field that varies with position
V = 0 and E = 0 are independent
midway between +q and −q: V = 0 with E strong. Midway between two +q: E = 0 with V = 4kq/d, nowhere near zero
Equipotentials
ΔV = 0 along one, so W = qΔV = 0 whatever the path length. Field lines pierce them at 90°, always
Energy conservation
ΔK + ΔU = 0, so ΔK = −qΔV = q(V start − V end). One line, no trajectory needed
Which way a charge goes
a POSITIVE charge accelerates toward lower V. A NEGATIVE charge accelerates toward HIGHER V. Both move toward lower U, because U = qV flips with the sign of q
Nonuniform field
the acceleration is not constant, so v² = v₀² + 2as does not apply. Energy does not care: use qΔV
Conductor in equilibrium
E = 0 inside, so V is CONSTANT inside, pinned at the surface value kQ/R. Flat is not the same claim as zero