Connect voltage, interaction energy and electric work

Compare locations using energy per charge

The positive source has a higher electric potential nearby than far away. At r = 0.20 m its potential is 89.9 V; at 0.40 m it is 44.95 V. A volt (V) is one joule per coulomb. Potential is a scalar: it has a signed value but no arrow direction. Electric field is a vector and has units N/C.

Predict whether interaction energy increases or decreases when q moves between these locations. Make one prediction for +2 nC and a separate one for −2 nC. Keeping q fixed during each comparison is essential.

From potential to energy

For this isolated point source, V = kQ/r, with zero potential chosen at infinite separation. Q is +2 nC, r is distance in meters, and k = 8.99 × 10⁹ N m²/C². The pair interaction energy is U = qV = kQq/r, in joules (J). Unlike field magnitude, potential falls as 1/r, so doubling separation halves it rather than quartering it.

For q = +2 nC, U changes from +0.1798 µJ at 0.20 m to +0.0899 µJ at 0.40 m. For q = −2 nC, it changes from −0.1798 µJ to −0.0899 µJ: it increases. A negative number becoming less negative is an increase. Negative interaction energy relative to infinity does not mean negative kinetic energy.

U is the interaction energy of this pair, counted once. Adding q times the potential from Q and Q times the potential from q would count the same pair twice. This model does not include the charges’ self-energies.

Account for work and its sign

Define ΔU = U final − U initial. With the source held fixed, the electric field does work W field = −ΔU as q changes position. For a slow transfer with no change in kinetic energy, the external agent does W external = ΔU. The two works sum to zero under that assumption. If a charge were released freely instead, kinetic energy could change; the present display does not simulate that motion.

Moving +2 nC from 0.20 to 0.40 m gives ΔU = −0.0899 µJ, W field = +0.0899 µJ and W external = −0.0899 µJ. Electric repulsion helps outward motion, while the external agent restrains it. Moving −2 nC over the same interval gives ΔU = +0.0899 µJ, W field = −0.0899 µJ and W external = +0.0899 µJ. An external agent must work against attraction.

Use the work ledger to check both signs. At the reference separation 0.20 m, every work difference is zero even though V and, for nonzero q, U are nonzero. The reference for a change is not the same as the chosen zero of potential at infinity.

What changes when the charge slider moves?

Changing q selects a new system, with its own same-charge reference energy. It is not a simulated charging process. You cannot interpret differences between two slider charges as mechanical work along a trajectory. At q = 0, U and both works are zero, while the source V and E remain nonzero.

Adding a constant to the potential reference changes numerical potential and the corresponding qV reference, but leaves potential differences, forces and fixed-charge work differences unchanged. Voltage differences are the useful quantities for work.

Connect potential slope to field

Along this radial axis, E = −dV/dr: field is the negative spatial slope of potential, measured in V/m, equivalent to N/C. For V = kQ/r the derivative is −kQ/r², so E = +kQ/r². Similarly F = −dU/dr at fixed q. These derivatives describe variation with position, not change with time.

Transfer to a specified voltage difference

A charge q = −2 nC moves from potential 60 V to 20 V. Determine ΔV, ΔU, field work and quasistatic external work. No point-charge distance calculation is needed.

Check the electric-work transfer

ΔV = −40 V. ΔU = qΔV = (−2 × 10⁻⁹)(−40) = +80 × 10⁻⁹ J = +0.08 µJ. Field work is −0.08 µJ and quasistatic external work is +0.08 µJ. A negative charge moving to lower potential gains interaction energy.

OpenStax: potential and potential difference relates voltage to work, and calculating electric potential develops point-charge potential.