A contour map joins points of equal height; equipotentials join points of equal electric potential. Together with the field lines, they give the complete picture of a field, and they connect the field to the voltage of a battery.
Why it works
Move a charge along an equipotential: $\Delta V = 0$, so the field does no work, which means the force has no component along the move. So $\vec E$ is always perpendicular to the equipotentials. Move a positive charge along the field and the field does positive work, so its potential energy, and the potential, drop: $\vec E$ points from high to low potential. In a uniform field, the drop over a distance $d$ along the field is $\Delta V = Ed$, so $E = \Delta V/d$ and $1\,\text{V/m} = 1\,\text{N/C}$.
Key idea
Field lines cross equipotentials at right angles and point toward lower potential. Equipotentials are crowded where the field is strong: $E = \dfrac{\Delta V}{d}$ (exactly so in a uniform field, such as between parallel plates).
Worked examples
Worked example: Between plates
Plates $0.50\,\text{cm}$ apart are connected to a $120\,\text{V}$ battery. Find the field.
- $E = \dfrac{\Delta V}{d} = \dfrac{120\,\text{V}}{0.0050\,\text{m}} = 2.4 \times 10^{4}\,\text{V/m}$, pointing from the positive plate to the negative plate.
Answer: $2.4 \times 10^{4}\,\text{V/m}$
Worked example: Reading a map
Equipotential lines are drawn every $10\,\text{V}$ and are $2.0\,\text{cm}$ apart, with the values falling toward the east. Find the field.
- $E = \dfrac{10\,\text{V}}{0.020\,\text{m}} = 500\,\text{V/m}$.
- It points toward lower potential: east, at right angles to the lines.
Answer: $500\,\text{V/m}$, east
Worked example: Zero potential, nonzero field
At the midpoint of a $+3\,\text{nC}$ and a $-3\,\text{nC}$ charge, is the field zero?
- The potentials cancel: $V = kQ/r - kQ/r = 0$.
- The fields do not: both point toward the negative charge, so they add.
- The field depends on how fast $V$ changes, and here $V$ drops steeply toward the negative charge.
Answer: No: $V = 0$, but $E \ne 0$
Common mistake
Thinking the field is zero wherever the potential is zero. The field is about the slope of the potential, not its value, just as a hillside at sea level can still be steep. Check: look at how closely the equipotentials are spaced around the point, not at the number on the one through it.