Electric Force, Field & Potential

Electric Field

The electric field at a point is the force per unit positive test charge: $\vec{E} = \frac{\vec{F}}{q_0}$

Field due to a point charge: $E = \frac{kQ}{r^2}$

• Direction: away from positive charges, toward negative charges.
• Superposition: the total field is the vector sum of individual fields.

Electric Field Lines

• Start on positive charges, end on negative charges.
• Density of lines represents field strength.
• Lines never cross.
• Perpendicular to conducting surfaces.

Electric Potential (Voltage)

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

• Scalar quantity (no direction).
• Potential difference: $\Delta V = V_B - V_A = -\frac{W_{\text{field}}}{q}$

Relationship Between Field and Potential

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

• $\vec{E}$ points from high to low potential.
• Equipotential surfaces are perpendicular to field lines.
• No work is done moving a charge along an equipotential.

Electric Potential Energy

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

Conductors in Electrostatic Equilibrium

• $E = 0$ inside the conductor.
• All excess charge resides on the surface.
• The surface is an equipotential.
• $E$ is perpendicular to the surface.

Problem: A $+4\ \mu\text{C}$ charge is at the origin. Find the electric field and potential at $r = 0.5\ \text{m}$.

Solution:

$E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(4 \times 10^{-6})}{(0.5)^2} = \frac{35{,}960}{0.25} = 143{,}840\ \text{N/C}$

$V = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(4 \times 10^{-6})}{0.5} = 71{,}920\ \text{V}$

The field points radially outward (away from the positive charge).

• Electric field ($\vec{E}$) is force per unit charge; points away from + and toward −.
• Electric potential ($V$) is a scalar; $V = kQ/r$ for a point charge.
• $E$ points from high to low potential; $E = -\Delta V / \Delta x$.
• Equipotential surfaces are perpendicular to field lines.
• Inside a conductor in equilibrium: $E = 0$, the surface is an equipotential.