Electric Potential

The electric potential at a point is the work done per unit positive charge in bringing a charge from infinity to that point. It is measured in volts, where $1\mathrm{V}=1\mathrm{J}{\mathrm{C}}^{-1}$, and infinity is defined as zero potential.

A charge $Q$ at a point of potential $V$ has electric potential energy $E_p=QV$. The work done moving a charge between two points is the charge times the potential difference.

Near a positive charge the potential is positive, since work must be done to push a positive charge closer. Near a negative charge it is negative.

For a point charge, $V=kQ/r$. It falls as $1/r$, not $1/r^2$ — more gradually than field strength.

Because potential is a $\text{scalar}$, the potential at a point due to several charges is simply the algebraic sum of each one. Signs matter, but directions do not.

Two charges produce potential energy $E_p=\frac{k{Q}_1{Q}_2}{r}$, the work needed to assemble them from infinite separation.

Field strength and potential are linked by $E=-\frac{\Delta V}{\Delta r}$: the field is the negative of the potential gradient. The minus sign means the field points from high potential to low — the direction a positive charge would naturally move.

For a uniform field this is just $E=V/d$ again, since potential falls steadily between the plates.

On a graph of $V$ against $r$, the field strength at any point is the negative of the gradient. Where the graph is steepest, the field is strongest.

An $\text{equipotential}$ is a line or surface joining points of equal potential. Moving a charge along one requires no work, because $\Delta V=0$.

Equipotentials are always $\text{perpendicular}$ to field lines. If they were not, the field would have a component along the equipotential and work would be done moving along it — a contradiction.

For a point charge, equipotentials are concentric spheres, getting further apart with distance since $V$ falls as $1/r$. For parallel plates they are equally spaced planes parallel to the plates.