Electricity-And-Magnetism · Unit 3 · Video 2 · Interactive Practice

Superposition: Adding Coulomb Forces as Vectors

IKey Formulas

Formula	Name	What it says
$\vec{\mathbf{F}}_{ij} = \frac{1}{4\pi\varepsilon_0}\frac{q_i q_j}{r_{ij}^2}\,\hat{\mathbf{r}}_{ij}$	Coulomb force of $q_i$ on $q_j$	$\hat{\mathbf{r}}_{ij}$ points from $q_i$ to $q_j$ ; a negative $q_i q_j$ reverses it (attraction)
$\vec{\mathbf{F}}_3 = \vec{\mathbf{F}}_{13} + \vec{\mathbf{F}}_{23}$	Superposition, three charges	Each pair force is computed as if the other charge were absent
$\vec{\mathbf{F}}_j = \sum_{i=1,\, i\neq j}^{N} \vec{\mathbf{F}}_{ij}$	Superposition, $N$ charges	Sum over every other charge; valid for charges in fixed positions
$\|\vec{\mathbf{F}}\| = \sqrt{F_x^2 + F_y^2}$ , $\ \tan\varphi = \frac{F_y}{F_x}$	Magnitude and direction	The signs of $F_x$ and $F_y$ pick the quadrant of $\varphi$

Key Insight: Add components, not magnitudes. On the video's square, $202.5\ \text{N} + 405\ \text{N} = 607.5\ \text{N}$ , but the net force on $q_3$ is only about $298\ \text{N}$ , because the $x$ components partly cancel.

IIVisualization 1 — Superposition at Any Position

Each pair force depends only on its own two charges; the net force on $q_3$ is their vector sum.

IIIVisualization 2 — The Video's Square, Component by Component

Along $x$ the two forces partly cancel, which is why $405\ \text{N}$ and $202.5\ \text{N}$ make only $298\ \text{N}$ .

Step 1

Step 1 — Pair forces, each computed alone

|\vec{\mathbf{F}}_{23}| = \frac{(9.0\times10^{9})(6.0\times10^{-6})(3.0\times10^{-6})}{(2.0\times10^{-2})^2}\ \text{N} = 405\ \text{N}

|\vec{\mathbf{F}}_{13}| = \frac{(9.0\times10^{9})(6.0\times10^{-6})(3.0\times10^{-6})}{2\,(2.0\times10^{-2})^2}\ \text{N} = 202.5\ \text{N}

Step 2 — Write each force in $\hat{\mathbf{i}}$ , $\hat{\mathbf{j}}$ components

\vec{\mathbf{F}}_{13} = 202.5\ \text{N}\cdot\tfrac{\sqrt{2}}{2}\,(\hat{\mathbf{i}} + \hat{\mathbf{j}}) = (143.2\,\hat{\mathbf{i}} + 143.2\,\hat{\mathbf{j}})\ \text{N}

\vec{\mathbf{F}}_{23} = -405\,\hat{\mathbf{i}}\ \text{N}\quad(\text{attraction reverses }\hat{\mathbf{r}}_{23} = \hat{\mathbf{i}})

Step 3 — Add the components

F_{3,x} = 143.2\ \text{N} - 405\ \text{N} = -261.8\ \text{N}, \qquad F_{3,y} = 143.2\ \text{N}

Step 4 — Magnitude, then the angle in the right quadrant

|\vec{\mathbf{F}}_3| = \sqrt{261.8^2 + 143.2^2}\ \text{N} \approx 298\ \text{N} \approx 3.0\times10^{2}\ \text{N}

\tan\varphi = \frac{143.2}{-261.8} \;\Rightarrow\; \tan^{-1} \text{ gives } -28.7\degree, \text{ but } F_{3,x} < 0,\ F_{3,y} > 0

\varphi = -28.7\degree + 180\degree = 151.3\degree

IVVisualization 3 — Predict the Direction

Charge signs fix each pair force's direction; the signs of $F_x$ and $F_y$ then fix the quadrant of $\vec{\mathbf{F}}_3$ .

Charges

VQuiz Questions

Problem 1 · Three Charges on a Line

Given: On the $x$ axis, $q_1 = +2.0\ \mu\text{C}$ sits at $x = 0$ , $q_3 = +1.0\ \mu\text{C}$ at $x = 0.10\ \text{m}$ , and $q_2 = -4.0\ \mu\text{C}$ at $x = 0.30\ \text{m}$ . Use $k = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find the net force on $q_3$ .

✅ Correct!

q_1

pushes

q_3

toward

+x

and

q_2

pulls it toward

+x

, so the two pair forces add:

1.8\ \text{N} + 0.90\ \text{N} = 2.7\ \text{N}

❌ Not quite.

q_2

is negative, so it attracts

q_3

toward itself, which is the

+x

direction. Both forces point

+x

; they add rather than cancel.

❌ Check the direction.

q_1

repels

q_3

away from the origin (

+x

), and

q_2

x = 0.30\ \text{m}

attracts

q_3

toward it (

+x

❌ Check $r_{23}$ . The distance is the separation

0.30\ \text{m} - 0.10\ \text{m} = 0.20\ \text{m}

, not the coordinate of

q_2

❌ Not quite. Compute each pair force on its own with its own distance, give it a direction from the signs, then add.

Show solution

Force of $q_1$ on $q_3$ ( $r_{13} = 0.10\ \text{m}$ , both positive, so repulsion, toward $+x$ ):

|\vec{\mathbf{F}}_{13}| = \frac{(9.0\times10^{9})(2.0\times10^{-6})(1.0\times10^{-6})}{(0.10)^2}\ \text{N} = 1.8\ \text{N}

Force of $q_2$ on $q_3$ ( $r_{23} = 0.20\ \text{m}$ ). Here $\hat{\mathbf{r}}_{23} = -\hat{\mathbf{i}}$ (from $q_2$ toward $q_3$ ), and $q_2 q_3 < 0$ reverses it to $+\hat{\mathbf{i}}$ , an attraction toward $q_2$ :

|\vec{\mathbf{F}}_{23}| = \frac{(9.0\times10^{9})(4.0\times10^{-6})(1.0\times10^{-6})}{(0.20)^2}\ \text{N} = 0.90\ \text{N}

\vec{\mathbf{F}}_3 = (1.8 + 0.90)\,\hat{\mathbf{i}}\ \text{N} = 2.7\,\hat{\mathbf{i}}\ \text{N}

Problem 2 · Perpendicular Pair Forces

Given: $q_3 = +2.0\ \mu\text{C}$ at the origin, $q_1 = +4.0\ \mu\text{C}$ at $(-0.30\ \text{m},\ 0)$ and $q_2 = -3.0\ \mu\text{C}$ at $(0,\ 0.30\ \text{m})$ . Use $k = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find the net force on $q_3$ (angle $\varphi$ measured from the $+x$ axis).

✅ Correct!

\vec{\mathbf{F}}_3 = (0.80\,\hat{\mathbf{i}} + 0.60\,\hat{\mathbf{j}})\ \text{N}

: a 3-4-5 triangle with

|\vec{\mathbf{F}}_3| = 1.0\ \text{N}

❌ That adds magnitudes.

0.80\ \text{N} + 0.60\ \text{N}

is right only for parallel forces; these are perpendicular, so the magnitude is

\sqrt{0.80^2 + 0.60^2}\ \text{N}

❌ The forces are not opposite. One points along

+x

and the other along

+y

; neither subtracting nor adding the magnitudes gives the vector sum.

❌ Check the direction of $\vec{\mathbf{F}}_{13}$ .

q_1

and

q_3

are both positive, so

q_1

pushes

q_3

away from itself, toward

+x

, not toward

q_1

❌ Not quite. Find each pair force, write it in components, then add.

Show solution

Both distances are $0.30\ \text{m}$ .

|\vec{\mathbf{F}}_{13}| = \frac{(9.0\times10^{9})(4.0\times10^{-6})(2.0\times10^{-6})}{(0.30)^2}\ \text{N} = 0.80\ \text{N}, \text{ repulsion: } +0.80\,\hat{\mathbf{i}}\ \text{N}

|\vec{\mathbf{F}}_{23}| = \frac{(9.0\times10^{9})(3.0\times10^{-6})(2.0\times10^{-6})}{(0.30)^2}\ \text{N} = 0.60\ \text{N}, \text{ attraction: } +0.60\,\hat{\mathbf{j}}\ \text{N}

|\vec{\mathbf{F}}_3| = \sqrt{0.80^2 + 0.60^2}\ \text{N} = 1.0\ \text{N}, \qquad \tan\varphi = \frac{0.60}{0.80} \Rightarrow \varphi = 36.9\degree

Both components are positive, so the first-quadrant angle from $\tan^{-1}$ is the right one.

Problem 3 · A 3-4-5 Rectangle

Given: $q_1 = +5.0\ \mu\text{C}$ at the origin, $q_2 = -8.0\ \mu\text{C}$ directly above it at $(0,\ 3.0\ \text{cm})$ , and $q_3 = +1.0\ \mu\text{C}$ to the right of $q_2$ at $(4.0\ \text{cm},\ 3.0\ \text{cm})$ . Use $k = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find the magnitude and direction of the net force on $q_3$ .

What is $|\vec{\mathbf{F}}_3|$ ?

What is the angle $\varphi$ from the $+x$ axis?

✅ Correct!

\vec{\mathbf{F}}_3 = (-30.6\,\hat{\mathbf{i}} + 10.8\,\hat{\mathbf{j}})\ \text{N}

: about

32\ \text{N}

, up and to the left at

160.6\degree

❌ That adds magnitudes.

18\ \text{N} + 45\ \text{N} = 63\ \text{N}

ignores that the

x

components point in opposite directions and partly cancel.

❌ The sign of $q_2$ was dropped.

q_2

is negative, so it attracts

q_3

toward

-x

; treating it as a repulsion flips

\vec{\mathbf{F}}_{23}

+x

❌ Wrong quadrant. That is what

\tan^{-1}(10.8/(-30.6))

returns, but

F_x < 0

and

F_y > 0

put the force in the second quadrant: add

180\degree

❌ This is not a square. The diagonal from

q_1

q_3

has

\hat{\mathbf{r}}_{13} = 0.8\,\hat{\mathbf{i}} + 0.6\,\hat{\mathbf{j}}

, not

45\degree

; compute the components before the angle.

❌ Check the magnitude. Use

r_{13} = 5.0\ \text{cm}

and

r_{23} = 4.0\ \text{cm}

, add the components, then take

\sqrt{F_x^2 + F_y^2}

❌ Check the angle. Use

\tan\varphi = F_y/F_x

and check the quadrant against the signs of

F_x

and

F_y

Show solution

Distances and unit vectors. $r_{23} = 4.0\ \text{cm}$ with $\hat{\mathbf{r}}_{23} = \hat{\mathbf{i}}$ ; $r_{13} = \sqrt{4.0^2 + 3.0^2}\ \text{cm} = 5.0\ \text{cm}$ with $\hat{\mathbf{r}}_{13} = 0.8\,\hat{\mathbf{i}} + 0.6\,\hat{\mathbf{j}}$ .

Pair forces.

\vec{\mathbf{F}}_{13} = \frac{(9.0\times10^{9})(5.0\times10^{-6})(1.0\times10^{-6})}{(0.050)^2}\,(0.8\,\hat{\mathbf{i}} + 0.6\,\hat{\mathbf{j}})\ \text{N} = (14.4\,\hat{\mathbf{i}} + 10.8\,\hat{\mathbf{j}})\ \text{N}

\vec{\mathbf{F}}_{23} = \frac{(9.0\times10^{9})(-8.0\times10^{-6})(1.0\times10^{-6})}{(0.040)^2}\,\hat{\mathbf{i}}\ \text{N} = -45\,\hat{\mathbf{i}}\ \text{N}

Add components. $F_{3,x} = 14.4 - 45 = -30.6\ \text{N}$ , $F_{3,y} = 10.8\ \text{N}$ .

|\vec{\mathbf{F}}_3| = \sqrt{30.6^2 + 10.8^2}\ \text{N} \approx 32.4\ \text{N} \approx 32\ \text{N}

Direction. $\tan^{-1}(10.8/(-30.6)) = -19.4\degree$ points down and to the right, but $F_x < 0$ and $F_y > 0$ , so $\varphi = -19.4\degree + 180\degree = 160.6\degree$ .

Problem 4 · Equilateral Triangle

Given: Two charges of $+2.0\ \mu\text{C}$ sit at the base corners of an equilateral triangle of side $0.10\ \text{m}$ , and $Q = +2.0\ \mu\text{C}$ sits at the apex. Use $k = 9.0\times10^{9}\ \text{N}\,\text{m}^2/\text{C}^2$ . Find the net force on $Q$ .

✅ Correct! The parts parallel to the base cancel; the perpendicular parts add to

2(3.6\ \text{N})\cos 30\degree \approx 6.2\ \text{N}

❌ That adds magnitudes. The two

3.6\ \text{N}

forces are

60\degree

apart, so only their components perpendicular to the base add fully.

❌ Check the angle. Each force makes

30\degree

with the perpendicular to the base (not

60\degree

), so its perpendicular part is

3.6\ \text{N}\cdot\cos 30\degree

❌ Only part cancels. The components parallel to the base cancel by symmetry, but both forces push

Q

away from the base.

Show solution

Each base charge is $0.10\ \text{m}$ from $Q$ , so each pair force has magnitude

\frac{(9.0\times10^{9})(2.0\times10^{-6})(2.0\times10^{-6})}{(0.10)^2}\ \text{N} = 3.6\ \text{N},

directed along the side, away from that base charge. Put the base on the $x$ axis with $Q$ above its midpoint: each force makes $30\degree$ with the $+y$ axis. The $x$ components, $\pm 3.6\sin 30\degree\ \text{N}$ , cancel. The $y$ components add:

F_y = 2\,(3.6\ \text{N})\cos 30\degree = 3.6\sqrt{3}\ \text{N} \approx 6.2\ \text{N}

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