Energy of system of charges due to their positions relative to each other.

$U = \text{potential energy of charge system}$

For two point charges: $U = k_e \frac{q_1 q_2}{r}$

Where r = separation distance

Units: Joules (J)

Sign conventions:

• Like charges: U > 0 (repulsive, requires work to bring together)
• Opposite charges: U < 0 (attractive, releases energy when brought together)

Electric force is conservative (path-independent work).

$W_{A \to B} = -\Delta U = U_A - U_B$

For electric field: $W = \int_A^B \vec{F} \cdot d\vec{l} = q\int_A^B \vec{E} \cdot d\vec{l}$

Key implications:

• Work independent of path
• Work = 0 around closed loop
• Potential energy defined relative to reference point

Total potential energy of N-charge system:

$U_{total} = \sum_{i<j} k_e \frac{q_i q_j}{r_{ij}}$

Sum over all unique pairs (each pair counted once).

For three charges: $U = k_e\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$

Physical meaning:

• Work required to assemble system from infinite separation
• Reference: U = 0 when all charges at infinity

Potential is potential energy per unit charge.

$V = \frac{U}{q_0}$

Where:

• U = potential energy
• $q_0$ = test charge

Units: Volts (V) = Joules/Coulomb (J/C)

Key properties:

• Scalar quantity (not a vector)
• Independent of test charge magnitude
• Reference point: V = 0 at infinity (convention)

Voltage between two points:

$\Delta V = V_B - V_A = \frac{U_B}{q} - \frac{U_A}{q}$

In terms of electric field: $\Delta V = -\int_A^B \vec{E} \cdot d\vec{l}$

For uniform field: $\Delta V = -Ed\cos\theta$

Physical meaning:

• Work required per unit charge to move between points
• Driving force in circuits
• Energy gained/lost per unit charge

For single point charge q at distance r:

$V = k_e \frac{q}{r}$

Units: Volts

Reference: V -> 0 as r -> ∞

Direction:

• V > 0 for positive charges
• V < 0 for negative charges

Total potential from multiple charges is scalar sum:

$V = \sum_{i=1}^{N} V_i = \sum_{i=1}^{N} k_e \frac{q_i}{r_i}$

Important: SCALAR addition (not vector!)

Procedure:

1. Calculate potential from each charge
2. Add as scalars (positive or negative numbers)
3. No direction considerations

Advantages over electric field:

• Easier (scalars vs vectors)
• No need to resolve components

Potential from continuous charge distribution:

$V = \int \frac{k_e\,dq}{r}$

Where:

• dq = infinitesimal charge element
• r = distance from dq to point of interest

Charge elements:

• Line: $dq = \lambda dl$, $V = \int \frac{k_e\lambda dl}{r}$
• Surface: $dq = \sigma dA$, $V = \int \frac{k_e\sigma dA}{r}$
• Volume: $dq = \rho dV$, $V = \int \frac{k_e\rho dV}{r}$

Easier than E-field because scalar integral.

Uniformly charged ring (radius R, charge Q, on axis at distance x):

$V(x) = \frac{k_e Q}{\sqrt{x^2 + R^2}}$

All charge elements at same distance $\sqrt{x^2 + R^2}$.

Uniformly charged disk (radius R, surface density σ, on axis at distance x):

$V(x) = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{x^2 + R^2} - |x|\right)$

For infinite disk (R -> ∞): $V = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{x^2 + R^2} - |x|\right)$

Electric field equals negative gradient of potential:

$\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)$

In one dimension (radial): $E = -\frac{dV}{dr}$

Physical meaning:

• E-field points in direction of decreasing potential
• E-field strength equals rate of potential change
• Positive E-field means potential decreasing in + direction

For uniform field: $E = -\frac{\Delta V}{\Delta x} = \frac{V}{d}$ (between parallel plates)

Equipotential: all points at same electric potential.

$V = \text{constant}$

Equipotential surfaces (3D):

• Surfaces where potential is constant
• Perpendicular to electric field lines
• No work required to move charge along equipotential

Equipotential lines (2D):

• Lines connecting points of equal potential
• Perpendicular to field lines at each point
• Never cross field lines

Relationship:

• E ⊥ equipotential surface
• Higher potential toward positive charges
• Lower potential toward negative charges

Find potential by integrating electric field:

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

For uniform field: $V_B - V_A = -Ed\cos\theta$

For radial field (point charge): $V(r) - V(r_0) = -\int_{r_0}^r E\,dr = -\int_{r_0}^r \frac{k_e q}{r^2} dr$

$V(r) = \frac{k_e q}{r} + C$ (C determined by reference)

Choosing reference (typically V(∞) = 0): $V(r) = \frac{k_e q}{r}$