Physic Labs

Electricity and magnetism

Electric potential and voltage

Electric potential is electric potential energy per unit charge; a potential difference determines the change in energy as charge moves.

An electric field can be described through the force on a test charge or through energy. The potential VV at a point is electric potential energy UU per coulomb of positive charge. The potential difference between two points gives the work done by the electric force per unit charge as it moves between them.

V=Uq,ΔV=VB−VA=−WA→Bq,E=−ΔVΔx (uniform field)V=\frac{U}{q},\qquad \Delta V=V_B-V_A=-\frac{W_{A\to B}}{q},\qquad E=-\frac{\Delta V}{\Delta x}\ (\text{uniform field})

Definition: Potential and potential difference

Potential and potential difference are measured in volts (V), where 1 V=1 J/C1\,\mathrm V=1\,\mathrm{J/C}. The numerical potential depends on the chosen zero; a difference between two points does not. The electric force does work WA→B=q(VA−VB)W_{A\to B}=q(V_A-V_B); for negative charge, track the sign of qq when finding the energy change.

Explore the field between parallel plates, potential variation with position, and a test charge’s potential energy.

Uniform field between parallel plates

Neglecting edge effects, the field between parallel plates separated by dd is approximately uniform: E=∣ΔV∣/dE=|ΔV|/d. It points from higher to lower potential. A charge qq changes its potential energy by ΔU=qΔVΔU=qΔV.

Example: Potential energy of a charge

A charge q=+2.0 μq=+2.0\,\muC moves from VA=120V_A=120 V to VB=70V_B=70 V. Find the change in potential energy and the work done by the electric force.

Solution

ΔV=−50\Delta V=-50 V, so ΔU=qΔV=−1.0×10−4\Delta U=q\Delta V=-1.0\times10^{-4} J. The electric force does W=−ΔU=+1.0×10−4W=-\Delta U=+1.0\times10^{-4} J of work.

The potential of a point charge at distance rr in vacuum is V=kQ/rV=kQ/r when zero is chosen at infinity; potential is a scalar, so source contributions add algebraically. For example, how much does a +3.0 μ+3.0\,\muC charge gain in potential energy when moving from 2020 V to 8080 V? ΔU=qΔV=(3.0×10−6)(60)=1.8×10−4\Delta U=q\Delta V=(3.0\times10^{-6})(60)=1.8\times10^{-4} J. A source must supply that energy to move the positive charge to higher potential when losses are absent.

Separating work from potential energy

Work by the electric force as a charge moves from AA to BB is WA→B=−ΔU=q(VA−VB)W_{A\to B}=-\Delta U=q(V_A-V_B). If a positive charge moves freely along the field, its potential energy decreases and the field does positive work. A negative charge feels force opposite the field, so track the charge sign rather than relying on the arrow alone. In an electrostatic field, work depends only on the endpoints, not the path.

Quick check

When charge qq moves through a potential difference ΔV\Delta V, its change in potential energy is:

For a uniform field between parallel plates, how are field magnitude, voltage difference, and separation related?

References

  1. Young, Freedman (2019). University Physics