Electric charge — properties, quantization, conservation: Worked examples

1.8.1 · D3 · Physics › Electromagnetism › Electric charge — properties, quantization, conservation

Scenario matrix

Charge ke baare mein har problem in cells mein se ek hoti hai. Columns hain teen properties; rows hain "flavour" of input jo problem tumhare haath mein deta hai.

Cell	Input ka Flavour	Kaunsi property test hoti hai	Example #
A	Single sign, bricks count karo	Quantization ($n=	q
B	Ek quoted charge jo illegal ho sakta hai	Quantization (integer test)	2
C	Do like signs add kiye	Additivity (same sign)	3
D	Ek positive aur ek negative saath (cancellation)	Additivity (signed sum)	4
E	Zero / degenerate input (ek neutral object, $q = 0$ )	Boundary par teeno	5
F	Charge shared by touching identical bodies	Conservation + symmetry	6
G	Charge shared by unequal conductors + repeated contacts	Conservation, no symmetry	7
H	Limiting / macroscopic feel (kyun $e$ invisible hai)	Quantization limit	8
I	Real-world word problem (current = charge per second)	Additivity in time	9
J	Exam twist (pair production bookkeeping)	Conservation with creation	10

Ab hum har cell ko hit karte hain. Dekho sign kaise har ek mein travel karta hai — yahi woh thread hai jis par poora topic tika hua hai.

Example 1 — Cell A: bricks count karo

Worked example

$- 3.2 μ C$ ke charge mein kitne electrons hain? Forecast: pehle order of magnitude guess karo. Ek coulomb mein $\sim 6 \times 1 0^{18}$ electrons hote hain, aur hamare paas us scale ka ek millionth hai. Toh lagbhag $1 0^{13}$ expect karo. Woh number pakad ke rakho.

Step 1 — Magnitude likho. $∣ q ∣ = 3.2 \times 1 0^{- 6} C$ . Yeh step kyun? Bricks count karna sirf kitna charge hai iske baare mein sochta hai, sign ke baare mein nahi. Sign sirf hume batata hai ki electrons add hue (yeh negative hai), kitne nahin.

Step 2 — Total charge ko ek brick se divide karo. $e = 1.6 \times 1 0^{- 19} C$ ke saath, $n = \frac{∣ q ∣}{e} = \frac{3.2 \times 1 0 ^{- 6}}{1.6 \times 1 0 ^{- 19}} = 2 \times 1 0^{13}$ Yeh step kyun? Yahi quantization relation $q = n e$ rearranged hai. Hum literally pooch rahe hain " $∣ q ∣$ mein kitni size- $e$ bricks fit hoti hain?"

Verify: $2 \times 1 0^{13}$ ek whole number hai ✓ (quantization respected). Yeh hamare forecast band $\sim 1 0^{13}$ mein hai ✓. Wapas multiply karo: $2 \times 1 0^{13} \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 6}$ C ✓.

Example 2 — Cell B: kya yeh charge legal bhi hai?

Worked example Ek student claim karta hai ki ek dust grain

$q = 4.0 \times 1 0^{- 19} C$ carry karta hai. Kya possible hai? Forecast: elementary charge $1.6 \times 1 0^{- 19}$ C hai. Kya $4.0$ , $1.6$ ka whole multiple hai? Divide karne se pehle guess karo.

Step 1 — Brick count compute karo. $n = \frac{4.0 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 2.5$ Yeh step kyun? Koi bhi real, free charge $q = n e$ hona chahiye jahan $n$ ek integer ho. Isliye hum test karte hain ki $n$ whole number aata hai ya nahi.

Step 2 — Verdict padho. $n = 2.5$ integer nahi hai ⇒ ek free object ke liye impossible hai. Yeh step kyun? Aadha electron transfer nahi ho sakta. Free world mein aadhi brick exist nahi karti (quarks jo $\pm e /3$ carry karte hain woh confined hain — dekho Atomic Structure).

Verify: nearest legal charges hain $n = 2 \Rightarrow 3.2 \times 1 0^{- 19}$ C aur $n = 3 \Rightarrow 4.8 \times 1 0^{- 19}$ C. Claimed value $4.0 \times 1 0^{- 19}$ do legal rungs ke beech hai, yeh confirm karta hai ki yeh forbidden hai ✓.

Example 3 — Cell C: do like signs add karna

Example 4 — Cell D: positive meets negative (cancellation)

Figure — Electric charge — properties, quantization, conservation

Figure s01 — signed charges kaise cancel hote hain (Cell D). Top row (cyan squares marked $+$ ) saat positive bricks $+ 7 e$ hain; bottom row (marked $-$ ) das negative bricks $- 10 e$ hain. $x$ -axis sirf brick position hai (left se right); koi $y$ -axis quantity nahi hai — do rows sirf "positive shelf" aur "negative shelf" hain. Thin white lines un saat pairs ko join karti hain jo annihilate hote hain. Bottom right par teen amber bricks woh unpaired negatives hain jo survive karti hain, toh net $- 3 e$ hai. Take-away: signed addition = opposite bricks ko pair off karo, aur jo bhi colour bachta hai woh final sign set karta hai.

Example 5 — Cell E: degenerate case ( $q = 0$ )

Worked example Ek copper ball

neutral hai. Kya usme koi charge nahi hai? $n$ kya hai? Dusre charges se door yeh kaunsi force feel karta hai? Forecast: "neutral" "empty" jaisa lagta hai, lekin ek metal charges se bhara hota hai. Guess: bahut saara charge, net zero.

Step 1 — Net charge. Neutral ka matlab $Q_{total} = 0$ hai. Yeh step kyun? Neutral ek statement hai sum ke baare mein, pieces ke baare mein nahi. Protons aur electrons equal numbers mein present hain (dekho Atomic Structure).

Step 2 — Net ka brick count. $n = \frac{∣ Q _{total} ∣}{e} = \frac{0}{e} = 0$ Yeh step kyun? $n = 0$ ek perfectly valid integer hai — quantization empty rung allow karta hai. Zero additive identity hai: ise add karne se kuch nahi badalta.

Step 3 — Door single charge se force. Net (monopole) Coulomb force net charge ke proportional hai, isliye $Q_{total} = 0$ ke saath leading force zero hai (dekho Coulomb's Law). Yeh step kyun? Coulomb's law net charge ke saath scale karta hai; zero net charge dominant term ko kill karta hai.

Step 4 — Edge case (over-claim "no force" mat karo). Ek nonuniform field mein, ek neutral conductor phir bhi polarise ho jaata hai: uske apne electrons shift hote hain taaki near face source ke opposite charge wala ho, ek chhoti induced-dipole attraction deta hai. Exactly isliye ek charged rod neutral kaagaz ke tukdon ko pick up karta hai. Yeh step kyun? "Zero net charge" sirf monopole force ko kill karta hai, nonuniform field mein weaker induced force ko nahi. Ise cover karna scenario ko honestly close karta hai.

Verify: $n = 0$ ek integer hai ✓. Equal protons aur electrons: e.g. $1 0^{23}$ each deta hai $1 0^{23} (+ e) + 1 0^{23} (- e) = 0$ ✓. Degenerate case kisi bhi rule ko todhe bina handle kiya ✓.

Example 6 — Cell F: identical spheres touch (symmetry)

Worked example Identical metal spheres

$+ 12 e$ aur $- 4 e$ carry karti hain. Woh touch karte hain, phir alag ho jaate hain. Har ek par charge? Forecast: conservation total fix karta hai; identical geometry ise evenly split karta hai. Computing se pehle guess karo.

Step 1 — Conservation total deta hai. $Q_{total} = (+ 12 e) + (- 4 e) = + 8 e (hamesha fixed)$ Yeh step kyun? Touch karne se electrons do spheres ke beech flow kar sakte hain, lekin isolated pair net charge gain ya lose nahi kar sakta — yahi conservation action mein hai.

Yeh phrase "time mein constant rehta hai" compactly likhi ja sakti hai. Maano $Q_{total}$ total charge hai aur $t$ time hai (seconds mein). Symbol $\frac{d Q _{total}}{d t}$ ka matlab hai "woh rate jis par $Q_{total}$ change hoti hai jab $t$ aage badhta hai" — kitne coulombs per second total gain ya lose hota hai. Conservation kehta hai woh rate exactly zero hai: $\frac{d Q _{total}}{d t} = 0 ⟺ Q_{total} kabhi nahi badlata.$

Step 2 — Symmetry ise split karta hai. $q_{A} = q_{B} = \frac{+ 8 e}{2} = + 4 e$ Yeh step kyun? Identical conductors equal potential par pahunchte hain, aur equal geometry at equal potential ⇒ equal charge (kyun woh touch karte hain: Methods of Charging).

Verify: $q_{A} + q_{B} = + 4 e + 4 e = + 8 e = Q_{total}$ ✓ (conservation closes). $+ 4$ ek integer hai ✓.

Example 7 — Cell G: unequal conductors, phir ek repeated contact

Worked example Sphere X (radius

$2 R$ ) $+ 9 e$ hold karta hai; sphere Y (radius $R$ , initially neutral) ise ek baar touch karta hai, phir remove ho jaata hai. Har ek ka charge find karo. Phir ek doosra Y-jaisa identical sphere Z (radius $R$ bhi, neutral) Y ko touch karta hai aur alag ho jaata hai. Z find karo. Forecast: unequal spheres evenly split nahi karte — bada wala zyada share rakhta hai, radius ke proportion mein. Toh X ko Y se zyada milna chahiye. Guess karo X ko roughly double Y milega.

Step 1 — Pehle contact ke liye conservation. Total $= + 9 e + 0 = + 9 e$ , aur yeh touch karne ke dauran fixed hai. Yeh step kyun? Wahi conservation rule: isolated pair ka total change nahi ho sakta.

Step 2 — Radius se unequal split (key new idea). Contact mein spheres ke liye, charge distribute hota hai taaki dono equal potential par pahunchein; ek sphere ka charge uski radius ke proportional hota hai. Radii $2 R$ (X) aur $R$ (Y) ke saath, shares $2 : 1$ ratio mein hain: $q_{X} = \frac{2}{3} (9 e) = 6 e, q_{Y} = \frac{1}{3} (9 e) = 3 e$ Yeh step kyun? Yahi woh hai jo Example 6 ki symmetry ko todta hai. Sirf tab jab radii equal hoti hain split $50 : 50$ hoti hai. Yahan $2 : 1$ radius ratio ek $2 : 1$ charge ratio force karta hai. (Dono shares yahan whole electrons par land karte hain, isliye quantization abhi stressed nahi hai.)

Step 3 — Doosra contact (Y aur identical neutral Z ke beech). Ab Y (radius $R$ ) aur Z (radius $R$ ) equal hain, isliye pehle wali symmetry rule apply hoti hai. Total $= 3 e + 0 = 3 e$ . Ideal continuous-fluid answer $1.5 e$ each dega — lekin woh legal charge nahi hai ( $1.5$ integer nahi hai). Reality ise whole electrons distribute karke resolve karti hai: ek sphere $2 e$ ke saath end hoti hai, doosri $1 e$ ke saath. Kaun sa sphere extra electron leta hai yeh symmetry se fix nahi hota — yeh contact ke moment ki random microscopic details se decide hota hai (thermal jiggling, tiny asymmetries), isliye kai trials mein yeh coin-flip hai. Hum sirf ek outcome label karte hain: maano $q_{Y} = 2 e, q_{Z} = 1 e$ ; mirror outcome $q_{Y} = 1 e, q_{Z} = 2 e$ equally valid hai. Yeh step kyun? Yahan do lessons collide karte hain: equal radii chahti hain equal share, lekin ek odd electron count ko halve nahi kiya ja sakta. Quantization smooth split ko override karta hai, aur leftover single electron ek side par randomly jaata hai — koi physical rule nahi hai jo Y ko Z se prefer kare.

Verify: pehla split conserve karta hai: $6 e + 3 e = 9 e$ ✓, aur ratio $6 : 3 = 2 : 1$ radius ratio se match karta hai ✓. Doosra split either labelling ke liye conserve karta hai: $2 e + 1 e = 3 e$ ✓. Har charge ( $6 e, 3 e, 2 e, 1 e$ ) $e$ ka integer multiple hai ✓.

Example 8 — Cell H: quantization kyun chhup jaata hai (the limit)

Worked example Tumhare phone battery lagbhag

$Q = 1 C$ circuit se push karta hai. Ek electron is total ko kitne fraction se change karta hai? Charge "smooth" kyun lagta hai? Forecast: ek electron ek coulomb mein se — expect karo ek absurdly tiny fraction.

Step 1 — Fractional step size. $\frac{e}{Q} = \frac{1.6 \times 1 0 ^{- 19}}{1} = 1.6 \times 1 0^{- 19}$ Yeh step kyun? Quantization tabhi visible hoti hai jab ek brick pure mein ek bada slice ho. Yahan ek brick total ka $\sim 1 0^{- 19}$ hai — bilkul negligible.

Step 2 — Limit interpret karo. Jaise jaise electrons ki sankhya bahut badi hoti hai, rung $n$ aur $n + 1$ ke beech relative gap $1/ n \to 0$ hota hai. Staircase ek ramp jaisi dikhti hai. Yeh step kyun? Exactly isliye parent note kehta hai $e$ "daily life mein invisible" hai — same reason sand door se smooth fluid jaisi lagti hai.

Verify: ek coulomb mein electrons ki sankhya (ise $N$ kaho, ek pure coulomb ka brick-count) $N = Q / e = 1/ e \approx 6.24 \times 1 0^{18}$ hai ✓ (wahi $q = n e$ relation Example 1 ki tarah, bas $q = 1$ C ke saath). Ek brick add karne se fractional change $1/ N \approx 1.6 \times 1 0^{- 19}$ hota hai ✓ — undetectable.

Figure s02 — large $n$ par quantization chhup jaati hai (Cell H). Dono panels ek hi cheez plot karte hain: horizontal axis $n$ hai, bricks ki sankhya (ek integer count); vertical axis charge hai $e$ ke units mein. Amber line true staircase $q = n e$ hai (charge har step par ek whole brick se jump karta hai); cyan dots (left) aur cyan dashed line (right) ideal $q = n$ trend mark karte hain. Left panel, small $n$ : amber steps values ke relative tall hain, isliye quantization obvious hai. Right panel, large $n$ : wahi amber staircase, $n$ tak $200$ tak zoom out kiya, near-smooth ramp mein flatten ho jaata hai kyunki har step ka relative size $1/ n$ zero ki taraf shrink karta hai. Physics dono panels mein identical hai — sirf scale badla.

Example 9 — Cell I: real-world word problem (charge in time)

Worked example Ek wire

$2 A$ (amperes) ka steady current $5 seconds$ tak carry karta hai. Kitna charge flow hota hai, aur kitne electrons? Forecast: current "charge per second" hai, isliye total charge hona chahiye current $\times$ time $= 10$ C.

Step 1 — Current symbol naam karo. Maano $I$ current ko represent karta hai: wire ke ek point se har second kitna charge pass hota hai, amperes (A) mein measure kiya. Definition se $1 A = 1 C per second$ . Toh yahan $I = 2 A$ ka matlab hai har second do coulombs cross karte hain. Yeh step kyun? Letter $I$ current ka standard symbol hai; ise formula mein use karne se pehle hum ise uske meaning se attach karna chahiye.

Step 2 — Current se charge. Kyunki har second $I$ coulombs deliver karta hai aur $t$ seconds pass hote hain, $Q = I t = 2 A \times 5 s = 10 C$ Yeh step kyun? Time mein additivity: har second $2$ C deliver karta hai, aur $5$ alag seconds simply add hote hain: $2 + 2 + 2 + 2 + 2 = 10$ . Wahi signed-sum rule, time mein spread.

Step 3 — Electrons mein convert karo. $n = \frac{Q}{e} = \frac{10}{1.6 \times 1 0 ^{- 19}} = 6.25 \times 1 0^{19}$ Yeh step kyun? Jo bricks pass hue unhe count karo, quantization phir use karke.

Verify: $I t$ ke units $A \cdot s = C$ hain ✓. Written form $6.25 \times 1 0^{19}$ electrons ki ek whole number hai: $6.25 \times 1 0^{19} = 625 \times 1 0^{17} = 62, 500, 000, 000, 000, 000, 000$ (ek 20-digit integer; decimal $6.25$ sirf $\times 1 0^{19}$ shorthand hai, fractional electron nahi) ✓. Yeh parent note ke " $6.25 \times 1 0^{18}$ per coulomb" fact ka bhi $10 \times$ hai ✓.

Example 10 — Cell J: exam twist (pair production)

Worked example Ek gamma-ray photon (charge

$0$ ) ek nucleus ke paas ek electron aur ek positron mein convert hota hai. Baad mein total charge kya hai, aur yeh empty vacuum mein kyun nahi ho sakta? Forecast: photon ka zero charge hai; jo bhi appear ho woh bhi zero sum hona chahiye. Toh expect karo $- e$ aur $+ e$ saath mein.

Step 1 — Charge bookkeeping. Pehle: photon, $q = 0$ . Baad mein: electron $(- e)$ + positron $(+ e)$ : $Q_{after} = (- e) + (+ e) = 0 = Q_{before}$ Yeh step kyun? Conservation kehta hai net charge change nahi ho sakta; charge sirf equal-and-opposite pairs mein create ho sakta hai. Zero in, zero out.

Step 2 — Twist: nucleus kyun zaruri hai. Charge kahin bhi balance hota hai, lekin free space mein ek akela photon do massive particles banate waqt energy aur momentum simultaneously conserve nahi kar sakta. Ek nearby nucleus recoil momentum absorb karta hai. Yeh step kyun? Yeh classic trap hai: students sochte hain ki charge conservation akele ise permit karta hai. Yeh zaruri hai lekin sufficient nahi — momentum conservation extra gatekeeper hai.

Verify: $(- e) + (+ e) = 0$ ✓ photon ke $0$ se match karta hai ✓. Net electron-type count $0$ se badla (ek electron, ek anti-electron) ✓ — conservation intact.

Recall Scenario checklist (khud test karo)

Ek single charge diya ho, electrons kaise count karoge? ::: $n = ∣ q ∣/ e$ ; sign sirf batata hai added/removed. Jawaab mein $2.5 e$ charge aata hai — kya conclude karte ho? ::: Yeh free particle ke liye illegal hai; $n$ integer hona chahiye. Do identical spheres $+ 12 e$ aur $- 4 e$ touch karte hain — har ek ke paas kya bachega? ::: $+ 4 e$ (total $+ 8 e$ symmetry se equally shared). Sphere X ( $2 R$ , $+ 9 e$ ) ko neutral Y ( $R$ ) ek baar touch karta hai — kaise split hoga? ::: Radius ratio $2 : 1$ se, $6 e$ aur $3 e$ milenge. Y ( $3 e$ ) identical neutral Z ko touch karne ke baad, $1.5 e$ each kyun nahi hoga? ::: $1.5$ legal charge nahi hai; whole electrons $2 e$ aur $1 e$ force karte hain, extra electron ek random side par jaata hai. $2 A$ for $5 s$ mein kitna charge deliver hota hai? ::: $Q = I t = 10 C$ . Ek akela photon vacuum mein pair-produce kyun nahi kar sakta? ::: Charge balance hota hai lekin energy AUR momentum dono ek recoil body ke bina conserve nahi ho sakte. Kya ek neutral conductor charged rod ke paas koi force feel karta hai? ::: Haan — nonuniform field mein ek chhoti induced-dipole attraction, chahe net charge zero ho.

Active-recall flashcards

- 3.2 μ

C mein kitne electrons hote hain?

2 \times 1 0^{13}

Kya

q = 4.0 \times 1 0^{- 19}

C allowed hai?

Nahi;

n = 2.5

integer nahi hai.

+ 7 e

aur

- 10 e

ka net?

- 3 e

+ 12 e

aur

- 4 e

wali identical spheres touch karein — har ek ke paas?

+ 4 e

Unequal spheres

2 R

(

+ 9 e

) aur

R

(neutral) touch karein — split?

6 e

aur

3 e

(radius ratio

2 : 1

se).

Y ke paas

3 e

hai aur woh identical neutral Z ko touch karta hai — result?

2 e

aur

1 e

(na ki

1.5 e

each; odd electron ek random side par jaata hai).

2

A mein

5

s ka charge?

10

C (

6.25 \times 1 0^{19}

electrons).

Photon se pair production ke baad total charge?

0

(ek

- e

aur ek

+ e