Section 02
The Idea, Step by Step
Start simple: a charge reaches out into empty space
Rub a balloon on your hair and hold it near a thin stream of water — the water
bends toward the balloon without anything touching it. The balloon got a little
extra electric charge, and that charge has an invisible "reach" around it. Get
closer and the pull is stronger; back away and it fades fast. That invisible
reach is the electric field: a map of how hard, and in which
direction, a charge would push or pull another charge placed nearby. A second
idea rides alongside it — the potential — which is like the
"height" of an electric hill. Charges roll "downhill" from high potential to
low, exactly the way a ball rolls down a slope.
Build it up: two quantities and one number
A point charge $q$ sets up a field whose strength is $E = kq/r^2$ and a
potential $V = kq/r$, where $r$ is your distance from the charge and
$k \approx 9.0\times10^{9}$ in SI units. Notice the field falls off as
$1/r^2$ (twice as far means a quarter the field) while the potential
falls off more gently as $1/r$. Put in the sim's starting numbers — a
$q = 5\text{ nC}$ charge, viewed from $r = 0.10\text{ m}$:
Worked example
$$V=\frac{kq}{r}=\frac{(9.0\times10^{9})(5\times10^{-9})}{0.10}\approx 450\text{ V},\quad E=\frac{kq}{r^2}\approx 4500\text{ V/m}$$
The field and the potential are not two unrelated things — the field is just
how steeply the potential changes as you move.
Go deeper: the field is the slope of the potential
Precisely, $\vec{E} = -\dfrac{dV}{dr}$ (in full 3D, $\vec{E}=-\nabla V$): the
field points "downhill," in the direction the potential drops fastest, and its
size is the steepness of that drop. Surfaces where $V$ is constant are
equipotentials; because there's no slope along them, $\vec{E}$
always crosses them at right angles and no work is needed to slide a charge
along one. Energy comes from differences in potential: a charge $q_0$
sitting where the potential is $V$ has energy $U = q_0 V$, so moving it through a
voltage drop $\Delta V$ releases $q_0\,\Delta V$ of energy. Between parallel
plates the potential drops steadily, giving the tidy uniform field
$E = V/d$. In the panel above, the Charge q and Distance r
sliders set the point-charge field and potential, while Voltage V₀ and
Plate sep d drive the uniform-field (Energy) view.
Try this in the sim above
First, in E-Field mode, halve Distance r from
0.10 m to 0.05 m and watch the E readout jump by about four times — that's the
inverse-square law in action, while V only doubles. Next, flip Charge q
negative and see every field line reverse to point inward. Finally, switch to
the Energy mode and shrink Plate sep d: the same
voltage squeezed into a smaller gap makes $E=V/d$ shoot up — the reason tiny
gaps in capacitors and microchips reach enormous field strengths.
Section 03
Equations & Derivation
Electric Field & Potential
$$E = -\frac{dV}{dr},\quad V = \frac{kq}{r},\quad U = qV = \frac{kqq_0}{r}$$
Gauss's Law
$$\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0},\quad E = \frac{\sigma}{\varepsilon_0}\text{ (conductor surface)}$$
Uniform Field (parallel plates)
$$E = \frac{V}{d},\quad U = qEd = qV,\quad \text{work: }W = q(V_1-V_2)$$
Symbol Definitions
| Symbol | Quantity | SI Unit |
|---|
| $E$ | Electric field strength | V m⁻¹ = N C⁻¹ |
| $V$ | Electric potential | V = J C⁻¹ |
| $U$ | Electric potential energy | J |
| $\sigma$ | Surface charge density | C m⁻² |
| $\varepsilon_0 = 8.85\times10^{-12}$ | Permittivity of vacuum | C² N⁻¹ m⁻² |
| $\Phi_E$ | Electric flux | N m² C⁻¹ |
1
Electric field. $\vec{E}=-\nabla V$ — field points in direction of decreasing potential. For a point charge: $E=kq/r^2$. The field is a vector; potential is a scalar (easier to compute by superposition).
2
Gauss's Law. $\oint\vec{E}\cdot d\vec{A}=Q_{\text{enc}}/\varepsilon_0$. The electric flux through any closed surface equals the enclosed charge divided by $\varepsilon_0$. Enormously powerful for symmetric charge distributions: sphere, cylinder, infinite plane.
3
Equipotential surfaces. Surfaces of constant $V$. $\vec{E}$ is always perpendicular to equipotentials. No work done moving charge along an equipotential. Inside a conductor: $V=$const (whole conductor is an equipotential).
4
Work and energy. Work done by electric field: $W=q(V_A-V_B)=q\Delta V$. Moving a charge from high to low potential releases energy (like rolling downhill). Potential difference in volts = energy in eV per unit charge.
Ref: Halliday, Resnick & Walker 10th Ed., Ch. 22–24; Griffiths — Introduction to Electrodynamics (4th Ed.), Ch. 2.
Section 05
Common Misconceptions
❌ Electric field lines show the path of a positive charge.
✅ Field lines show the direction of force on a positive test charge at rest. A charge's trajectory depends on its initial velocity — it follows a field line only if starting from rest (and even then, non-straight lines require curved trajectories).
📖 HRW 10th Ed., §22-4.
❌ A strong electric field means high electric potential.
✅ No relationship. The field is $E=-dV/dr$ — it measures the rate of change of potential, not its magnitude. A region of very high constant potential has $E=0$ (no change). Lightning strikes near ground level where potential changes rapidly, not where potential is highest.
📖 HRW 10th Ed., §24-1.
❌ Gauss's Law only works for spherical charge distributions.
✅ Gauss's Law is universally true for any closed surface and any charge distribution. It's most useful for calculating E when the charge distribution has enough symmetry (spherical, cylindrical, planar) that E is constant on the Gaussian surface.
📖 HRW 10th Ed., §23-1 to §23-5.
Misconception research: McDermott — Tutorials in Introductory Physics; Arons — A Guide to Introductory Physics Teaching.