Unit 1: Electrostatics

15–25% of AP Exam · 7 Topics · ~25 Practice Questions

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1.1 Electric Charge & Coulomb's Law

Electric charge is a fundamental property of matter. There are two types: positive and negative. Like charges repel; opposite charges attract. Charge is quantized in units of the elementary charge $e = 1.60 \times 10^{-19} \text{ C}$ and is conserved — the net charge of an isolated system never changes.

Conductors (metals, ionic solutions) allow charge to flow freely because they have mobile electrons. Insulators (glass, rubber, plastic) bind electrons tightly, preventing macroscopic charge flow.

Coulomb's Law:

$$F = k \frac{|q_1 q_2|}{r^2}, \quad k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9 \text{ N·m}^2/\text{C}^2$$ $$\vec{F}_{12} = k \frac{q_1 q_2}{r^2} \hat{r}_{12}$$

Coulomb's law gives the force between two point charges. The force is directed along the line connecting them: attractive for opposite signs, repulsive for like signs. It obeys superposition: the net force on a charge is the vector sum of forces from all other charges individually.

Exam Tip When using superposition, calculate each force vector separately using $\vec{F} = k\frac{q_1 q_2}{r^2}\hat{r}$, then sum component-wise. Don't add magnitudes directly — forces are vectors.

Common Mistake The $\hat{r}$ in Coulomb's law points from the source charge to the test charge. Getting the direction wrong is the most common error on AP free-response.

Practice

Q1.1: Net Force on a Charge

Three point charges lie along the x-axis: $q_1 = +3.0\ \mu\text{C}$ at $x = 0$, $q_2 = -5.0\ \mu\text{C}$ at $x = 2.0\ \text{cm}$, and $q_3 = +4.0\ \mu\text{C}$ at $x = 5.0\ \text{cm}$. Find the net electric force (magnitude and direction) on $q_2$.

Show Solution

Force from $q_1$ on $q_2$: $F_{12} = k\frac{|q_1 q_2|}{r_{12}^2} = (8.99\times10^9)\frac{(3.0\times10^{-6})(5.0\times10^{-6})}{(0.020)^2} = 337\ \text{N}$ (attractive, toward $q_1$: $-x$ direction)

Force from $q_3$ on $q_2$: $F_{32} = k\frac{|q_3 q_2|}{r_{32}^2} = (8.99\times10^9)\frac{(4.0\times10^{-6})(5.0\times10^{-6})}{(0.030)^2} = 200\ \text{N}$ (attractive, toward $q_3$: $+x$ direction)

Net force: $F_{\text{net}} = +200 - 337 = -137\ \text{N}$ → 137 N in the −x direction (toward q₁)

Practice

Q1.2: Equilibrium Position

Two point charges $+Q$ and $+4Q$ are fixed 12 cm apart. Where should a third charge be placed so that it experiences zero net electric force? Does the sign of the third charge matter?

Show Solution

For zero net force: $k\frac{Qq}{x^2} = k\frac{4Qq}{(12-x)^2}$ where $x$ is distance from $+Q$.

Cancel $kQq$: $\frac{1}{x^2} = \frac{4}{(12-x)^2} \implies (12-x)^2 = 4x^2 \implies 12-x = 2x \implies x = 4\ \text{cm}$

4.0 cm from +Q, on the line between them. Any sign works. Forces scale with charge product — both flip direction but still cancel.

1.2 Electric Fields

The electric field $\vec{E}$ is the force per unit charge on a small positive test charge:

$$\vec{E} = \frac{\vec{F}}{q} \quad \text{or} \quad \vec{F} = q\vec{E}$$

For a point charge $Q$: $\vec{E} = k\frac{Q}{r^2}\hat{r}$. Field points away from + charges, toward − charges. Units: N/C = V/m.

Electric field lines start on + charges, end on − charges, never cross, denser where field is stronger. The $\vec{E}$ vector is tangent to field lines.

Key field configurations:

Point charge: radial, $E \propto 1/r^2$
Dipole: two equal opposite charges; $E \propto 1/r^3$ at large distances
Uniform field: parallel lines, constant magnitude (between parallel plates)
Infinite line: $E = \frac{2k\lambda}{r}$, radial, $E \propto 1/r$
Infinite sheet: $E = \frac{\sigma}{2\varepsilon_0}$, uniform, perpendicular to sheet

Exam Tip Know how to sketch field lines for point charges, dipoles, and parallel plates. Line density = field strength.

Practice

Q1.3: Field from Two Charges

Charges $+q$ and $-2q$ are fixed 6.0 cm apart. Find $\vec{E}$ at the midpoint.

Show Solution

At midpoint, $r = 3.0\ \text{cm}$. Field from $+q$ points away (toward $-2q$). Field from $-2q$ points toward it (same direction!).

$E_{\text{net}} = k\frac{q}{(0.03)^2} + k\frac{2q}{(0.03)^2} = k\frac{3q}{0.0009} = \frac{3kq}{9\times10^{-4}}$

$\vec{E} = \frac{3kq}{(0.03)^2}$ directed toward the −2q charge.

1.3 Gauss's Law

$$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

Gauss's law: net electric flux through any closed surface equals enclosed charge divided by $\varepsilon_0$.

Strategy: (1) Choose Gaussian surface matching the symmetry. (2) Ensure $E$ is constant and ⟂ or parallel on each face. (3) Find $Q_{\text{enc}}$. (4) Solve.

Standard results (must memorize):

Spherical shell: $E = 0$ inside ($rR$)
Solid sphere (uniform ρ): $E = kQr/R^3$ inside; $E = kQ/r^2$ outside
Infinite line: $E = \lambda/(2\pi\varepsilon_0 r) = 2k\lambda/r$
Infinite sheet: $E = \sigma/(2\varepsilon_0)$ — independent of distance!
Parallel plates: $E = \sigma/\varepsilon_0$ between; $E = 0$ outside

Common Mistake Gauss's law always holds but is only useful when symmetry exists (spherical, cylindrical, planar). Don't force it on asymmetric distributions.

Practice

Q1.4: Gauss — Solid Sphere

A solid insulating sphere of radius $R = 5.0\ \text{cm}$ has uniform $\rho = 2.0\ \mu\text{C/m}^3$. Find $E$ at (a) $r = 3.0\ \text{cm}$ and (b) $r = 8.0\ \text{cm}$.

Show Solution

(a) Inside: $E(4\pi r^2) = \frac{\rho(4/3)\pi r^3}{\varepsilon_0} \implies E = \frac{\rho r}{3\varepsilon_0} = \frac{(2\times10^{-6})(0.03)}{3(8.85\times10^{-12})} = 2.26 \times 10^3\ \text{N/C}$

(b) Outside: $Q_{\text{tot}} = \rho(4/3)\pi R^3 = 1.047 \times 10^{-9}\ \text{C}$; $E = kQ/r^2 = 1.47 \times 10^3\ \text{N/C}$

Practice

Q1.5: Conducting Shell + Point Charge

Point charge $+Q$ at center of a neutral thick conducting shell (inner radius $a$, outer $b$). Find $E$ everywhere and induced charges.

Show Solution

$r < a$: $E = kQ/r^2$ (point charge field). $a < r < b$: $E = 0$ (inside conductor). $r > b$: $E = kQ/r^2$ (net charge = +Q).

Induced: −Q on inner surface (r=a), +Q on outer surface (r=b).

1.4 Electric Potential & Energy

Electric potential energy of two point charges: $U = k\frac{q_1 q_2}{r} = qV$

Electric potential (voltage) = potential energy per unit charge: $V = U/q$

For a point charge: $V = k\frac{Q}{r}$ (with $V = 0$ at $r \to \infty$)

Potentials add as scalars: $V_{\text{total}} = \sum_i k\frac{q_i}{r_i}$ (unlike fields, which add as vectors)

Electron-Volt (eV)

The electron-volt is the energy gained/lost when one elementary charge moves through 1 volt of potential difference:

$$1\text{ eV} = (1.60 \times 10^{-19}\text{ C})(1\text{ V}) = 1.60 \times 10^{-19}\text{ J}$$

eV is the natural energy unit for atomic/particle physics. To convert: multiply eV by e to get joules. Common conversions: 1 keV = 10³ eV, 1 MeV = 10⁶ eV.

Work-Energy Theorem

Energy conservation for a charge in an electric field:

$$\Delta K + \Delta U = 0 \quad \Rightarrow \quad \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = -q\Delta V$$

A positive charge accelerates toward lower potential (V decreases). A negative charge accelerates toward higher potential. Work done by the field: $W = -\Delta U = -q\Delta V$.

Equipotential Surfaces

Key properties:

$\vec{E}$ is always perpendicular to equipotential surfaces
No work is done moving charge along an equipotential ($\Delta V = 0$)
Closer equipotential spacing = stronger E-field
Conductors in equilibrium are equipotential volumes
Point charge: equipotentials are concentric spheres
Uniform field: equipotentials are parallel planes

Relationship between E and V

$E_x = -dV/dx$ or $\vec{E} = -\nabla V$. E points toward steepest decrease of V.

From potential energy: $\vec{F} = -\nabla U = -q\nabla V$. Force on a charge always points toward lower potential energy.

For a uniform field: $E = \frac{|\Delta V|}{d}$ (field magnitude = potential difference / separation).

Potential of Continuous Distributions

For continuous charge distributions, integrate the scalar potential:

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

Key results:

Charged ring (on axis): $V = \frac{kQ}{\sqrt{x^2+a^2}}$
Charged disk (on axis): $V = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{R^2+x^2} - |x|\right)$
Line of charge (off-axis): $V = \frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{\text{ref}}{r}\right)$

Strategy: compute V first (scalar integral), then differentiate to find E: $\vec{E} = -\nabla V$.

Exam Tip Finding E from V(x) via $E_x = -dV/dx$ is a frequent free-response task. Practice differentiating potential functions. Also: when asked for speed of a charged particle, use $\frac{1}{2}mv^2 = |q\Delta V|$.

Practice

Q1.6: Potential at Midpoint

$q_1 = +5.0\ \mu\text{C}$ and $q_2 = -3.0\ \mu\text{C}$ are 4.0 cm apart. Find V at the midpoint.

Show Solution

$r = 2.0\ \text{cm}$. $V_1 = k(5\times10^{-6})/0.02 = 2.25\times10^6\ \text{V}$; $V_2 = k(-3\times10^{-6})/0.02 = -1.35\times10^6\ \text{V}$

$V_{\text{net}} = 9.0 \times 10^5\ \text{V}$

Practice

Q1.7: E from V(x)

$V(x) = 3x^2 - 12x + 5$ volts (x in meters). Find $\vec{E}(x)$ and where $E = 0$.

Show Solution

$E_x = -dV/dx = -(6x - 12) = 12 - 6x$. $E = 0$ at $x = 2.0\ \text{m}$.

Practice

Q1.7b: Electron Accelerated Through Potential Difference

An electron ($m_e = 9.11\times10^{-31}\ \text{kg}$) is accelerated from rest through 500 V. Find its final speed in m/s and the energy gained in eV.

Show Solution

Energy gained: $\Delta K = |q\Delta V| = e(500) = 500\ \text{eV} = 8.0\times10^{-17}\ \text{J}$.

$\frac{1}{2}mv^2 = 8.0\times10^{-17} \implies v = \sqrt{\frac{2(8.0\times10^{-17})}{9.11\times10^{-31}}} = 1.33\times10^7\ \text{m/s}$ (about 4.4% of c).

Practice

Q1.7c: Work to Assemble Charges

Three point charges $+q$, $+q$, and $-q$ are brought from infinity to the vertices of an equilateral triangle of side a. How much work is required to assemble this configuration? (Express in terms of k, q, a)

Show Solution

Bring first +q: W₁ = 0 (no other charges present).

Bring second +q: W₂ = kq²/a.

Bring -q: it interacts with BOTH +q charges: W₃ = -kq²/a + (-kq²/a) = -2kq²/a.

W_total = 0 + kq²/a - 2kq²/a = -kq²/a. Negative work means the system releases energy — the configuration is bound.

1.5 Continuous Charge Distributions

For continuous distributions: $dq = \lambda\ d\ell$ (linear), $dq = \sigma\ dA$ (surface), $dq = \rho\ dV$ (volume).

$$d\vec{E} = k\frac{dq}{r^2}\hat{r}, \quad \vec{E} = \int k\frac{dq}{r^2}\hat{r} \qquad dV = k\frac{dq}{r}, \quad V = \int k\frac{dq}{r}$$

Key results:

Charged rod: $E = k\frac{Q}{x(x+L)}$ on axis
Charged ring: $E = k\frac{Qx}{(x^2+a^2)^{3/2}}$, $V = k\frac{Q}{\sqrt{x^2+a^2}}$ on axis
Charged disk: $E = \frac{\sigma}{2\varepsilon_0}\left[1 - \frac{x}{\sqrt{x^2+R^2}}\right]$ on axis

Exam Tip Always exploit symmetry. For a ring, perpendicular components cancel. Use V (scalar integral) when computing field direction is hard — then use $\vec{E} = -\nabla V$.

Practice

Q1.8: Ring of Charge

A ring of radius $R = 10\ \text{cm}$ carries $Q = 5.0\ \mu\text{C}$. Find $E$ and $V$ at a point 20 cm along the axis.

Show Solution

$E_x = k\frac{Qx}{(x^2+R^2)^{3/2}} = 8.04\times10^5\ \text{N/C}$. $V = k\frac{Q}{\sqrt{x^2+R^2}} = 2.01\times10^5\ \text{V}$.

1.6 Conductors in Equilibrium

When a conductor is in electrostatic equilibrium:

$\vec{E} = 0$ inside (otherwise free charges would move)
All excess charge on the surface
$\vec{E}$ at surface is ⟂ with magnitude $E = \sigma/\varepsilon_0$
Charge density highest at sharpest curvature
Entire conductor is equipotential

Charge distribution on irregular conductors: Surface charge density $\sigma$ is inversely proportional to the local radius of curvature: $\sigma \propto 1/R$. Sharp points accumulate high charge density and produce strong local fields — this is how lightning rods work.

Shielding: E = 0 inside a cavity within a conductor, regardless of external fields.

Induction: A charged object redistributes charge in a conductor. Grounding + removing ground before removing the charged object leaves net opposite charge.

Common Mistake E = 0 inside a conductor ≠ V = 0. V is constant inside, equal to the surface potential.

Practice

Q1.9: Conducting Sphere

A solid conducting sphere of radius $R$ carries $+Q$. Describe charge location, E(r), V(r). Sketch both.

Show Solution

Charge all on surface. $E = 0$ ($rR$). $V = kQ/R$ ($r\le R$); $V = kQ/r$ ($r>R$). Identical to point charge for $r>R$.

1.7 Capacitors

Capacitance: $C = Q/V$ (farads). Parallel-plate: $C = \varepsilon_0 A/d$. Field: $E = \sigma/\varepsilon_0 = V/d$.

Energy Stored in a Capacitor

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

Energy density of the electric field:

$$u_E = \frac{U}{\text{volume}} = \frac{1}{2}\varepsilon_0 E^2$$

This is a general result — the energy per unit volume stored in ANY electric field. For a parallel-plate capacitor, $U = u_E \cdot Ad = \frac{1}{2}\varepsilon_0 E^2 Ad$.

Dielectric (κ)

Dielectric (κ): $C = \kappa C_0$. With fixed Q (battery disconnected): V drops to $V_0/\kappa$, E fields weaken, U decreases (energy goes into work pulling dielectric in).

With fixed V (battery connected): Q increases to $\kappa Q_0$, more energy is stored, battery supplies additional charge.

Combinations

Series: $1/C_{\text{eq}} = \sum 1/C_i$ (same Q). Parallel: $C_{\text{eq}} = \sum C_i$ (same V).

Critical Series: same CHARGE. Parallel: same VOLTAGE. Reversing these is the #1 capacitor error.

Practice

Q1.10: Parallel-Plate Capacitor

Plates area $200\ \text{cm}^2$, separation $0.50\ \text{mm}$, 12 V battery. Find C, Q, E, U.

Show Solution

$C = \varepsilon_0 A/d = 354\ \text{pF}$; $Q = CV = 4.25\ \text{nC}$; $E = V/d = 2.4\times10^4\ \text{V/m}$; $U = \frac{1}{2}CV^2 = 2.55\times10^{-8}\ \text{J}$

Practice

Q1.11: Capacitor Network

$C_1=2.0$, $C_2=3.0$, $C_3=6.0\ \mu\text{F}$ with 24 V. $C_2 \parallel C_3$, then in series with $C_1$. Find $C_{\text{eq}}$ and Q, V on each.

Show Solution

$C_{23} = 9.0\ \mu\text{F}$. $C_{\text{eq}} = 1.636\ \mu\text{F}$. $Q_{\text{total}} = 39.3\ \mu\text{C} = Q_1$. $V_1 = 19.65\ \text{V}$. $V_{23} = 4.35\ \text{V}$. $Q_2 = 13.1\ \mu\text{C}$, $Q_3 = 26.1\ \mu\text{C}$.

Practice

Q1.12: Dielectric — Battery Disconnected

$C_0 = 10\ \mu\text{F}$ charged to 9 V, then disconnected. Dielectric $\kappa = 4$ inserted. Find new C, V, Q, U.

Show Solution

Q fixed at $90\ \mu\text{C}$. $C = 40\ \mu\text{F}$. $V = Q/C = 2.25\ \text{V}$ (drops!). $U = Q^2/(2C) = 1.01\times10^{-4}\ \text{J}$ (decreased — energy went into mechanical work pulling the dielectric in).

Practice

Q1.12b: Energy Density

A parallel-plate capacitor with A = 500 cm², d = 0.20 mm is charged to 100 V. Find: (a) the energy stored and (b) the energy density u_E inside.

Show Solution

$C = \varepsilon_0 A/d = (8.85\times10^{-12})(0.05)/(2\times10^{-4}) = 2.21\ \text{nF}$. $U = \frac{1}{2}CV^2 = 1.11\times10^{-5}\ \text{J}$.

$E = V/d = 100/(2\times10^{-4}) = 5\times10^5\ \text{V/m}$. $u_E = \frac{1}{2}\varepsilon_0 E^2 = 0.5(8.85\times10^{-12})(2.5\times10^{11}) = 1.11\ \text{J/m}^3$.

Check: $U = u_E \times Ad = 1.11 \times (0.05)(2\times10^{-4}) = 1.11\times10^{-5}\ \text{J}$. ✓

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