# UY1: Electric Potential

The potential V at any point in an electric field is the potential energy U per unit charge associated with a test charge $q_{0}$ at that point:

$$V = \frac{U}{q_{0}}$$

$$\Delta U = q_{0} \Delta V$$

The SI unit of electric potential is volt.

$$1 V = 1 \, J C^{-1}$$

The work done per unit charge by the electric force when a charged body moves from a to b is equal to the potential at a minus the potential at b.

$$W_{a \rightarrow b} = \, – (U_{b} – U_{a})$$

\begin{aligned} \frac{W_{a \rightarrow b}}{q_{0}} &= \, – \left( \frac{U_{b}}{q_{0}} – \frac{U_{a}}{q_{0}} \right) \\ &= \, – (V_{b} – V_{a}) \\ &= V_{a} – V_{b} \\ &= V_{ab} \end{aligned}

where $V_{ab}$ is the potential of a with respect to b.

### Example: Oppositely Charged Parallel Plates

Recall that the electric potential energy of a charge in uniform electric field is:

$$U = q_{0}Ey$$

Since $V = \frac{U}{q_{0}}$,

$$V = Ey$$

\begin{aligned} V_{ab} &= V_{a}-V_{b} \\ &= E(y_{a}-y_{b}) \\ &= Ed \end{aligned}

Hence, you get the “potential gradient”:

$$E = \frac{V_{ab}}{d}$$

Recall that $E = \frac{\sigma}{\epsilon_{0}}$ for oppositely charged parallel plate of infinite size, where $\sigma$ is the surface charge density of the plate. You get an expression for the surface charge density:

$$\sigma = \epsilon_{0} \frac{V_{ab}}{d}$$

The above equation is useful for solving problems related to parallel plates.

### Electric Potential Of A Point Charge

Recall that: $W_{a \rightarrow b} = \int\limits_{a}^{b} \vec{F}.d\vec{l} = \int\limits_{a}^{b} q_{0}\vec{E}.d\vec{l}$ and $\frac{W_{a \rightarrow b}}{q_{0}} = V_{a} – V_{b} = \int\limits_{a}^{b} \vec{E}.d\vec{l}$

Hence, the potential V due to a single point charge q:

\begin{aligned} V &= \frac{U}{q_{0}} \\ &= \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r} \end{aligned}

where r is the distance from the point charge q to the point at which the potential is evaluated.

In general, moving with the direction of $\vec{E}$ means moving in the direction of decreasing V, and moving against the direction of \vec{E} means moving in the direction of increasing V. 