Electric Potential

Overview

Electric potential describes the potential energy per unit charge at a point in an electric field. It provides an alternative way to analyze electric phenomena using energy rather than force.

Electric Potential Energy

Work done by external force to move charge $q$ from A to B:

$$\Delta U = U_B - U_A = -W_{\text{field}} = W_{\text{external}}$$

For point charges:

$$U = \frac{kq_1 q_2}{r}$$

Electric Potential (Voltage)

Definition

Potential energy per unit charge:

$$V = \frac{U}{q}$$

Unit: Volt (V) = J/C

Potential Difference

$$\Delta V = V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r}$$

Work to move charge $q$ through potential difference:

$$W = q\Delta V$$

Potential Due to Point Charge

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

• Positive charge: positive potential
• Negative charge: negative potential
• Zero at infinity (reference point)

Superposition of Potentials

Total potential from multiple charges:

$$V_{\text{total}} = V_1 + V_2 + V_3 + \cdots = k\sum\frac{q_i}{r_i}$$

Note: Potentials are scalars (add algebraically, not vectorially)

Relationship Between E and V

$$E = -\frac{dV}{dr} \quad \text{(in 1D)}$$ $$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$$

For uniform field:

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

Equipotential Surfaces

• Surfaces where $V$ is constant
• Electric field is perpendicular to equipotential surfaces
• No work done moving charge along equipotential
• Conductors are equipotential surfaces

Potential Energy of Systems

Two Point Charges

$$U = \frac{kq_1 q_2}{r}$$

Three Point Charges

$$U = k\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$$

General System

$$U = k \sum_{i<j} \frac{q_i q_j}{r_{ij}}$$

Electron Volt

A convenient unit of energy:

$$1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$$

Energy gained by electron accelerated through 1 V

Potential of Continuous Distributions

$$V = k\int \frac{dq}{r}$$

Uniformly Charged Ring (on axis)

$$V = \frac{kQ}{\sqrt{x^2 + R^2}}$$

Uniformly Charged Disk (on axis)

$$V = 2k\sigma\pi\left(\sqrt{x^2 + R^2} - x\right)$$

Examples

Example 1: Potential from Point Charge

Find potential at 2 m from a 5 μC charge.

$$V = \frac{kq}{r} = \frac{8.99 \times 10^9 \times 5 \times 10^{-6}}{2} = 22{,}475 \text{ V} \approx 22.5 \text{ kV}$$

Example 2: Work to Move Charge

How much work to move a 3 μC charge from $r = 1$ m to $r = 0.5$ m from a 4 μC charge?

$$W = q\Delta V = q(V_f - V_i)$$ $$W = 3 \times 10^{-6} \times \left(\frac{kq}{0.5} - \frac{kq}{1}\right)$$ $$W = 3 \times 10^{-6} \times 8.99 \times 10^9 \times 4 \times 10^{-6} \times (2 - 1)$$ $$W = 0.108 \text{ J}$$

Example 3: Potential Energy of System

Three charges: $q_1 = 2$ μC, $q_2 = -3$ μC, $q_3 = 4$ μC at corners of equilateral triangle (side = 1 m).

$$U = k\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$$ $$U = 8.99 \times 10^9 \times 10^{-12} \times (-6 + 8 - 12)$$ $$U = 8.99 \times 10^{-3} \times (-10) = -0.090 \text{ J}$$

Example 4: Field from Potential

$V(x) = 3x^2 - 2x + 5$ (in volts). Find $E$ at $x = 2$ m.

$$E = -\frac{dV}{dx} = -(6x - 2) = -(6 \times 2 - 2) = -10 \text{ V/m}$$

Field is 10 V/m in the negative x direction.

Example 5: Parallel Plates

Parallel plates are 5 mm apart with potential difference 200 V. Find field.

$$E = \frac{V}{d} = \frac{200}{0.005} = 40{,}000 \text{ V/m} = 40 \text{ kV/m}$$

Example 6: Electron Acceleration

An electron is accelerated through 1000 V. Find final speed (starting from rest).

$$KE = qV = eV = 1000 \text{ eV} = 1.6 \times 10^{-16} \text{ J}$$ $$\frac{1}{2}mv^2 = 1.6 \times 10^{-16}$$ $$v = \sqrt{\frac{2 \times 1.6 \times 10^{-16}}{9.11 \times 10^{-31}}} = 1.87 \times 10^7 \text{ m/s}$$

Key Points

• Potential is a scalar (easier to work with than vector field)
• Work depends only on endpoints, not path
• Field points from high to low potential
• Conductors in equilibrium have constant potential