Electric Potential and Potential Energy: MDCAT Physics notes
Electric Potential and Potential Energy MDCAT notes: V = W/q, potential difference, the volt, electron volt, U = kq₁q₂/r and energy and power calculations.
Electric potential energy
When a charge is moved against an electric field, work is done on it and stored as electric potential energy. For two point charges separated by $r$:
$$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} = \frac{kq_1q_2}{r}$$
Note the $r$ (not $r^2$) in the denominator. $U$ is positive for like charges and negative for unlike charges. Potential energy belongs to the system of charges, so it changes if either charge changes.
Potential difference
The potential difference between two points is the work done (or change in potential energy) per unit positive charge in moving it from one point to the other, keeping it in equilibrium (no gain in kinetic energy):
$$\Delta V = V_B - V_A = \frac{W_{AB}}{q_0} = \frac{\Delta U}{q_0}$$
Electric potential (absolute potential)
The electric potential at a point is the work done per unit positive charge in bringing it from infinity (where $V = 0$) to that point:
$$V = \frac{W}{q_0}$$
- Potential is a scalar.
- Unit: volt (V) = J C$^{-1}$.
- Potential is a property of the source, like the field. Replacing a 1 C test charge by a 3 C charge at the same point leaves $V$ unchanged but triples the potential energy $qV$.
- Positive charge flows naturally from higher to lower potential. Potential difference is what drives a current.
Work, energy and power
$$W = qV \qquad P = \frac{W}{t} = \frac{q\Delta V}{t}$$
Example 1: moving a 4 C charge between two points needs 1000 J. The potential difference is $1000/4 = 250$ V.
Example 2: a 0.05 C charge moves from a point at 30 V to one at 70 V in 4 s. Work $= 0.05\times40 = 2$ J; power $= 2/4 = 0.5$ W.
The electron volt
One electron volt is the energy gained (or lost) by one electron when it moves through a potential difference of one volt:
$$1\ \text{eV} = 1.6\times10^{-19}\ \text{C}\times1\ \text{V} = 1.6\times10^{-19}\ \text{J}$$
It is a unit of energy, not of charge, current or potential. A particle of charge $ne$ moved through $V$ volts gains $nV$ eV, that is $nV\times1.6\times10^{-19}$ J. For example, charge $3e$ through 40 V gains 120 eV $= 120\times1.6\times10^{-19}$ J.
| Quantity | Definition | Unit | Type |
|---|---|---|---|
| Electric field $E$ | Force per unit charge | N C$^{-1}$ or V m$^{-1}$ | Vector |
| Potential $V$ | Work per unit charge | V = J C$^{-1}$ | Scalar |
| Potential energy $U$ | $qV$ | J or eV | Scalar |
Key formulas
- $V = W/q$ and $\Delta V = \Delta U/q$
- $U = kq_1q_2/r$
- $W = qV$, $P = qV/t$
- 1 eV $= 1.6\times10^{-19}$ J
Common MDCAT traps
- $V = W/q$ gives potential, not field or power.
- Potential energy uses $1/r$; force and field use $1/r^2$.
- Changing the test charge changes the potential energy, not the potential or the field.
- Electron volt is energy; watch for options that call it charge or power.
- For power, use the potential difference between the two points, then divide by time.
Quick revision
- Potential difference = change in potential energy per unit charge.
- Absolute potential is measured from infinity, where $V = 0$.
- SI unit of potential difference is the volt.
- 1 V = 1 J C$^{-1}$.
- Charges flow from higher to lower potential.