Electric Potential

Electric potential ($V$) is the electric potential energy per unit charge. It is a scalar quantity that describes the "electric pressure" at a point in space, independent of any test charge placed there.

Definition

The electric potential at a point is the work needed to bring a unit positive charge from infinity to that point:

$$V = \frac{U}{q} = \frac{W_{\infty \to point}}{q}$$

Where:

Symbol Meaning Units
$V$ Electric potential Volts (V)
$U$ Electric potential energy Joules (J)
$q$ Test charge Coulombs (C)

Unit: 1 Volt = 1 Joule/Coulomb

Potential Due to a Point Charge

$$V = \frac{kQ}{r}$$

Where:

  • $k = 8.99 \times 10^9$ N·m²/C²
  • $Q$ = source charge
  • $r$ = distance from the charge

Key Properties

Property Description
Sign Positive for $+Q$, negative for $-Q$
Zero point $V = 0$ at $r = \infty$
Dependence $V \propto \frac{1}{r}$ (not $\frac{1}{r^2}$)
Scalar No direction — just magnitude and sign

Potential from Multiple Charges

Since potential is a scalar, we simply add the contributions:

$$V_{total} = \sum_i \frac{kq_i}{r_i}$$

No vector addition needed! This is much simpler than adding electric fields.

Relationship to Electric Field

Integral Form

$$V_b - V_a = -\int_a^b \vec{E} \cdot d\vec{l}$$

Differential Form

$$E = -\frac{dV}{dr}$$

The electric field points in the direction of decreasing potential.

Key Insight

  • Field lines point from high potential to low potential
  • Positive charges move from high $V$ to low $V$
  • Negative charges move from low $V$ to high $V$

Equipotential Surfaces

Surfaces where $V$ is constant everywhere.

Properties

  • $\vec{E}$ is perpendicular to equipotential surfaces
  • No work is done moving a charge along an equipotential
  • Field lines cross equipotentials at right angles

Examples

Source Equipotential Surfaces
Point charge Concentric spheres
Line charge Coaxial cylinders
Plane sheet Parallel planes

Examples

Example 1: Potential from Proton

Find $V$ at $r = 1.00$ cm from a proton.

$$V = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(1.6 \times 10^{-19})}{0.01} = 1.44 \times 10^{-7} \text{ V}$$

Example 2: Potential Difference

Find $\Delta V$ between points at $r_1 = 1.00$ cm and $r_2 = 2.00$ cm from a proton.

$$\Delta V = V_1 - V_2 = kQ\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$

$$\Delta V = (8.99 \times 10^9)(1.6 \times 10^{-19})\left(\frac{1}{0.01} - \frac{1}{0.02}\right) = 7.19 \times 10^{-8} \text{ V}$$

Example 3: Multiple Charges

Potential at point P from rectangular arrangement:

$$V_P = \frac{kq_1}{r_1} + \frac{kq_2}{r_2} + \frac{kq_3}{r_3} + ...$$

Just sum the scalar values!

Advantages of Using Potential

  1. Scalar quantity — No vector components
  2. Easy superposition — Simple algebraic sum
  3. Direct connection to energy — $U = qV$
  4. Practical measurements — Voltmeters measure potential difference

Electron Volt (eV)

A convenient unit of energy in atomic physics:

$$1 \text{ eV} = e \times (1 \text{ V}) = 1.6 \times 10^{-19} \text{ J}$$

The energy gained by an electron accelerated through 1 Volt potential difference.

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