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1.8.3Electromagnetism

Superposition principle for forces

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WHAT is the superposition principle?

The key words are vector sum (not scalar — directions matter) and independent (charge ii's force does not get modified by the presence of charge jj).

WHY is this true? (first-principles reasoning)

WHY it could have been false: if forces "saturated" (a charge could only feel so much total force) or if one charge screened another nonlinearly, superposition would break. It doesn't, to extraordinary precision, in classical electrostatics — so we adopt it as a principle.

HOW to derive the net force (from scratch)

We never just "plug in." We build it.

Step 1 — Coulomb's law for ONE pair. The force on q0q_0 at position r⃗0\vec{r}_0 due to qiq_i at r⃗i\vec{r}_i: F⃗0i=14πε0 q0qi∣r⃗0−r⃗i∣2 r^0i,r^0i=r⃗0−r⃗i∣r⃗0−r⃗i∣\vec{F}_{0i} = \frac{1}{4\pi\varepsilon_0}\,\frac{q_0 q_i}{|\vec{r}_0 - \vec{r}_i|^2}\,\hat{r}_{0i}, \qquad \hat{r}_{0i} = \frac{\vec{r}_0 - \vec{r}_i}{|\vec{r}_0 - \vec{r}_i|} Why this step? The unit vector r^0i\hat{r}_{0i} points from the source ii toward the target 00, so a repulsion (like signs) automatically pushes q0q_0 away. Direction handled by geometry, not by hand.

Step 2 — Apply superposition. Add the contributions:   F⃗net=q04πε0∑i=1nqi (r⃗0−r⃗i)∣r⃗0−r⃗i∣3  \boxed{\;\vec{F}_{\text{net}} = \frac{q_0}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i\,(\vec{r}_0-\vec{r}_i)}{|\vec{r}_0-\vec{r}_i|^3}\;} Why the cube? We wrote r^0i=(r⃗0−r⃗i)/∣r⃗0−r⃗i∣\hat{r}_{0i} = (\vec{r}_0-\vec{r}_i)/|\vec{r}_0-\vec{r}_i|, which contributes one extra power of distance in the denominator, turning r2r^2 into r3r^3 while the numerator carries the direction.

Step 3 — Resolve into components. Because it's a vector sum: Fx=∑iF0icos⁡θi,Fy=∑iF0isin⁡θiF_x = \sum_i F_{0i}\cos\theta_i, \qquad F_y = \sum_i F_{0i}\sin\theta_i ∣F⃗net∣=Fx2+Fy2,tan⁡ϕ=FyFx|\vec F_{\text{net}}| = \sqrt{F_x^2 + F_y^2}, \qquad \tan\phi = \frac{F_y}{F_x} Why components? You cannot add magnitudes of forces pointing in different directions. Components let you add along each axis safely, then recombine.

Figure — Superposition principle for forces

Worked Example 1 — Two charges, both on the x-axis

Charges q1=+3 μCq_1 = +3\,\mu C at x=0x=0 and q2=−2 μCq_2=-2\,\mu C at x=4 mx=4\,\text{m}. Find force on q0=+1 μCq_0=+1\,\mu C at x=2 mx=2\,\text{m}. Let k=14πε0=9×109k=\frac{1}{4\pi\varepsilon_0}=9\times10^9.

  • From q1q_1: distance =2=2 m, like signs ⇒ repulsion ⇒ pushes q0q_0 in +x. F01=k(3×10−6)(1×10−6)22=6.75×10−3 N (+x)F_{01}=\frac{k(3\times10^{-6})(1\times10^{-6})}{2^2}=6.75\times10^{-3}\,\text{N (+x)} Why +x? q1q_1 is to the left and repels, so it shoves q0q_0 rightward.
  • From q2q_2: distance =2=2 m, opposite signs ⇒ attraction ⇒ pulls q0q_0 toward q2q_2 (right) ⇒ +x. F02=k(2×10−6)(1×10−6)22=4.5×10−3 N (+x)F_{02}=\frac{k(2\times10^{-6})(1\times10^{-6})}{2^2}=4.5\times10^{-3}\,\text{N (+x)} Why +x again? q2q_2 sits to the right and attracts, pulling q0q_0 rightward.
  • Net: both same direction ⇒ add: F=11.25×10−3 N=11.25 mNF=11.25\times10^{-3}\,\text{N}=11.25\,\text{mN} along +x.

Worked Example 2 — Right-angle (2D vector sum)

q1=+1 μCq_1=+1\,\mu C at origin, q2=+1 μCq_2=+1\,\mu C at (0,3)(0,3) m. Find force on q0=+1 μCq_0=+1\,\mu C at (4,0)(4,0) m.

  • From q1q_1: r=4r=4 m, repulsion along +x. F01=k(1)(1)×10−1216=5.625×10−4 N⇒(5.625×10−4, 0)F_{01}=\frac{k(1)(1)\times10^{-12}}{16}=5.625\times10^{-4}\,\text{N} \Rightarrow (5.625\times10^{-4},\,0)
  • From q2q_2: separation vector (4,0)−(0,3)=(4,−3)(4,0)-(0,3)=(4,-3), magnitude 55 m. F02=k×10−1225=3.6×10−4 NF_{02}=\frac{k\times10^{-12}}{25}=3.6\times10^{-4}\,\text{N} Direction = unit vector (4,−3)/5=(0.8,−0.6)(4,-3)/5=(0.8,-0.6). Why this unit vector? It points from source q2q_2 to target q0q_0 — repulsion pushes outward along it. F⃗02=3.6×10−4(0.8,−0.6)=(2.88×10−4, −2.16×10−4)\vec F_{02}=3.6\times10^{-4}(0.8,-0.6)=(2.88\times10^{-4},\,-2.16\times10^{-4})
  • Sum components: Fx=(5.625+2.88)×10−4=8.505×10−4,Fy=−2.16×10−4F_x=(5.625+2.88)\times10^{-4}=8.505\times10^{-4},\quad F_y=-2.16\times10^{-4} ∣F⃗∣=8.5052+2.162×10−4=8.78×10−4 N,ϕ=arctan⁡−2.168.505=−14.3∘|\vec F|=\sqrt{8.505^2+2.16^2}\times10^{-4}=8.78\times10^{-4}\,\text{N},\quad \phi=\arctan\frac{-2.16}{8.505}=-14.3^\circ

Why arctan? The angle below +x axis records the tilt caused by q2q_2's downward pull-component.

Common Mistakes (Steel-manned)

Recall Feynman: explain it to a 12-year-old

Imagine you're standing in a crowd and three friends are pushing or pulling you with ropes. Friend A doesn't care that friends B and C exist — they pull you with exactly the strength they'd use if alone. To find where you actually move, you add up all three pulls like arrows: a pull north and a pull east together send you northeast. That "add the arrows" rule is the superposition principle.

Active Recall

What does the superposition principle state for forces?
The net force on a charge is the vector sum of the individual Coulomb forces from every other charge, each computed as if it were the only source.
Is superposition a scalar or vector operation?
Vector — directions must be added via components, not magnitudes.
Does a third charge alter the pairwise force between two charges?
No. Each pair force is independent; the third charge only adds its own separate force term.
Why does the vector Coulomb form have r3r^3 in the denominator?
Writing r^=(r⃗0−r⃗i)/∣r⃗0−r⃗i∣\hat r = (\vec r_0-\vec r_i)/|\vec r_0-\vec r_i| adds one power of distance, turning r2r^2 into r3r^3 while the numerator carries direction.
Which way does r^0i\hat r_{0i} point for the force ON q0q_0?
From the source charge qiq_i toward the target q0q_0.
When can you simply add force magnitudes?
Only when all forces are collinear (along the same line).
Is superposition derivable from Coulomb's law alone?
No — linearity in source charges is a separate experimental fact that Coulomb's law happens to obey.

Connections

Concept Map

depends only on

direction from

justifies

combined with

net force is

summed formula gives

extra power of r

cannot add magnitudes

recombine via

could fail if

not observed so

Coulomb's law for one pair

Pair q0 and qi

Unit vector r-hat toward target

Linearity - experimental fact

Superposition principle

Vector sum of forces

F net equation with r cubed

Resolve into x and y components

Magnitude and angle phi

Saturation or screening

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, superposition principle ka matlab bahut simple hai: agar kisi charge q0q_0 ke aas-paas kai saare charges hain, to har charge use alag-alag, independently force lagata hai — jaise baaki charges hain hi nahi. Coulomb's law sirf do charges ke beech ka force batata hai, usme teesre charge ka koi role nahi. To total force nikaalne ke liye hum saare individual forces ko vector mein add karte hain, scalar mein nahi.

Yeh "vector mein add" waali baat sabse important hai. Agar do forces alag-alag direction mein hain (jaise ek +x aur ek tirchi), to tum unke numbers seedhe jod nahi sakte. Pehle dono ko xx aur yy components mein todo (Fcos⁡θF\cos\theta, Fsin⁡θF\sin\theta), phir har axis ke components alag-alag add karo, aur last mein Pythagoras se magnitude aur arctan se angle nikaalo. Yahi 80/20 trick hai — agar components sahi nikaal liye, baaki sab apne aap ho jaata hai.

Kyun matter karta hai? Kyunki real life aur exams mein hamesha multiple charges hote hain. Superposition ke bina tum sirf 2-charge problems solve kar paate. Aur ek dhyaan rakhne wali baat: teesra charge kabhi do charges ke beech ka force "kam-zyada" nahi karta — woh bas apna alag force add karta hai. Isko yaad rakhne ka mantra: "Each alone, then add as arrows" — pehle har force akela nikaalo, phir arrows ki tarah jodo.

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Connections

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