The potential energy of a charge depends on that charge. Divide it out, just as we did to get the electric field from the force, and you get a property of the place itself: the electric potential, measured in volts.
Why it works
If a charge $q$ at some point has potential energy $U$, then $V = U/q$ is the energy each coulomb would have there. Near a point charge, $U = kQq/r$, so $V = kQ/r$. Potential is a scalar, so for several charges you just add numbers, signs included. When a charge moves between two points, its potential energy changes by $q\Delta V$; if only electric forces act, its kinetic energy changes by the opposite amount.
Key idea
$V = \dfrac{U}{q}$, in volts ($1\,\text{V} = 1\,\text{J/C}$). Near a point charge, $V = k\dfrac{Q}{r}$. A charge crossing a potential difference $\Delta V$ has $\Delta U = q\Delta V$ and $\Delta K = -q\Delta V$. One electron volt, $1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}$, is the energy an electron gains crossing $1\,\text{V}$.
| Quantity | Vector or scalar? | Depends on the test charge? | Unit |
|---|---|---|---|
| Force $\vec F$ | vector | yes | N |
| Field $\vec E = \vec F/q$ | vector | no | N/C = V/m |
| Potential energy $U$ | scalar | yes | J |
| Potential $V = U/q$ | scalar | no | V = J/C |
Worked examples
Worked example: Near a point charge
Find the potential $0.30\,\text{m}$ from a $+6.0\,\text{nC}$ charge.
- $V = k\dfrac{Q}{r} = (8.99 \times 10^{9})\dfrac{6.0 \times 10^{-9}}{0.30} = 180\,\text{V}$.
Answer: $180\,\text{V}$
Worked example: Two charges
A $+4.0\,\text{nC}$ charge is at $x = 0$ and a $-2.0\,\text{nC}$ charge at $x = 0.40\,\text{m}$. Find the potential at $x = 0.20\,\text{m}$.
- Scalars add: $V = (8.99 \times 10^{9})\left(\dfrac{4.0 \times 10^{-9}}{0.20} + \dfrac{-2.0 \times 10^{-9}}{0.20}\right)$.
- $V = (8.99 \times 10^{9})(1.0 \times 10^{-8}) = 89.9\,\text{V}$. No directions needed.
Answer: $89.9\,\text{V}$
Worked example: An electron gun
An electron starts from rest and crosses a potential difference of $1000\,\text{V}$. Find its kinetic energy and speed.
- $K = e\Delta V = (1.60 \times 10^{-19}\,\text{C})(1000\,\text{V}) = 1.60 \times 10^{-16}\,\text{J} = 1000\,\text{eV}$.
- $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(1.60 \times 10^{-16})}{9.11 \times 10^{-31}}} = 1.87 \times 10^{7}\,\text{m/s}$, about 6% of the speed of light.
Answer: $1000\,\text{eV}$; $1.87 \times 10^{7}\,\text{m/s}$
Common mistake
Mixing up potential with potential energy, or with the field. Potential $V$ is energy per coulomb and a scalar; potential energy $U = qV$ depends on the charge; the field $\vec E$ is a vector and depends on how $V$ changes from place to place. Check the units: volts, joules, or newtons per coulomb.