Electric Field

AP Physics 2· AP Physics 2 CED — Electric Force, Field, and Potential· 14 min read

1. Definition of Electric Field★★☆☆☆⏱ 3 min

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. This sub-topic makes up 4-6% of the AP Physics 2 exam, appearing in both multiple-choice and free-response questions as a standalone concept or foundation for larger problems connecting to potential, capacitors, and charged particle motion.

📘 Definition

Electric Field

$E$

If a test charge $q_{0}$ experiences an electric force $F_{e}$ at a point, the electric field at that point is defined as $E = \frac{F _{e}}{q _{0}}$ . It is a property of the source charge distribution, independent of the test charge used to measure it.

Example:

A 1 C test charge experiencing 5 N of electric force is in a 5 N/C electric field.

The SI unit of electric field is newtons per coulomb (N/C), which is equivalent to volts per meter (V/m), the unit more commonly used when working with electric potential. This abstraction of field allows us to analyze electrostatic interactions without knowing the size of the test charge, and is foundational for all further work in electrostatics and DC circuits.

2. Point Charge Fields and Superposition★★★☆☆⏱ 4 min

From Coulomb's law, we can derive the electric field from a point source charge $Q$ by dividing the force on a test charge by the test charge magnitude.

E = \frac{k|Q|}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{|Q|}{r^2}

Where $k \approx 9 \times 1 0^{9} N \cdotp m^{2} / C^{2}$ for AP problems, and $ϵ_{0}$ is the permittivity of free space. The direction of $E$ follows a simple rule: it points away from positive source charges (a positive test charge is repelled) and towards negative source charges (a positive test charge is attracted).

For multiple point charges, the total electric field at a point is the vector sum of the individual electric fields from each charge, a rule called the superposition principle. Because electric field is a vector, you must break fields into components before adding, then recombine to get the net field's magnitude and direction.

📐 Worked Example

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

1
Calculate distance from each charge to the point of interest:
2
$r_1 = |1 - 0| = 1 \text{ m}, \quad r_2 = |3 - 1| = 2 \text{ m}$
3
Find the magnitude and direction of each individual field:
4
$E_1 = \frac{k Q_1}{r_1^2} = \frac{(9 \times 10^9)(2 \times 10^{-9})}{1^2} = 18 \text{ N/C}$
5
Since $Q_{1}$ is positive, the field points right (+x direction) away from $Q_{1}$ . Next, for $Q_{2}$ :
6
$E_2 = \frac{k |Q_2|}{r_2^2} = \frac{(9 \times 10^9)(8 \times 10^{-9})}{2^2} = 18 \text{ N/C}$
7
Since $Q_{2}$ is negative, the field points towards $Q_{2}$ , which is also the +x direction here.
8
Add the fields: both are along the +x axis, so net field is:
9
$E_{net} = E_1 + E_2 = 18 + 18 = 36 \text{ N/C}$
10
The net field points in the +x direction along the x-axis.

3. Uniform Electric Fields★★★☆☆⏱ 3 min

A uniform electric field has the same magnitude and direction at all points in a region. The most common AP exam example is the field between two parallel charged plates with equal and opposite charge, separated by distance $d$ , with potential difference $Δ V$ across them.

For an infinite charged plate, the electric field magnitude is constant. Fields from the two plates add between the plates and cancel outside, resulting in a uniform field with magnitude:

E = \frac{\Delta V}{d}

Direction of the uniform field always points from the positively charged plate to the negatively charged plate, moving from high potential to low potential. A charged particle in a uniform electric field experiences a constant force $F = q E$ , so it has constant acceleration, allowing you to use kinematics to analyze its motion, similar to 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 $m_{e} = 9.11 \times 1 0^{- 31} kg$ , charge $e = 1.6 \times 1 0^{- 19} C$ ) starts from rest at the negative plate and accelerates towards the positive plate. What is its acceleration magnitude?

1
Convert separation to SI units:
2
$d = 2 \text{ cm} = 0.02 \text{ m}$
3
Calculate electric field strength between the plates:
4
$E = \frac{\Delta V}{d} = \frac{120 \text{ V}}{0.02 \text{ m}} = 6000 \text{ N/C}$
5
Calculate the magnitude of the electric force on the electron:
6
$F = |q| E = (1.6 \times 10^{-19} \text{ C})(6000 \text{ N/C}) = 9.6 \times 10^{-16} \text{ N}$
7
Use Newton's second law to find acceleration:
8
$a = \frac{F}{m_e} = \frac{9.6 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 1.05 \times 10^{15} \text{ m/s}^2$

4. Gauss's Law for Symmetric Charge Distributions★★★★☆⏱ 4 min

Gauss's law relates the net electric flux through a closed Gaussian surface to the net charge enclosed by that surface. Electric flux $Φ_{E}$ measures the number of electric field lines passing through the closed surface. For symmetric cases where $E$ is constant and perpendicular to the entire surface, $Φ_{E} = E A$ , where $A$ is the total surface area of the Gaussian surface.

\Phi_E = \frac{Q_{enclosed}}{\epsilon_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 AP-tested result is that the electric field inside any conductor at electrostatic equilibrium is always zero, because all excess charge resides on the outer surface, so the enclosed charge inside the conductor material is zero.

📐 Worked Example

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

1
Choose a spherical Gaussian surface of radius $r = 0.05 m$ concentric with the insulating sphere. By symmetry, $E$ is constant and perpendicular to the surface everywhere, so total flux is:
2
$\Phi_E = E (4 \pi r^2)$
3
Calculate the enclosed charge. Since charge is uniformly distributed, enclosed charge scales with volume:
4
$Q_{enclosed} = Q_{total} \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = Q_{total} \frac{r^3}{R^3}$
5
Apply Gauss's law:
6
$E (4 \pi r^2) = \frac{Q_{total} r^3}{\epsilon_0 R^3}$
7
Simplify and substitute values:
8
$E = \frac{Q_{total} r}{4 \pi \epsilon_0 R^3} = k \frac{Q_{total} r}{R^3} = (9 \times 10^9) \frac{(1 \times 10^{-6})(0.05)}{(0.1)^3} = 4.5 \times 10^5 \text{ N/C}$
9
The field is directed radially outward from the center.

✓ Quick check

Test your understanding of electric field direction with this AP-style multiple choice question:

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?
- Parallel to the side between the two adjacent charges, pointing towards the empty corner
- Along the diagonal of the square, pointing away from the center of the square towards the empty corner
- Along the diagonal of the square, pointing towards the center of the square from the empty corner
- Perpendicular to the diagonal, pointing parallel to the side opposite the empty corner
Reveal answer
1 —
Each positive charge produces a field pointing away from itself at the empty corner. Perpendicular components from the two adjacent charges cancel, leaving all components adding along the diagonal away from the center, so this is the correct answer.

5. Common Pitfalls

Wrong move:

Assigning electric field direction based on test charge sign instead of source charge sign

Why:

Students confuse the definition $E = F / q_{0}$ , and flip direction when using a negative test charge

Correct move:

Electric field is a property of the source. Direction is defined for a positive test charge: away from positive sources, towards negative sources regardless of the actual test charge used.

Wrong move:

Adding electric field magnitudes as scalars when fields point in opposite directions

Why:

Students forget electric field is a vector and just add magnitudes, leading to a result double the correct value

Correct move:

Assign positive/negative signs to directions along each axis, add signed components, then find the magnitude of the resultant net field.

Wrong move:

Using $E = k Q / r^{2}$ for a point inside a uniformly charged insulating sphere

Why:

Students memorize the point charge formula and apply it everywhere, forgetting enclosed charge is less than total charge inside an insulating sphere

Correct move:

For uniformly charged insulating spheres, use $E = k Q r / R^{3}$ inside the sphere, and $E = k Q / r^{2}$ only outside the sphere.

Wrong move:

Claiming electric field is zero inside any hollow sphere, regardless of whether it is conducting or insulating

Why:

Students confuse the conducting sphere zero-field result with hollow insulating spheres that have charge distributed on the surface or volume

Correct move:

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, use Gauss's law to calculate enclosed charge to find the field.

Wrong move:

Using $E = Δ V / d$ to find the electric field between two point charges

Why:

Students mix up formulas for uniform and non-uniform electric fields

Correct move:

Use $E = Δ V / d$ only for uniform fields between parallel plates, and use superposition of point charge fields for point charge systems.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Core Definition	$E = \frac{F _{e}}{q _{0}}$	Direction defined for positive test charge; units N/C = V/m
Point Charge Electric Field	$E = \frac{k ∣ Q ∣}{r ^{2}} = \frac{∣ Q ∣}{4 π ϵ _{0} r ^{2}}$	Direction: away from +Q, towards -Q; only applies outside the charge distribution
Superposition of Fields	$E_{n e t} = \sum_{i = 1}^{n} E_{i}$	Add as vectors; break into components for multiple charges
Uniform Field Between Parallel Plates	$E = \frac{Δ V}{d}$	Only applies to uniform fields; direction from + plate to - plate
Gauss's Law	$Φ_{E} = \frac{Q _{e n c l ose d}}{ϵ _{0}}$	Only simplifies calculation for highly symmetric charge distributions
Inside Uniform Insulating Sphere	$E = \frac{k Q r}{R ^{3}}$	$R$ = sphere radius, $r$ = distance from center, $Q$ = total charge
Electric Field Inside Conductor	$E = 0$	Applies only to conductors at electrostatic equilibrium
Force on Charge in Field	$F = q E$	Direction same as E for +q, opposite for -q

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Net field from two point charges
2022 · FRQ
Gauss's law for charged sphere

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. Mastering the vector nature of electric field and superposition is critical for understanding capacitance, electric circuits, and more advanced electrostatic applications that appear frequently on the AP Physics 2 exam. Building on Gauss's law, you will also explore how charge behaves in conductors at equilibrium, a common exam topic.