Electric potential and field patternsEdexcel A-Level Physics: Revision notes

The electric potential $V$ at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point. It is a scalar, measured in volts, and is taken as zero at infinity. Potentials from several charges add as ordinary numbers, with their signs.

The field strength is the potential gradient:

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

The minus sign shows that the field points in the direction in which the potential decreases. Where the potential changes quickly with distance the field is strong. A point can have zero potential and still have a field.

Between two large parallel plates with a potential difference $V$ and separation $d$, the field is uniform in the central region:

$E = \frac{V}{d}$

The unit of $E$ can be written V m⁻¹ or N C⁻¹.

• Field lines are parallel, equally spaced and at right angles to the plates, pointing from the positive plate to the negative plate
• Equipotentials are equally spaced planes parallel to the plates, at right angles to the field lines
• Near the edges the lines bulge outwards and the field is not uniform
• The force on a charge $Q$ in the field is $F = QE$

For a point charge $Q$ (or a spherical charge treated as a point charge at its centre) the potential at distance $r$ is

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

The potential is positive near a positive charge and negative near a negative charge, and it varies as $1/r$. The field strength varies as $1/r^2$, so doubling $r$ halves $V$ but quarters $E$.

• Field lines are radial, pointing away from a positive charge and towards a negative charge
• Equipotentials are concentric spheres, drawn as circles, always at right angles to the field lines
• Equipotentials get further apart as $r$ increases, showing that the field is weakening

Field lines and equipotentials always meet at right angles. This is because moving a charge along an equipotential involves no change in potential energy, so no work is done, and so there can be no component of force along the surface.

To sketch a pattern:

• Draw field lines leaving positive charges and ending on negative charges, never crossing
• The closer the field lines, the stronger the field
• Draw equipotentials at right angles to the lines, with equal potential steps

For equal and opposite charges the plane halfway between them is an equipotential at 0 V.

A sphere carrying +8.0 nC is treated as a point charge. Find the potential and field strength 0.30 m from it.

Potential: $V = \dfrac{8.0\times10^{-9}}{4\pi\times 8.85\times10^{-12}\times 0.30} = 2.4\times10^{2}$ V.

Field strength: $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{V}{r} = \dfrac{240}{0.30} = 8.0\times10^{2}$ N C⁻¹, directed away from the sphere.

At 0.60 m the potential is 120 V and the field strength is 200 N C⁻¹.