Coulomb's law — force, comparison with gravity: Question bank

1.8.2 · D5 · Physics › Electromagnetism › Coulomb's law — force, comparison with gravity

Notation aur pictures jo pehle chahiye hongi

Neeche kai traps us notation par hinge karte hain jo parent note mein jaldi se use hui. Aao use pictures ke saath pin karte hain taaki koi bhi symbol raaz na rahe jab woh aaye.

Figure — Coulomb's law — force, comparison with gravity

Figure dekho: violet arrow hai $r_{21}$ jo 1 se 2 tak ja raha hai, aur chhota magenta arrow hai $\overset{r}{^}_{21}$ — same direction, length 1. "Force kis taraf point karta hai" ke baare mein sab kuch is chhote arrow se decide hota hai.

True ya false — justify karo

r

ko double karne se Coulomb force half ho jaati hai.

False. Law

1/ r^{2}

ke hisaab se jaata hai,

1/ r

ke nahi;

r

double karne par force

1/ 2^{2} = 1/4

se multiply hoti hai. "Twice as far → half as strong" linear intuition hai, lekin electric influence

4 π r^{2}

area wale ek sphere par spread hoti hai, toh yeh square ke hisaab se thin hoti hai.

Agar dono charges negative hain, toh unke beech force negative (attractive) hai.

False. Do negatives repel karte hain. Scalar form mein hum

∣ q_{1} q_{2} ∣

use karte hain, toh magnitude positive hai; product

(-) (-) = +

signal karta hai like charges → repulsion. "Negative force" apne aap mein koi cheez nahi hai — direction vector form mein rehti hai, magnitude ke sign mein nahi.

Do electrons ke liye

F_{e} / F_{g}

mein

r^{2}

tabhi cancel hota hai jab woh bahut door hon.

False. Yeh hamesha cancel hota hai, har distance par.

F_{e}

aur

F_{g}

dono identical

1/ r^{2}

carry karte hain, toh unka ratio ek pure constant hai (

\approx 4.2 \times 1 0^{42}

) jo

r

se independent hai.

Coulomb's law do charged metal spheres par exactly apply hota hai, size se parwaah nahi.

Saamaanyatah false. Yeh point charges ke liye exact hai, ya spherically symmetric charge ke liye jo centre par concentrated hai. Do bade spheres paas rakhne par ek doosre ka charge redistribute ho jaata hai (induction), symmetry break ho jaati hai, toh simple formula approximate ho jaata hai.

Jab objects bade hote hain tab gravity electric force se zyada strong hoti hai.

Misleading. Gravity bade scale par dominate isliye nahi karti ki woh zyada strong ho jaati hai, balki isliye ki bade bodies electrically neutral hote hain (equal

+

aur

-

), toh unki bahut zyada strong electric forces cancel ho jaati hain, sirf kamzor gravity uncancelled rehti hai.

k

ka bada hona (

8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}

) aur

G

ka tiny hona (

6.67 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}

) akele prove karta hai ki electricity gravity se zyada strong hai.

Jaisa kaha gaya hai woh false hai. Dono constants alag units carry karte hain (per C² vs per kg²), toh aap bare numbers directly compare nahi kar sakte. Honest comparison mein same particles ke real charges aur masses daalo — tabhi electricity

\sim 1 0^{42}

se jeetti hai.

Newton's third law Coulomb forces par fail hota hai kyunki ek charge bada ho sakta hai.

False.

F_{12} = - F_{21}

hamesha: 2 se 1 par force, 1 se 2 par force ke barabar size mein hoti hai aur opposite direction mein, chahe

q_{1} \neq = q_{2}

ho. Dono same

∣ q_{1} q_{2} ∣

aur same

r

share karte hain.

Ek charge apni khud ki electric force feel karta hai.

False. Coulomb's law do charges ke beech force ke baare mein hai. Ek charge khud ko push nahi karta; aapko hamesha ek doosra (source) charge chahiye.

Error dhundo

F = k \frac{q _{1} q _{2}}{r ^{2}}

ne mujhe

- 5.4 N

diya, toh force

- 5.4

newtons hai."

Scalar magnitude formula mein

∣ q_{1} q_{2} ∣

use karna zaroori hai; ek magnitude negative nahi ho sakti. Minus sign ne bataya ki charges unlike hain (attractive), lekin yeh ek direction conclusion hai, alag se read kiya — negative newton value nahi.

"Charges hain

3 μ C

aur

2 μ C

, toh maine

3

aur

2

plug in kiye."

Units error. Coulomb's law ko SI chahiye:

3 μ C = 3 \times 1 0^{- 6} C

1 0^{- 6}

bhoolna answer ko

1 0^{12}

se inflate kar deta hai.

"Woh

10 cm

apart hain, toh

r^{2} = 1 0^{2} = 100

r

ke liye wrong units. Pehle convert karo:

10 cm = 0.1 m

, toh

r^{2} = 0.01 m^{2}

. Centimetres mein square karna baad ki sab cheez corrupt kar deta hai.

\overset{r}{^}_{21}

2 se 1 ki taraf point karta hai, kyunki yeh charge 2 par hai."

Reversed. Jaise upar define kiya gaya,

r_{21} = r_{2} - r_{1}

1 se 2 ki taraf point karta hai (source → target); subscript order

21

padhta hai "on 2 due to 1." Yeh ulta samajhna har attract/repel conclusion flip kar deta hai.

"Dono charges ko half karna aur distance ko half karna force ko unchanged chodta hai."

Arithmetic sahi hai lekin yahan yaad karne ki koi rule nahi hai. Yeh hua ki do alag effects — charges

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

aur distance

1/ (\frac{1}{2})^{2} = \times 4

— in particular factors ke liye exactly

1

par multiply ho gaye. Yeh un chosen numbers ka ek lucky coincidence hai, principle nahi: koi aur factors lo (kaho charge quarter karo, distance half karo) aur force change ho jaayegi. Ise kabhi "yeh cancel ho jaata hai" ke roop mein mat yaad karo.

"Coulomb force ke liye charges ko act karne ke liye contact mein hona zaroori hai."

Poora point yahi hai ki yeh distance par force hai — koi contact zaroori nahi. Charges ek doosre ko empty space ke paar "feel" karte hain (via electric field, woh invisible arrow-map jo har charge apne around set up karta hai).

"Teen charges ke saath, main sirf do forces ki magnitudes add karta hoon."

Force ek vector hai. Ek se zyada source ke saath aapko forces ko vectors ke roop mein add karna hoga — Superposition Principle — direction respect karte hue (neeche figure dekho); scalar addition tabhi kaam karta hai jab sab forces same direction mein point karein.

Figure mein, ek test charge source A se force $F_{A}$ aur source B se $F_{B}$ feel karta hai. Correct total hai tip-to-tail resultant $F_{A} + F_{B}$ (orange), jo $∣ F_{A} ∣ + ∣ F_{B} ∣$ se chhota hai kyunki dono arrows alag directions mein point karte hain. Magnitudes add karna overcount kar deta.

Why questions

Coulomb's law mein

1/ r^{2}

kyun hai na ki

1/ r

1/ r^{3}

Charge ka influence ek expanding sphere ki surface par uniformly spread hota hai jiska area

4 π r^{2}

hai (neeche sphere figure dekho). Total influence conserve karna use area badhne ke saath thinner spread karta hai, exactly

1/ r^{2}

deta hai — 3D space ki geometry exponent choose karti hai.

Figure mein "influence lines" ki same fixed number ek charge se nikal kar do spheres ko pierce karti hain. Bade sphere par wahi same lines $4 π r^{2}$ area par spread hain, toh unki density — aur isliye force — $1/ r^{2}$ ke hisaab se fall karti hai. Yeh picture exactly Gauss's Law ka seed hai, jo kisi bhi closed surface se un lines ko count karta hai.

k

ko sirf ek akele number ki jagah

\frac{1}{4 π ε _{0}}

kyun likha jaata hai?

4 π

ek unit sphere ka surface area factor hai (wahi

4 π r^{2}

spreading picture se). Ise

k

mein build karna baad ke laws ko — khaaskar Gauss's Law, jo ek closed surface par field-line "flux" sum karta hai — clean aur factor-free banata hai, yahan messier-looking constant worth hai.

Electric force dono push aur pull kyun kar sakti hai jabki gravity sirf pull karti hai?

Charge do signs mein aata hai (

+

aur

-

); like signs repel karte hain, unlike attract karte hain. Mass sirf ek sign mein aata hai (koi negative mass nahi hai), toh gravity sirf attract hi kar sakti hai.

Vector form mein signed charges kyun use karte hain lekin scalar form mein magnitudes?

Vector form

F_{21} = k \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21}

mein,

q_{1} q_{2}

ka sign automatically arrow flip kar deta hai (

\overset{r}{^}_{21}

ke along ya against) repel vs attract ke liye. Scalar form sirf size report karta hai, toh

∣ q_{1} q_{2} ∣

leta hai aur aap direction haath se decide karte ho.

Do protons ke beech nucleus mein electric force ek "problem" kyun hai jise nature ko solve karna pada?

Nuclear separations par protons ke beech Coulomb repulsion enormous hoti hai (subatomic speck par hundreds of newtons). Kuch zyada strong — strong nuclear force — isko overcome karna zaroori hai nucleus ko ek saath hold karne ke liye.

Everyday objects ke beech bahut badi electric force unhe kyun nahi taar deti?

Ordinary matter electrically neutral hai: har atom ka positive nucleus uske electrons se balance hota hai. Colossal forces internally present hain lekin almost perfectly cancel ho jaate hain, sirf tiny residual effects chodkar.

Same pair ke liye electric aur gravitational force compare karte waqt distance "matter nahi karta" kyun?

Dono forces identically

1/ r^{2}

ke hisaab se scale karti hain, toh jab aap ek ko doosre se divide karte ho toh

r^{2}

terms exactly cancel ho jaate hain, ek distance-independent ratio chhodkar.

Edge cases

Coulomb force kya hogi agar ek charge exactly zero ho?

Zero.

F \propto q_{1} q_{2}

, toh ek neutral (uncharged) object ko kisi charge se Coulomb force nahi milti — field ko act karne ke liye kuch nahi hota.

Jab

r \to 0

(charges saath laaye jaayein) toh predicted force ka kya hoga?

Formula

F \to \infty

deta hai, kyunki

1/ r^{2}

blow up ho jaata hai. Physically yeh point-charge idealisation ke breakdown ka signal hai: real charges ka size hota hai aur doosri forces intervene karti hain infinity reach hone se pehle.

r \to \infty

(charges bahut door) par force kya hai?

Yeh zero ki taraf jaati hai lekin kabhi reach nahi karti.

1/ r^{2}

fast shrink hota hai, toh large distance par force negligible ho jaati hai — phir bhi formally yeh har finite separation par nonzero hai (infinite range).

Agar do charges equal aur opposite hain (jaise

+ q

aur

- q

), toh kya har ek par force alag hai?

Nahi. Newton's third law se dono forces magnitude mein equal aur direction mein opposite hain. Dono

∣ q_{1} q_{2} ∣ = q^{2}

aur same

r

share karte hain, toh koi "jeetta" nahi.

Kya dono charges ko negative banana (same sizes) force ki magnitude ko dono positive ke compare mein change karta hai?

Nahi. Magnitude sirf

∣ q_{1} ∣∣ q_{2} ∣

par depend karti hai, toh

(+ q, + q)

aur

(- q, - q)

identical magnitudes dete hain — dono repulsive. Sirf sign pattern (like vs unlike), kaunsa specific sign nahi, attraction vs repulsion set karta hai.

Kya Coulomb's law ko "shield" kiya ja sakta hai jaise gravity ko nahi kiya ja sakta?

Haan. Kisi region ko conductors se gherne ya opposite charges arrange karne se ek electric field cancel ya block ho sakti hai, toh electric forces shieldable hain — gravity shield karne ka koi jaana-mana tarika nahi hai, kyunki mass sirf ek sign mein aata hai.

Agar aap ek charge double karo lekin uska sign flip karo, toh force ka kya hota hai?

Uski magnitude double ho jaati hai (kyunki

∣ q ∣

double hua) aur uski direction reverse ho jaati hai (like→unlike ya vice versa). Dono effects ek saath hote hain: zyada strong aur opposite kind (attract↔repel).

Recall Traps ka ek-line summary

Squared not linear distance ::: force $\propto 1/ r^{2}$ , toh $r$ double → force quarter. Magnitudes vs signs ::: $∣ q_{1} q_{2} ∣$ size ke liye, sign of product attract/repel ke liye. Point charges only ::: points ya spherical symmetry ke liye exact, warna integrate karo. Neutrality, not weakness ::: gravity cosmically dominate karti hai kyunki charges cancel ho jaate hain, electricity weak hai isliye nahi. $k$ vs $G$ ::: alag units (per C² vs per kg²), toh bare numbers compare nahi hote.