Electrostatics | AP - Wyatt's Notes

Electric Charge

Charge is a fundamental property of matter. There are two types: positive and negative.

The elementary charge is $e = 1.602 \times 10^{-19}$ C.
Charge is quantized: $q = ne$ for integer $n$ .
Charge is conserved: the net charge in an isolated system is constant.
Conductors allow free movement of charge; insulators do not.

Coulomb”s Law

The electrostatic force between two point charges is:

\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\hat{r}_{12}

Where $\epsilon_0 = 8.854 \times 10^{-12}$ C $^2$ /N $\cdot$ M $^2$ is the permittivity of free space and $k = \dfrac{1}{4\pi\epsilon_0} = 8.99 \times 10^9$ N $\cdot$ M $^2$ /C $^2$ .

Superposition Principle

For a system of $n$ point charges, the net force on charge $q_0$ is:

\vec{F} = q_0 \sum_{i=1}^{n} \frac{1}{4\pi\epsilon_0} \frac{q_i}{|\vec{r} - \vec{r}_i|^2} \hat{r}_{0i}

This is a vector sum; each pair interacts independently.

Continuous Charge Distributions

For a continuous charge distribution, replace the sum with an integral:

\vec{F} = \int \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2}\hat{r}

Where $dq$ depends on the geometry:

Linear: $dq = \lambda\, dl$ (charge per unit length)
Surface: $dq = \sigma\, dA$ (charge per unit area)
Volume: $dq = \rho\, dV$ (charge per unit volume)

Example: Force from a uniformly charged rod A rod of length $L$ carries total charge $Q$ distributed uniformly. Find the force on a point charge $q$ Placed along the rod’s axis at distance $a$ from one end.

The linear charge density is $\lambda = Q/L$ . Place the rod along the $x$ -axis from $x = 0$ to $x = L$ And the point charge at $x = L + a$ .

F = \int_0^L \frac{1}{4\pi\epsilon_0} \frac{q \lambda\, dx}{(L + a - x)^2} = \frac{q\lambda}{4\pi\epsilon_0} \int_0^L \frac{dx}{(L+a-x)^2}

Let $u = L + a - x$ , $du = -dx$ :

F = \frac{q\lambda}{4\pi\epsilon_0} \int_{L+a}^{a} \frac{-du}{u^2} = \frac{q\lambda}{4\pi\epsilon_0} \left[\frac{1}{u}\right]_a^{L+a} = \frac{q\lambda}{4\pi\epsilon_0}\left(\frac{1}{a} - \frac{1}{L+a}\right) = \frac{qQ}{4\pi\epsilon_0\, a(L+a)}

Electric Field

The electric field at a point is the force per unit charge:

\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\hat{r}

The field is defined at every point in space regardless of whether a test charge is present.

Electric Field of a Continuous Distribution

\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r^2}\hat{r}

Example: Electric field on the axis of a charged ring A ring of radius $R$ carries total charge $Q$ . Find the electric field at distance $x$ along its axis.

By symmetry, the perpendicular components cancel. Only the axial component survives:

DE_x = \frac{1}{4\pi\epsilon_0} \frac{dq}{R^2 + x^2} \cdot \frac{x}{\sqrt{R^2 + x^2}} = \frac{1}{4\pi\epsilon_0} \frac{x\, dq}{(R^2 + x^2)^{3/2}}

E_x = \frac{1}{4\pi\epsilon_0} \frac{x}{(R^2 + x^2)^{3/2}} \int dq = \frac{1}{4\pi\epsilon_0} \frac{Qx}{(R^2 + x^2)^{3/2}}

At the center ( $x = 0$ ): $E = 0$ As expected by symmetry.

Example: Electric field of an infinite line of charge For an infinite line with linear charge density $\lambda$ Use cylindrical symmetry. Place the line along The $z$ -axis. A segment $dz$ at the origin produces a field with perpendicular component:

DE_\perp = \frac{1}{4\pi\epsilon_0} \frac{\lambda\, dz}{z^2 + r^2} \cdot \frac{r}{\sqrt{z^2 + r^2}}

Integrating from $z = -\infty$ to $\infty$ :

E = \frac{\lambda r}{4\pi\epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{(z^2 + r^2)^{3/2}} = \frac{\lambda r}{4\pi\epsilon_0} \cdot \frac{2}{r^2} = \frac{\lambda}{2\pi\epsilon_0 r}

Electric Field Lines

Field lines point away from positive charges and toward negative charges.
The density of lines is proportional to the field magnitude.
Lines never cross.
Lines begin on positive charges and end on negative charges (or extend to infinity).

Gauss’s Law

Gauss’s law relates the electric flux through a closed surface to the enclosed charge:

\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}{\epsilon_0}

Choosing a Gaussian Surface

Choose a surface where $\vec{E}$ is either constant and parallel to $d\vec{A}$ Or perpendicular to $d\vec{A}$ (contributing zero flux). Common choices exploit symmetry: spheres, cylinders, and boxes.

Applications of Gauss’s Law

Field of an Infinite Plane (surface charge density $\sigma$ )

Choose a cylindrical Gaussian surface piercing the plane. The flux through the curved sides is zero ( $\vec{E} \perp d\vec{A}$ ). The flux through each flat end is $EA$ .

2EA = \frac{\sigma A}{\epsilon_0} \implies E = \frac{\sigma}{2\epsilon_0}

This result is independent of distance from the plane.

Field of a Uniformly Charged Sphere (total charge $Q$ Radius $R$ )

Outside ( $r > R$ ): Choose a spherical Gaussian surface of radius $r$ .

\oint \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \implies E = \frac{Q}{4\pi\epsilon_0 r^2}

This is identical to a point charge.

Inside ( $r < R$ ): Only charge within radius $r$ is enclosed. If the charge density is uniform, $\rho = Q / (\tfrac{4}{3}\pi R^3)$ So $Q_{\text{enc} = \rho \cdot \tfrac{4}{3}\pi r^3 = Q r^3/R^3$ .

E \cdot 4\pi r^2 = \frac{Qr^3}{\epsilon_0 R^3} \implies E = \frac{Qr}{4\pi\epsilon_0 R^3}

The field inside grows linearly with $r$ and reaches its maximum at $r = R$ .

Field of an Infinite Cylindrical Shell (radius $R$ Linear charge density $\lambda$ )

Outside ( $r > R$ ): Cylindrical Gaussian surface of radius $r$ Length $L$ .

\oint \vec{E} \cdot d\vec{A} = E \cdot 2\pi r L = \frac{\lambda L}{\epsilon_0} \implies E = \frac{\lambda}{2\pi\epsilon_0 r}

Inside ( $r < R$ ): $Q_{\text{enc} = 0$ So $E = 0$ .

Example: Non-conducting sphere with non-uniform charge density A sphere of radius $R$ has charge density $\rho(r) = \rho_0 (1 - r/R)$ for $0 \le r \le R$ . Find $E$ Inside and outside.

$Q_{\text{enc}(r) = \int_0^r \rho(r') \cdot 4\pi r'^2\, dr' = 4\pi\rho_0 \int_0^r \left(r'^2 - \frac{r'^3}{R}\right) dr'$

Q_{\text{enc}(r) = 4\pi\rho_0 \left[\frac{r^3}{3} - \frac{r^4}{4R}\right]

Inside ( $r \le R$ ):

E = \frac{\rho_0}{\epsilon_0}\left(\frac{r}{3} - \frac{r^2}{4R}\right)

Outside ( $r > R$ ): First find $Q_{\text{total} = 4\pi\rho_0 [R^3/3 - R^3/4] = 4\pi\rho_0 R^3/12 = \pi\rho_0 R^3/3$ .

E = \frac{\pi\rho_0 R^3}{12\epsilon_0 r^2}

Electric Potential

The electric potential difference between two points is the work done per unit charge:

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

The potential at a distance $r$ from a point charge $q$ (choosing $V(\infty) = 0$ ):

V = \frac{q}{4\pi\epsilon_0 r}

Relation Between Field and 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: $E = -\dfrac{dV}{dx}$ .

Potential of Continuous Distributions

V = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r}

Example: Potential on the axis of a charged disk A disk of radius $R$ has surface charge density $\sigma$ . Find the potential at distance $x$ along the Axis.

Divide the disk into rings of radius $r$ and width $dr$ . Each ring has charge $dq = \sigma \cdot 2\pi r\, dr$ .

The distance from a ring at radius $r$ to the point at height $x$ is $\sqrt{r^2 + x^2}$ .

V = \frac{1}{4\pi\epsilon_0} \int_0^R \frac{\sigma \cdot 2\pi r\, dr}{\sqrt{r^2 + x^2}} = \frac{\sigma}{2\epsilon_0} \left[\sqrt{r^2 + x^2}\right]_0^R

V = \frac{\sigma}{2\epsilon_0}\left(\sqrt{R^2 + x^2} - |x|\right)

Electric Potential Energy

For a system of point charges, the total potential energy is:

U = \frac{1}{2}\sum_{i=1}^{n} q_i V_i

Where $V_i$ is the potential at the location of $q_i$ due to all other charges. For two point charges:

U = \frac{q_1 q_2}{4\pi\epsilon_0 r}

Example: Energy to assemble a square of charges Four charges $q$ are placed at the corners of a square of side $a$ . Find the total potential energy.

There are $\binom{4}{2} = 6$ pairs. Four pairs are side-by-side (distance $a$ ), and two are diagonal (distance $a\sqrt{2}$ ).

U = \frac{4 q^2}{4\pi\epsilon_0 a} + \frac{2 q^2}{4\pi\epsilon_0 a\sqrt{2}} = \frac{q^2}{4\pi\epsilon_0 a}\left(4 + \sqrt{2}\right)

Capacitance

Capacitance is defined as:

C = \frac{Q}{V}

Parallel Plate Capacitor

Two plates of area $A$ separated by distance $d$ :

C = \frac{\epsilon_0 A}{d}

Derivation: The field between infinite parallel plates is $E = \sigma/\epsilon_0 = Q/(\epsilon_0 A)$ . The potential difference is $V = Ed = Qd/(\epsilon_0 A)$ So $C = Q/V = \epsilon_0 A/d$ .

Cylindrical Capacitor

Inner radius $a$ Outer radius $b$ Length $L$ .

By Gauss’s law, the field at radius $r$ ( $a \le r \le b$ ) is $E = \lambda/(2\pi\epsilon_0 r)$ where $\lambda = Q/L$ .

V = -\int_b^a E\, dr = \frac{\lambda}{2\pi\epsilon_0} \int_a^b \frac{dr}{r} = \frac{\lambda}{2\pi\epsilon_0}\ln\frac{b}{a}

C = \frac{Q}{V} = \frac{\lambda L}{V} = \frac{2\pi\epsilon_0 L}{\ln(b/a)}

Spherical Capacitor

Inner radius $a$ Outer radius $b$ .

E = \frac{Q}{4\pi\epsilon_0 r^2} \quad (a \le r \le b)

V = -\int_b^a \frac{Q}{4\pi\epsilon_0 r^2}\, dr = \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right)

C = \frac{Q}{V} = \frac{4\pi\epsilon_0}{\frac{1}{a} - \frac{1}{b}} = \frac{4\pi\epsilon_0 ab}{b - a}

For an isolated sphere ( $b \to \infty$ ): $C = 4\pi\epsilon_0 a$ .

Energy Stored in a Capacitor

U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV

Derivation from work: The work to add charge $dq$ to a capacitor at potential $V = q/C$ is $dU = V\, dq = (q/C)\, dq$ .

U = \int_0^Q \frac{q}{C}\, dq = \frac{Q^2}{2C}

Energy Density of the Electric Field

U_E = \frac{1}{2}\epsilon_0 E^2

For a parallel plate capacitor: $U = \frac{1}{2}\epsilon_0 E^2 \cdot Ad = u_E \cdot \text{volume$ .

Capacitors in Combination

Series: $\dfrac{1}{C_{\text{eq}} = \sum \dfrac{1}{C_i}$

Parallel: $C_{\text{eq} = \sum C_i$

Dielectrics

A dielectric is an insulating material placed between capacitor plates. When an external field $\vec{E}_0$ Is applied, the dielectric polarizes, creating an opposing field $\vec{E}_p$ . The net field is reduced:

E = \frac{E_0}{\kappa}

Where $\kappa$ is the dielectric constant ( $\kappa \ge 1$ ).

Effect on Capacitance

With a dielectric filling the gap:

C = \kappa C_0

Where $C_0$ is the capacitance without the dielectric.

Bound Charge and Gauss’s Law with Dielectrics

The electric displacement field is:

\vec{D} = \epsilon_0 \vec{E} + \vec{P} = \epsilon_0 \kappa \vec{E}

Gauss’s law for $\vec{D}$ :

\oint \vec{D} \cdot d\vec{A} = Q_{\text{free, enc}

Example: Parallel plate capacitor with partial dielectric A parallel plate capacitor has plate area $A$ and separation $d$ . A dielectric of thickness $t < d$ and Constant $\kappa$ is inserted. Find the capacitance.

Treat the gap as two capacitors in series: one with dielectric ( $t$ ) and one air-filled ( $d - t$ ).

\frac{1}{C} = \frac{t}{\kappa \epsilon_0 A} + \frac{d - t}{\epsilon_0 A} = \frac{t + \kappa(d - t)}{\kappa \epsilon_0 A}

C = \frac{\kappa \epsilon_0 A}{\kappa d - (\kappa - 1)t}

Intuition

Electrostatics is about how stationary charges create fields and exert forces. The key idea is that charges don’t act at a distance — they create an electric field that fills space, and other charges respond to that field.

Coulomb’s law intuition: The force between two charges follows an inverse-square law, just like gravity. But unlike gravity, electric charges come in two signs (positive and negative), so the force can be attractive or repulsive. The $1/r^2$ dependence means doubling the distance quarters the force.

Gauss’s law intuition: Gauss’s law says the total electric flux through any closed surface equals the enclosed charge divided by $\epsilon_0$ . It’s always true, but only useful when symmetry lets you pull $E$ out of the integral. For a sphere, use a sphere. For a line, use a cylinder. For a plane, use a flat-ended cylinder. The right Gaussian surface turns a hard integral into simple algebra.

Electric potential intuition: Potential is the “height” of the electric landscape. Positive charges roll downhill (toward lower potential), negative charges roll uphill. The relationship $\vec{E} = -\nabla V$ means the electric field points in the direction of steepest decrease in potential — like water flowing downhill.

Capacitor intuition: A capacitor stores energy in the electric field between its plates. The energy density is $u = \frac{1}{2}\epsilon_0 E^2$ — the field itself carries energy. A dielectric increases capacitance because it polarizes, reducing the internal field and allowing more charge at the same voltage.

Common Pitfalls

Confusing electric field and electric force. $\vec{E} = \vec{F}/q_0$ . The field exists independently of any test charge. The force depends on the charge placed in the field.
Forgetting that Gauss’s law gives the total flux, not the field directly. You must exploit symmetry to pull $E$ out of the integral. Gauss’s law is always true, but it is only useful when symmetry allows you to determine $\vec{E}$ .
Incorrect sign in the potential integral. $V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}$ . The negative sign is essential. When you move against the field, the potential increases.
Confusing potential and potential energy. $U = qV$ . Potential is a property of the field; potential energy depends on the charge placed in the field.
Using the wrong Gaussian surface. Choose the surface that matches the symmetry of the charge distribution. For a point charge or sphere, use a sphere. For a line or cylinder, use a cylinder. For a plane, use a cylinder with flat ends parallel to the plane.
Ignoring conductor behavior in electrostatics. Inside a conductor in equilibrium, $\vec{E} = 0$ and all excess charge resides on the surface. The surface is an equipotential.
Incorrectly handling series and parallel capacitors. In series, the charge on each capacitor is the same. In parallel, the voltage across each capacitor is the same.
Forgetting the factor of $1/2$ in potential energy of a charge distribution. The energy to assemble $n$ charges is $\frac{1}{2}\sum q_i V_i$ Not $\sum q_i V_i$ . Without the factor of $1/2$ each pair is counted twice.

Practice Questions

Three point charges $q_1 = 2\,\mu\text{C$$q_2 = -3\,\mu\text{C$$q_3 = 4\,\mu\text{C$ are placed at the corners of an equilateral triangle of side 0.5 m. Find the net force on $q_1$ .
A uniformly charged rod of length $L = 2$ m carries charge $Q = 8\,\mu\text{C$ . Find the electric field at a point 1 m from one end along the perpendicular bisector of the rod.
A solid insulating sphere of radius $R = 0.1$ m has charge density $\rho = 10^{-6}$ C/m $^3$ . Find the electric field at (a) $r = 0.05$ m and (b) $r = 0.2$ m from the center.
Derive the potential on the perpendicular bisector of a uniformly charged rod of length $L$ and total charge $Q$ .
A spherical capacitor has inner radius 2 cm and outer radius 5 cm. The space between is filled with a dielectric of $\kappa = 3$ . Find the capacitance.
A parallel plate capacitor with $C = 10\,\mu\text{F$ is charged to $100$ V. A dielectric with $\kappa = 4$ is inserted while the battery remains connected. Find the new charge on the plates and the change in stored energy.
A conducting sphere of radius $a$ is surrounded by a conducting spherical shell of inner radius $b$ and outer radius $c$ . The inner sphere has charge $+Q$ and the outer shell has charge $-2Q$ . Find the electric field in all regions and the potential at $r = a$ .
Calculate the work required to bring four charges of $+1\,\mu\text{C$ from infinity to the corners of a square of side 1 m.

Question 9: AP Exam-Style -- Field of a non-uniformly charged cylinder

An infinitely long solid cylinder of radius $R$ has volume charge density $\rho(r) = \rho_0 r/R$ for $0 \le r \le R$ . Find the electric field (a) inside ( $r < R$ ) and (b) outside ( $r > R$ ) the cylinder.

Answer

(a) Inside ( $r < R$ ): Use a cylindrical Gaussian surface of radius $r$ and length $L$ .

$Q_{\text{enc} = \int_0^r \rho(r') \cdot 2\pi r' L\, dr' = \frac{2\pi \rho_0 L}{R} \int_0^r r'^2\, dr' = \frac{2\pi \rho_0 L r^3}{3R}$

$E \cdot 2\pi r L = \frac{Q_{\text{enc}}{\epsilon_0} = \frac{2\pi \rho_0 r^3 L}{3R\epsilon_0}$

$E = \frac{\rho_0 r^2}{3R\epsilon_0}$

(b) Outside ( $r > R$ ):

$Q_{\text{total} = \frac{2\pi \rho_0 R^3 L}{3R} = \frac{2\pi \rho_0 R^2 L}{3}$

$E \cdot 2\pi r L = \frac{2\pi \rho_0 R^2 L}{3\epsilon_0}$

$E = \frac{\rho_0 R^2}{3\epsilon_0 r}$

Question 10: AP Exam-Style -- Potential and field from a charged arc

A thin rod is bent into a semicircular arc of radius $R$ with total charge $Q$ uniformly distributed. Find (a) the electric field at the center of the semicircle and (b) the electric potential at the Center.

Answer

Let the arc span from $\theta = -\pi/2$ to $\theta = \pi/2$ . The linear charge density is $\lambda = Q/(\pi R)$ .

(a) By symmetry, the field points along the axis of symmetry (let us call it the $y$ -direction, with The arc opening to the right). A charge element $dq = \lambda R\, d\theta$ at angle $\theta$ produces:

$dE_y = \frac{1}{4\pi\epsilon_0}\frac{dq}{R^2}\sin\theta = \frac{\lambda}{4\pi\epsilon_0 R}\sin\theta\, d\theta$

$E_y = \frac{\lambda}{4\pi\epsilon_0 R}\int_{-\pi/2}^{\pi/2}\sin\theta\, d\theta = \frac{\lambda}{4\pi\epsilon_0 R}[-\cos\theta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4\pi\epsilon_0 R}(1 - (-1)) = \frac{2\lambda}{4\pi\epsilon_0 R}$

$E = \frac{2Q}{4\pi^2\epsilon_0 R^2} = \frac{Q}{2\pi^2\epsilon_0 R^2}$

The $x$ -components cancel by symmetry.

(b) Every charge element is at distance $R$ from the center:

$V = \frac{1}{4\pi\epsilon_0}\int \frac{dq}{R} = \frac{Q}{4\pi\epsilon_0 R}$

flowchart TD
    A[6_Electrostatics] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the fundamental principles of electrostatics, including the key equations, experimental methods, and applications relevant to the specification.

Key concepts include:

fundamental principles and equations
SI units and dimensional analysis
mathematical modelling of physical phenomena
experimental techniques and measurement
applications to real-world problems

A strong understanding of these principles, combined with regular practice of quantitative problems and past paper questions, is essential for success in examinations.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

Circuits: Electrostatics provides the foundation for understanding how charges flow through circuits.
Magnetism: Electric and magnetic fields are unified in Maxwell’s equations — electrostatics is the static limit.
Work, Energy, and Power: Electric potential energy and work done by electric fields follow the same energy principles.
AP Calculus — Integrals: Electric potential and field from charge distributions are calculated using integrals, applying calculus to electrostatic problems.

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