One field. Two points. Any test charge.
ΔU = q[V(B) - V(A)] · Welectric = -ΔU
Ideal voltmeter
- V(A)
- V(B)
The meter reads red minus black. Moving a charge A → B uses B minus A.
- ΔV, A → B
- ΔU, A → B
- Electric work, A → B
Along the chosen path
P follows a prescribed path, not a force-driven trajectory. Both vector arrows show direction only: teal E and gold F = qE. The horizontal plot axis is fraction of path length, not time.
Different paths, same electrostatic work
Numerical field integral versus endpoint energy| Path | Length | ∫ E · dr | q ∫ E · dr | -qΔV |
|---|
An external agent moving the charge quasistatically does Wexternal = +ΔU, opposite to the electric work. Without that condition, external work also depends on the change in kinetic energy.
Charge counts; voltage sets energy per charge
e = 1.602176634 × 10-19 C, exactlyOnly an electron imbalance is counted. Positive net charge can mean electrons were removed; no proton transport is implied. The main macroscopic charge input is continuous, while this transfer control uses integer elementary charges.
A real source is not an ideal voltage source
Simple DC resistance model- Terminal voltage
- Total source current
- Without meter loading
Vterminal = ℰ - Ir. The meter is parallel to the load, so its current is included in I. This is a resistive steady-state model, not battery chemistry, discharge history or practical wiring guidance.
Reading the physics
Potential is the landscape; energy belongs to a charge
The sources determine V and E. A small test charge samples that environment without changing it. Two different test charges crossing the same endpoints see the same ΔV, but their energy changes differ: ΔU = qΔV. One volt is one joule per coulomb, not one joule for every particle.
Compare positive and negative q in the +5 V example. The potential plot stays fixed. The U plot flips when q changes sign and scales when its magnitude changes. A neutral test marker has no electric force or energy change, even when E and ΔV are nonzero.
Why field and work have their signs
Electric field points downhill in potential: E = -∇V. The force on a positive charge points with E; the force on a negative charge points against E. Either can lose potential energy while moving in its force direction. A negative charge can therefore move toward higher V while U decreases.
For the plate example E = 200 V/m and d = 0.050 m, A → B has ΔV = -10 V. A +2 nC test charge loses 20 nJ of potential energy. The electric field does +20 nJ of work. For the worked -4 μC example, ΔV = -3 V instead gives ΔU = +12 μJ and electric work -12 μJ.
A bent path does not change the answer
In an electrostatic field, ΔV = -∫AB E · dr. Only the field component along each segment contributes. A segment along an equipotential has zero work, even when the field is strong. The straight and waypoint paths can have different lengths and different intermediate energy profiles but the same net electric work.
The table independently integrates the field along each segment using ODEX and compares it with -qΔV. If a path enters a source exclusion disk, its integral and traversal are withheld, rather than drawing a fictitious passage through a point-charge singularity. A closed trip, made by putting B on A and using H, has zero net electrostatic work.
Zero potential does not mean zero field
In the dipole preset the perpendicular bisector has V = 0 before a reference shift, but E is nonzero. For two equal positive sources the midpoint instead has E = 0 but positive V. Potential is a scalar sum; electric fields add as vectors. Their zeros are different statements.
Point-source potentials use V(∞) = 0 before the common shift C. The point-source view is a two-dimensional slice of the physical three-dimensional fields: V = kΣQ/r and E = kΣQ r/r³. No softened force law is used. The small source disks are display exclusions, not conducting surfaces.
The voltmeter compares, it does not assign an absolute zero
The ideal meter reads V(red) - V(black). With red on A and black on B that is the negative of the A → B potential change. Reversing the leads changes the displayed sign without reversing the charge path or changing the field.
Add a common shift C: V(A), V(B) and U = qV change by their reference offsets, but the voltmeter reading, ΔV, ΔU, force and work do not. Reference zero is not automatically Earth. In the plate view the uniform field is an ideal local model, extended over the plotted patch; the two vertical rails mark a spacing d, not finite electrodes with a simulated edge field.
Electronvolts and sources of energy
One electronvolt is exactly 1.602176634 × 10-19 J. It is an energy unit, not a voltage. For an electron, ΔV = +3 V gives ΔU = -3 eV. If electric work alone becomes kinetic energy, the kinetic-energy gain is +3 eV; the prescribed-path marker itself does not simulate that acceleration.
A battery uses nonelectrostatic processes to provide energy per charge, its EMF. Internal resistance lowers terminal voltage under load; finite meter resistance can lower it further. Setting r = 0 gives the ideal-source limit. This lab does not model induction, changing magnetic flux, radiation, charge redistribution in conductors or test-charge back-reaction.