Electric potentialAQA A-Level Physics: Revision notes

The absolute electric potential V at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. By definition V = 0 at infinity.

$V = \frac{W}{Q}$

The unit is the volt (V = J C⁻¹). Potential is a scalar: it has size and sign but no direction. Near a positive charge it is positive, because work must be done to push a positive test charge towards it. Near a negative charge it is negative.

The electric potential difference between two points is the work done per unit charge in moving a charge between them.

The work done in moving a charge Q through a potential difference ΔV is

$W = Q\,\Delta V$

If the charge moves because of the field (for example a proton moving away from a positive charge), the work done by the field equals the gain in kinetic energy: $\tfrac{1}{2}mv^2 = Q\Delta V$.

Worked example. An electron is accelerated from rest through 2.0 kV. $\tfrac{1}{2}mv^2 = eV = 1.60 \times 10^{-19} \times 2000 = 3.2 \times 10^{-16}$ J, so $v = \sqrt{2 \times 3.2 \times 10^{-16} / 9.11 \times 10^{-31}} = 2.7 \times 10^7$ m s⁻¹.

An equipotential surface is a surface on which every point is at the same potential.

• No work is done moving a charge along an equipotential, because ΔV = 0 so $W = Q\Delta V = 0$.
• Field lines cross equipotentials at right angles.
• Around a point charge the equipotentials are concentric spheres.
• In a uniform field they are parallel planes at right angles to the field.

The field is strongest where equipotentials are closest together.

For a point charge Q, the potential at distance r is

$V = \frac{Q}{4\pi\varepsilon_0 r}$

It falls as 1/r, whereas field strength falls as 1/r². It takes the sign of Q.

Worked example. Q = +4.0 nC. At r = 0.50 m: $V = \dfrac{4.0 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times 0.50} = 72$ V. At r = 0.20 m, V = 180 V. A +1.0 nC charge moved from the first to the second point needs $W = Q\Delta V = 1.0 \times 10^{-9} \times 108 = 1.1 \times 10^{-7}$ J.

For an isolated point charge or sphere (r measured from the centre):

• The V–r graph is an inverse curve, falling from a high value near the charge towards zero at infinity.
• The E–r graph is an inverse-square curve, which falls more steeply at first.

The two are linked by

$E = -\frac{\Delta V}{\Delta r}$

so E is the negative of the gradient of the V–r graph. Rearranged, ΔV is the area under the E–r graph between two distances (up to the sign). The negative sign shows that V decreases in the direction of the field.

Worked example. In a uniform field of 500 V m⁻¹, moving 0.040 m along the field gives $|\Delta V| = 500 \times 0.040 = 20$ V.