Electric Field — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Definition of electric field, point charge electric field formula, uniform electric fields, superposition principle, electric field symmetry, and Gauss’s law applications for symmetric charge distributions.

You should already know: Coulomb’s law for electric force between point charges. Vector addition for multiple vector quantities. Basic flux concepts from earlier units.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Electric Field?

Electric field is a vector field that describes the force a test charge would experience at any point in space, independent of the properties of the test charge itself. The AP Physics 2 CED lists this sub-topic as accounting for approximately 4-6% of the total exam, making it one of the core components of Unit 3. Electric field concepts appear regularly in both multiple-choice (MCQ) and free-response (FRQ) sections: most often as standalone MCQ testing superposition or field direction, or as the foundation for larger FRQ connecting to potential, capacitors, or point charge motion. The core definition comes from electric force: if a test charge $q_0$ experiences an electric force ${\vec{F}}_e$ at a point, the electric field at that point is defined as $\vec{E}=\frac{{\vec{F}}_e}{q_0}$. The SI unit is newtons per coulomb (N/C), which is equivalent to volts per meter (V/m), a unit you will use more often when working with electric potential. Unlike electric force, which depends on the charge of the object interacting with the field, electric field is a fundamental property of the source charge distribution that created it. This abstraction allows us to analyze electrostatic interactions even when we do not know the size of the test charge, which is foundational for all further work in electrostatics and DC circuits.

2. Point Charge Electric Fields and Superposition

From Coulomb’s law, the force on a test charge $q_0$ from a source point charge $Q$ at distance $r$ is $F_e=\frac{k|Q{q}_0|}{r^2}$. Dividing by $q_0$ to isolate the field (the force per unit charge) gives the magnitude of the electric field from a point charge: $$E=\frac{k|Q|}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{|Q|}{r^2}$$ where $k=8.988\times 10^9{\text{ Ncdotpm}}^2/{\text{C}}^2\approx 9\times 10^9$ for AP problems, and ${ϵ}_0$ is the permittivity of free space. The direction of $\vec{E}$ follows a simple rule: it points away from positive source charges (since a positive test charge would be repelled) and towards negative source charges (since a positive test charge would be attracted). For multiple point charges, the total electric field at a point is the vector sum of the electric fields from each individual charge, a rule called the superposition principle. Because electric field is a vector, you must always break fields into x and y components before adding, then recombine to get the magnitude and direction of the net field.

Worked Example

Two point charges are placed on the x-axis: $Q_1=+2\text{ nC}$ at $x=0$, and $Q_2=-8\text{ nC}$ at $x=3\text{ m}$. Find the magnitude and direction of the net electric field at $x=1\text{ m}$.

1. Calculate distance from each charge to the point of interest: For $Q_1$, $r_1=|1-0|=1\text{ m}$. For $Q_2$, $r_2=|3-1|=2\text{ m}$.
2. Find the magnitude and direction of each individual field: $E_1=\frac{k{Q}_1}{r_1^2}=\frac{(9\times 10^9)(2\times 10^{-9})}{1^2}=18\text{ N/C}$. Since $Q_1$ is positive, the field points right (+x direction) away from $Q_1$. $E_2=\frac{k|Q_2|}{r_2^2}=\frac{(9\times 10^9)(8\times 10^{-9})}{2^2}=18\text{ N/C}$. Since $Q_2$ is negative, the field points towards $Q_2$, which is also +x direction here.
3. Add the fields: Both are along the +x axis, so $E_{net}=E_1+E_2=18+18=36\text{ N/C}$.
4. The net field points in the +x direction along the axis.

Exam tip: On AP MCQ, you can often eliminate wrong options just by checking direction before calculating magnitude. Always assign directions based on source charge sign first to cut down computation time.

3. Uniform Electric Fields

A uniform electric field has the same magnitude and direction at all points in the region of interest. The most common example tested on the AP exam is the field between two infinitely large parallel charged plates with equal and opposite charge, separated by a distance $d$, with potential difference $\Delta V$ across them. For an infinite charged plate, the electric field magnitude is constant, so the fields from the positive and negative plates add between the plates and cancel outside, resulting in a uniform field with magnitude: $$E=\frac{\Delta V}{d}$$ This is one of the most frequently used formulas on the AP exam, especially for problems connecting to electric potential, capacitors, and charged particle motion. The direction of the uniform field always points from the positively charged plate towards the negatively charged plate, moving from high potential to low potential. A charged particle placed in a uniform electric field experiences a constant electric force $F=qE$, so it has constant acceleration, meaning you can use kinematics to analyze its motion just like you would for a mass in a uniform gravitational field.

Worked Example

Two parallel plates are separated by 2 cm, with a 120 V battery connected across them. An electron (mass $m_e=9.11\times 10^{-31}\text{ kg}$, charge $e=1.6\times 10^{-19}\text{ C}$) starts from rest at the negative plate and accelerates towards the positive plate. What is its acceleration magnitude?

1. Convert units to SI: $d=2\text{ cm}=0.02\text{ m}$.
2. Calculate electric field strength between the plates: $E=\frac{\Delta V}{d}=\frac{120\text{ V}}{0.02\text{ m}}=6000\text{ V/m}=6000\text{ N/C}$.
3. Calculate the magnitude of the electric force on the electron: $F=|q|E=(1.6\times 10^{-19}\text{ C})(6000\text{ N/C})=9.6\times 10^{-16}\text{ N}$.
4. Use Newton’s second law to find acceleration: $a=\frac{F}{m_e}=\frac{9.6\times 10^{-16}}{9.11\times 10^{-31}}\approx 1.05\times 10^{15}{\text{ m/s}}^2$.

Exam tip: Remember that $E=\Delta V/d$ only applies to uniform electric fields. Never use this formula for non-uniform fields from point charges, as the potential varies non-linearly with distance there.

4. Gauss's Law for Symmetric Charge Distributions

Gauss’s law relates the net electric flux through a closed Gaussian surface to the net charge enclosed by that surface. Electric flux ${\Phi }_E$ measures the number of electric field lines passing through a closed surface, and for cases where the electric field is constant and perpendicular to the surface at all points, ${\Phi }_E=EA$, where $A$ is the total surface area of the Gaussian surface. Gauss’s law is written as: $${\Phi }_E=\frac{Q_{enclosed}}{{ϵ}_0}$$ Gauss’s law greatly simplifies calculating electric fields for highly symmetric charge distributions (spherical, infinite line, infinite plate) that would be tedious to sum via superposition. A key result of Gauss’s law that is heavily tested on the AP exam is that the electric field inside any conductor at electrostatic equilibrium is always zero, because all excess charge resides on the outer surface of the conductor, so the enclosed charge inside the conductor material is zero.

Worked Example

A solid insulating sphere of radius $R=0.1\text{ m}$ has a total charge of $+1\mu C$ uniformly distributed throughout its volume. Use Gauss's law to find the electric field at a distance $r=0.05\text{ m}$ from the center of the sphere.

1. Choose a spherical Gaussian surface of radius $r=0.05\text{ m}$ concentric with the insulating sphere. By symmetry, the electric field is constant and perpendicular to this surface everywhere, so total flux is ${\Phi }_E=E(4\pi r^2)$.
2. Calculate the enclosed charge: The charge is uniformly distributed, so enclosed charge scales with volume. $Q_{enclosed}=Q_{total}\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=Q_{total}\frac{r^3}{R^3}$.
3. Apply Gauss’s law: $E(4\pi r^2)=\frac{Q_{total}{r}^3}{{ϵ}_0{R}^3}$.
4. Simplify and substitute values: $E=\frac{Q_{total}r}{4\pi {ϵ}_0{R}^3}=k\frac{Q_{total}r}{R^3}=(9\times 10^9)\frac{(1\times 10^{-6})(0.05)}{(0.1{)}^3}=4.5\times 10^5\text{ N/C}$, directed radially outward from the center.

Exam tip: Gauss's law only gives you the electric field due to the enclosed charge. The electric field at a point inside an insulating charge distribution is not zero unless the enclosed charge is zero; this zero-field rule only applies to the interior of conductors at equilibrium.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Assigning the direction of the electric field based on the sign of the test charge instead of the source charge. Why: Students confuse the definition $\vec{E}=\vec{F}/q_0$, so they flip direction when they use a negative test charge. Correct move: Always remember electric field is a property of the source, so direction is defined for a positive test charge: away from positive sources, towards negative sources regardless of the actual test charge.
• Wrong move: Adding electric field magnitudes as scalars when the fields point in opposite directions. Why: Students forget electric field is a vector and just add magnitudes, leading to double the correct value. Correct move: Always assign positive/negative signs to directions along each axis, then add the signed components of the field before taking the magnitude of the resultant.
• Wrong move: Using $E=kQ/r^2$ for a point inside a uniformly charged insulating sphere. Why: Students memorize the point charge formula and apply it everywhere, forgetting the enclosed charge is less than the total charge inside an insulating sphere. Correct move: For uniformly charged insulating spheres, use $E=kQr/R^3$ inside the sphere and $E=kQ/r^2$ only outside.
• Wrong move: Claiming the electric field is zero inside any hollow sphere, regardless of whether the sphere is conducting or insulating. Why: Students confuse the conducting sphere result (zero field inside) with hollow insulating spheres that have charge distributed on the surface or throughout the volume. Correct move: The electric field is only zero inside a conducting sphere (or any conductor at equilibrium), where all charge moves to the outer surface. For hollow insulating spheres, calculate enclosed charge with Gauss's law to find the field.
• Wrong move: Using $E=\Delta V/d$ to find the electric field between two point charges. Why: Students mix up uniform and non-uniform field formulas. Correct move: Use $E=\Delta V/d$ only for uniform fields between parallel plates, and superposition of point charge fields for point charge systems.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

Three identical positive point charges $+Q$ are placed at three corners of a square with side length $s$. What is the direction of the net electric field at the empty corner of the square? A) Parallel to the side between the two adjacent charges, pointing towards the empty corner B) Along the diagonal of the square, pointing away from the center of the square towards the empty corner C) Along the diagonal of the square, pointing towards the center of the square from the empty corner D) Perpendicular to the diagonal, pointing parallel to the side opposite the empty corner

Worked Solution: Each positive charge produces an electric field pointing away from the charge at the empty corner. The two adjacent charges produce fields whose components perpendicular to the diagonal cancel, leaving only a component along the diagonal pointing away from the center. The third charge at the opposite corner also produces a field pointing along the diagonal in the same direction, away from itself towards the empty corner. All perpendicular components cancel, so the net field is along the diagonal pointing away from the center. Options A, C, and D are incorrect. The correct answer is B.

Question 2 (Free Response)

A long, straight conducting wire carries a uniform linear charge density $\lambda$ (charge per unit length, in C/m). (a) Use Gauss's law to derive an expression for the electric field at a distance $r$ from the center of the wire, for $r$ greater than the radius of the wire. State the direction of the field for positive $\lambda$. (b) If the wire has a linear charge density of $+2.5\times 10^{-6}\text{ C/m}$, calculate the electric field strength at a distance of 0.5 m from the wire. (c) A small point charge of mass $1.0\times 10^{-4}\text{ kg}$ and charge $+3.0\times 10^{-9}\text{ C}$ is placed 0.5 m from the wire. What magnitude acceleration will it experience? Is the acceleration towards or away from the wire?

Worked Solution: (a) By symmetry, the electric field is radial (perpendicular to the wire) and constant in magnitude at a fixed distance $r$. We choose a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the wire. Flux through the end caps is zero (E is parallel to the end caps), so all flux passes through the curved surface: ${\Phi }_E=E(2\pi rL)$. Enclosed charge is $Q_{enclosed}=\lambda L$. Apply Gauss's law: $E(2\pi rL)=\frac{\lambda L}{{ϵ}_0}$, so $E=\frac{\lambda }{2\pi {ϵ}_{0}r}=\frac{2k\lambda }{r}$. For positive $\lambda$, the field points radially outward. (b) Substitute values: $E=\frac{2(9\times 10^9)(2.5\times 10^{-6})}{0.5}=9.0\times 10^4\text{ N/C}$. (c) Force $F=qE=(3.0\times 10^{-9})(9.0\times 10^4)=2.7\times 10^{-4}\text{ N}$. Acceleration $a=\frac{F}{m}=\frac{2.7\times 10^{-4}}{1.0\times 10^{-4}}=2.7{\text{ m/s}}^2$. Both charges are positive, so the force is repulsive, and acceleration is away from the wire.

Question 3 (Application / Real-World Style)

In a laser printer, a uniform electric field is used to deflect charged ink particles towards the paper. The electric field in the deflection region has a strength of $1.2\times 10^5\text{ N/C}$ and is oriented vertically downward. A typical ink particle has a mass of $1.2\times 10^{-10}\text{ kg}$ and a charge of $-2.0\times 10^{-13}\text{ C}$. What is the net acceleration (magnitude and direction) of the ink particle in this field, accounting for gravity?

Worked Solution: The electric field points downward, so the force on the negative charge points upward. Magnitude of electric force: $F_e=|q|E=(2.0\times 10^{-13})(1.2\times 10^5)=2.4\times 10^{-8}\text{ N}$. Gravitational force points downward: $F_g=mg=(1.2\times 10^{-10})(9.8)\approx 1.18\times 10^{-9}\text{ N}$. Net force: $F_{net}=F_e-F_g=2.28\times 10^{-8}\text{ N}$ upward. Acceleration: $a=\frac{F_{net}}{m}=\frac{2.28\times 10^{-8}}{1.2\times 10^{-10}}=190{\text{ m/s}}^2$ upward. In context, this means the electric field easily overcomes gravity to deflect the negatively charged ink particles toward the positively charged paper, as required for printing.

7. Quick Reference Cheatsheet

8. What's Next

Electric field is the foundational concept for all further electrostatics in AP Physics 2. Next, you will connect electric field to electric potential and potential energy, using the relationship between E and V to map potential landscapes for different charge distributions. Without mastering the vector nature of electric field and how to calculate E for different charge distributions, you will not be able to correctly analyze potential, capacitance, or electric circuits later in the course. Electric field also is the basis for understanding electrostatic equilibrium of conductors, a common AP exam topic that connects to many other electrostatics concepts. In the bigger picture, electric field is one half of the electromagnetic field that underpins all of modern physics, from electromagnetism to quantum mechanics.

Next topics to study:

• Electric Potential
• Gauss's Law Applications
• Capacitance and Dielectrics
• Electric Force