A charge of -3.0\,μC sits where E = 200 N/C points north. Find the force on it. [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

What is the electric field, how does a point charge produce one, and how do field lines represent it?

Topic 8.3 Electric Fields: define the electric field as force per unit charge, calculate the field of point charges, and represent fields with field lines.

A calculus-based answer to AP Physics C E&M Topic 8.3, covering the electric field as force per charge, the field of a point charge, superposition of fields, field lines, and the field inside and around conductors.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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

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Quick answer

The electric field at a point is the electric force per unit positive test charge there: $\vec{E} = \dfrac{\vec{F}}{q}$ , in N/C. The field of a single point charge has magnitude $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q|}{r^2} = k\dfrac{|q|}{r^2}$ , pointing away from a positive charge and toward a negative one. Fields add by superposition: $\vec{E}_{net} = \sum_i \vec{E}_i$ . Field lines point in the field's direction, start on positive and end on negative charge, never cross, and are denser where the field is stronger. The force on a charge placed in a field is $\vec{F} = q\vec{E}$ , in the field's direction for positive $q$ and opposite for negative $q$ .

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What this topic is asking
The electric field
The field of a point charge
Superposition of fields
Field lines
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What this topic is asking

The College Board (Topic 8.3) wants you to define the electric field as the force per unit charge, compute the field of one or more point charges by superposition, and represent fields with field lines. The field is the central object of the whole course: it lets you describe what a charge distribution does to the space around it, independent of any test charge.

The electric field

The field is defined so that it describes the source charges alone: place any charge $q$ at a point and it feels a force $\vec{F} = q\vec{E}$ . For positive $q$ the force is along $\vec{E}$ ; for negative $q$ it is opposite. The test charge is imagined small enough not to disturb the sources.

The field of a point charge

A point charge $Q$ produces a radial field of magnitude

E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r^2} = k\frac{|Q|}{r^2}

directed away from $Q$ if positive and toward $Q$ if negative. This is Coulomb's law divided by the test charge: $E = F/q$ removes the test charge, leaving a property of the source and the location.

Superposition of fields

Because force superposes, so does field. The total field of several charges is the vector sum of their individual fields:

\vec{E}_{net} = \sum_i \vec{E}_i = \sum_i k\frac{q_i}{r_i^2}\,\hat{r}_i

Resolve each contribution into components, add the components, recombine. Symmetry often kills one component: on the perpendicular bisector of two equal charges, the components along the line cancel and only the perpendicular components survive.

Field lines

Field lines are a picture of the field:

The tangent to a line gives the field direction at that point.
Lines start on positive charge and end on negative charge (or run to or from infinity).
The density of lines is proportional to the field strength: closely spaced lines mean a strong field.
Lines never cross, because the field has one definite direction at each point.

Field on the axis of two equal charges

Two charges, each $+4.0$ nC, sit at $(\pm 3.0\text{ cm}, 0)$ . Find the electric field at $(0, 4.0\text{ cm})$ . Use $k = 8.99\times10^9$ N m squared per C squared.

step 1 Find the distance and each magnitude

Each charge is $r = \sqrt{(0.03)^2 + (0.04)^2} = 0.05$ m from the field point. $E_1 = k\dfrac{q}{r^2} = (8.99\times10^9)\dfrac{4.0\times10^{-9}}{(0.05)^2} = 1.44\times10^4$ N/C.

step 2 Use symmetry to cancel the horizontal parts

The two charges are mirror images across the y-axis, so their x-components are equal and opposite and cancel. Only the y-components survive, and they add.

step 3 Project onto the y-axis

$\cos\theta = \dfrac{y}{r} = \dfrac{0.04}{0.05} = 0.80$ . Each y-component is $E_1\cos\theta = (1.44\times10^4)(0.80) = 1.15\times10^4$ N/C.

step 4 Add the surviving components

$E = 2(1.15\times10^4) = 2.30\times10^4$ N/C, directed in $+y$ (away from the charges).

Try this

Q1. A $+5.0$ nC point charge is at the origin. Calculate the field magnitude $0.20$ m away ( $k = 8.99\times10^9$ ). [2 points]

Cue. $E = k\dfrac{q}{r^2} = (8.99\times10^9)\dfrac{5.0\times10^{-9}}{(0.20)^2} = 1.1\times10^3$ N/C, radially outward.

Q2. A charge of $-3.0\,\mu\text{C}$ sits where $\vec{E} = 200$ N/C points north. Find the force on it. [2 points]

Cue. $F = |q|E = (3.0\times10^{-6})(200) = 6.0\times10^{-4}$ N, directed south (opposite the field, since $q < 0$ ).

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 2022 (style)1 marksSection I (multiple choice). At a point in space the electric field is

400

N/C directed east. A charge of

-2.0\,\mu\text{C}

is placed there. The force on the charge is (A)

8.0\times10^{-4}

N east (B)

8.0\times10^{-4}

N west (C)

2.0\times10^{-4}

N east (D) zero. Justify your reasoning.

Show worked answer →

A 1-point MCQ on $\vec{F} = q\vec{E}$ . The answer is (B).

The force is $F = |q|E = (2.0\times10^{-6})(400) = 8.0\times10^{-4}$ N. Because the charge is negative, the force is opposite to the field, so it points west. The trap is choosing (A) and ignoring the sign of the charge.

AP 2024 (style)5 marksSection II (FRQ, quantitative). Two point charges sit on the x-axis:

+q

x = -a

and

+q

x = +a

. (a) Derive an expression for the electric field at a point on the y-axis a distance

y

from the origin. (b) Determine the direction of the field there. (c) State what the field becomes for

y \gg a

and explain why.

Show worked answer →

A 5-point FRQ on superposition of point-charge fields with calculus-style reasoning.

(a) Derivation (3 points): each charge is a distance $r = \sqrt{a^2 + y^2}$ from the field point, giving magnitude $E_1 = \dfrac{kq}{a^2 + y^2}$ . By symmetry the x-components cancel and the y-components add. Each y-component is $E_1\cos\theta$ with $\cos\theta = \dfrac{y}{\sqrt{a^2+y^2}}$ , so $E = 2\dfrac{kq}{a^2+y^2}\cdot\dfrac{y}{\sqrt{a^2+y^2}} = \dfrac{2kqy}{(a^2+y^2)^{3/2}}$ .
(b) Direction (1 point): along $+y$ , away from the charges.
(c) Limit (1 point): for $y \gg a$ , $E \approx \dfrac{2kqy}{y^3} = \dfrac{2kq}{y^2}$ , the field of a single point charge $2q$ at the origin, because from far away the pair looks like one combined charge.

Markers reward the distance and the $\cos\theta$ projection, the symmetry argument, and the far-field limit.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)