Calculating electric potential from a charge distribution

Electric potential V is potential energy U per unit test charge q: V = U/q, so U = qV. Its unit is the volt: 1 V = 1 J/C. In a fixed electrostatic field, a positive charge released from rest where E is nonzero initially accelerates toward lower V if only the electric force acts. It loses potential energy as it gains kinetic energy. This describes its initial motion, not every later part of its path.

For a point charge q, V = kq/r when zero potential is set at infinity. Here r is the distance from the charge, and k is about 8.99 × 10^9 N m²/C². Keep the sign of q: a negative source charge gives negative V with this reference. At r = 0 the point-charge formula is undefined.

For several fixed point charges, add their signed contributions: V = kΣ(qi/ri). Each ri is the distance from that charge to the point. At a square's center, all four distances are equal, so V = k(q1 + q2 + q3 + q4)/r. For square side a, r = a/√2. Convert charge to coulombs before calculating: 1 nC = 10^-9 C.

For a finite continuous distribution, replace the sum by V = k∫dq/r, with zero at infinity. The distance r runs from each charge element dq to the observation point. For a line with charge per length λ, dq = λ dl. At a ring's center every element is distance R away, so V = (k/R)∫dq = kQ/R, where Q is the ring's total signed charge.

For one fixed charge distribution, a new zero adds the same constant C everywhere: Vnew = Vold + C. Every difference VB - VA is unchanged. Choose a reference where potential is finite and defined, not at a point charge. For finite localized charges, zero at infinity is available. Ground can also be a reference. Choosing a reference does not mean physically grounding a conductor; a grounding connection can change its charge.

Each field component is minus the local slope of potential: Ex = -∂V/∂x, and likewise for y and z. Hold y and z fixed when finding the x slope. For V = A + bx, Ex = -b. The field points toward the steepest decrease of V. A spatial constant C has zero slope, so V and V + C give the same field everywhere. Since F = qE, the same charge with the same initial position and velocity has the same motion under this electric force.