United StatesPhysics 2Syllabus dot point

What is an electric field, and how does it tell a charge which way to move?

Topic 10.3 Electric Fields: define the electric field, calculate the field of a point charge, and represent fields with field lines and superposition.

A focused answer to AP Physics 2 Topic 10.3, covering the electric field as force per unit charge, the field of a point charge, field-line diagrams and their rules, superposition of fields, the uniform field between parallel plates, and fields in conductors, with full worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The electric field $\vec{E}$ at a point is the force per unit positive charge that a small test charge would feel there: $\vec{E} = \dfrac{\vec{F}}{q}$ , measured in N/C. A field exists around any charge whether or not a test charge is present; place a charge $q$ in it and the force is $\vec{F} = q\vec{E}$ (same direction as $\vec{E}$ for positive $q$ , opposite for negative). A point charge $Q$ produces a field $E = \dfrac{kQ}{r^2}$ pointing outward from a positive charge and inward toward a negative one. Field lines point in the field's direction, start on positive and end on negative charges, never cross, and are denser where the field is stronger. With several charges, the net field is the vector sum of each charge's field (superposition). Between two oppositely charged parallel plates the field is uniform, $E = \dfrac{V}{d}$ . The field is the bridge from charge to force: it carries the influence of a source across space.

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What this topic is asking
What the electric field is
The field of a point charge
Field lines, superposition and uniform fields
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What this topic is asking

The College Board (Topic 10.3) wants you to define the electric field as force per unit charge, calculate the field of a point charge, and represent fields with field lines and superposition, including the uniform field between parallel plates.

What the electric field is

The field concept replaces "action at a distance" with a local influence: a source charge fills the space around it with a field, and any other charge responds to the field at its own location. The field points the way a positive test charge would be pushed; a negative charge feels a force in the opposite direction. This is why knowing the field at a point immediately gives the force on any charge there, just multiply by the charge and flip the direction if it is negative.

The field of a point charge

The point-charge field is just Coulomb's law divided by the test charge: the $q_2$ drops out, leaving the field that $Q$ alone sets up. Its inverse-square falloff means the field weakens rapidly with distance. The direction rule, outward from positive and inward toward negative, is the same as the direction a positive test charge would be pushed.

Field lines, superposition and uniform fields

Field lines are the standard way to picture a field, and their rules are tested directly: lines point in the direction of $\vec{E}$ , begin on positive and end on negative charges, never cross, and are closer together where the field is stronger. For several charges, the net field at a point is the vector sum of the individual point-charge fields (superposition): compute each field, then add as vectors. A special and important case is the uniform field between two large, oppositely charged parallel plates, where the field is constant in magnitude and direction, $E = \dfrac{V}{d}$ (with $V$ the voltage across the plates and $d$ their separation). Inside a conductor in electrostatic equilibrium the field is zero, because free charges rearrange until they cancel any internal field. The strategic point is that the field organizes the whole unit: it comes directly from Coulomb's law (Topic 10.1), it determines the force on any charge ( $\vec{F} = q\vec{E}$ ), and it connects to the electric potential of Topic 10.5 (the field points from high to low potential). Reading a field-line diagram and adding fields as vectors are the core skills the exam rewards here.

Net field from two charges

A charge $+4.0$ microcoulombs sits at the origin and a charge $+4.0$ microcoulombs sits at $x = 0.40$ m. Find the magnitude and direction of the net field at the midpoint $x = 0.20$ m. Take $k = 8.99 \times 10^9$ N m squared per C squared.

step 1 Field from the first charge

At $0.20$ m from the origin charge: $E_1 = \dfrac{kQ}{r^2} = \dfrac{(8.99 \times 10^9)(4.0 \times 10^{-6})}{(0.20)^2} = 8.99 \times 10^5$ N/C, pointing in the $+x$ direction (away from the positive charge).

step 2 Field from the second charge

At $0.20$ m from the second charge: $E_2 = 8.99 \times 10^5$ N/C, pointing in the $-x$ direction (away from that positive charge, back toward the origin).

step 3 Add as vectors

The two fields are equal in magnitude and opposite in direction, so they cancel: $E_{net} = 0$ .

step 4 Interpret

Midway between two equal positive charges, the fields exactly oppose, giving zero net field. A positive test charge placed there would feel no net force, an unstable balance point.

Try this

Q1. State the direction of the electric field around an isolated negative point charge. [1 point]

Cue. Radially inward, toward the charge.

Q2. State the magnitude of the electric field inside a conductor in electrostatic equilibrium. [1 point]

Cue. Zero (free charges rearrange to cancel any internal field).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (short FRQ). A point charge

Q = +6.0

microcoulombs sits at the origin. Take

k = 8.99 \times 10^9

N m squared per C squared. (a) Calculate the magnitude of the electric field

0.50

m from the charge. (b) State the direction of the field at that point. (c) A charge

q = -2.0

microcoulombs is placed there. Calculate the magnitude and state the direction of the force it experiences.

Show worked answer →

A 6-point FRQ on the electric field of a point charge.

(a) Field magnitude (2 points): $E = \dfrac{kQ}{r^2} = \dfrac{(8.99 \times 10^9)(6.0 \times 10^{-6})}{(0.50)^2} = \dfrac{5.39 \times 10^4}{0.25} = 2.16 \times 10^5$ N/C.
(b) Direction (1 point): the field of a positive charge points radially outward, so it points away from $Q$ at that point.
(c) Force (3 points): $F = |q|E = (2.0 \times 10^{-6})(2.16 \times 10^5) = 0.43$ N. The charge is negative, so the force is opposite the field, pointing back toward $Q$ (attraction).

Markers reward the point-charge field, the outward direction for a positive source, and using $F = qE$ with the direction reversed for a negative charge.

AP 2023 (style)1 marksSection I (multiple choice). At a point in space the electric field points east. A negative charge is released from rest there. In which direction does it begin to move? (A) east (B) west (C) it stays at rest (D) north. Justify your reasoning.

Show worked answer →

A 1-point MCQ on field and force direction. The answer is (B).

The force on a charge is $\vec{F} = q\vec{E}$ . For a negative charge the force is opposite the field, so a field pointing east gives a force (and initial motion) pointing west. The trap is (A): only a positive charge moves along the field.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)