Exercises — Electric charge — properties, quantization, conservation

Rule: har free charge $q=ne$ hona chahiye jahan $n$ ek whole number ho. Isliye $\frac{|q|}{e}$ ka result whole aana chahiye.

• (a) $\frac{3.2\times 10^{-19}}{1.6\times 10^{-19}}=2$ ✓
• (b) $\frac{8.0\times 10^{-19}}{1.6\times 10^{-19}}=5$ ✓
• (c) $\frac{4.0\times 10^{-19}}{1.6\times 10^{-19}}=2.5$ ✗ — whole number nahi
• (d) $\frac{1.6\times 10^{-18}}{1.6\times 10^{-19}}=10$ ✓

Answer: (c) impossible hai. Aadha brick nahi ho sakta.

Electrons glass se nikal ke silk par chale gaye. Protons nucleus mein locked rehte hain aur rubbing mein move nahi karte. Negative electrons kho dene se glass mein electrons se zyada protons ho jaate hain → net positive. (Parent note mein dekho, "Where does charge physically come from?")

Galat. Conservation sirf isolated system ka total fix karta hai. Individual objects aazaadi se charge swap kar sakte hain — exactly yahi kaam karte hain do touching spheres. Sirf sum $\sum q$ constant rehta hai.

KYA: hum bricks count kar rahe hain. KYUN: positive charge ka matlab hai electrons hataaye gaye, aur har hataya gaya electron ek brick $e$ ka negative charge subtract karta hai, peeche $+e$ chodke. Count hai: $n=\frac{|q|}{e}=\frac{4.8\times 10^{-19}}{1.6\times 10^{-19}}=3.$ Answer: 3 electrons hataye gaye.

KYUN minus "extra" batata hai: ek negative object ne electrons gain kiye hain. $n=\frac{|q|}{e}=\frac{2\times 10^{-6}}{1.6\times 10^{-19}}=1.25\times 10^{13}.$ Answer: $1.25\times 10^{13}$ extra electrons.

Additivity, sign ke saath: $Q_{\text{total}}=(+5e)+(-3e)+(-9e)=-7e.$ Coulombs mein: $-7\times 1.6\times 10^{-19}=-1.12\times 10^{-18}\text{C}$. Answer: $-7e=-1.12\times 10^{-18}\text{C}$. (Kyunki charge ek scalar hai, humne sirf signed numbers add kiye — koi directions, koi vectors nahi.)

Step 1 — pehla touch (conservation): pehle total $=(+8e)+(-2e)=+6e$. Touch karne se sirf unke beech charge flow ho sakta hai; isolated pair ka total fixed rehta hai. Step 2 — symmetry: identical spheres equal potential par pahunchti hain, isliye equally share karti hain: $q_A=q_B=\frac{+6e}{2}=+3e.$ Step 3 — teesri (neutral) sphere A ko touch karti hai: ab A ($+3e$) aur neutral sphere ($0$) share karti hain. Total $=+3e+0=+3e$, do identical spheres ke beech equally split: $q_A=\frac{+3e}{2}=+1.5e.$ Dhyan rakho: $+1.5e$ $e$ ka integer multiple nahi hai! Kyunki $8e$ aur $2e$ ka matlab hai $8$ aur $2$ whole electrons, $6e$ ko $3e$ mein share karna theek hai, lekin $3e$ ko $1.5e$ mein share karna physically exact nahi hai — reality mein ek sphere $+2e$ (2 electrons kam) aur doosri $+1e$ leti hai, jis taraf bhi aakhri electron jaata hai. Idealised textbook symmetry mein hum $+1.5e$ ko average ke roop mein quote karte hain, lekin sachche instantaneous charges har ek $e$ ke integer multiples hone chahiye. Answer: pehle touch ke baad, $+3e$ each. Neutral teesri sphere ke baad, A ka ideal share $+1.5e$ hai (physically $+2e$ ya $+1e$).

$n=\frac{|q|}{e}=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}.$ Yeh six billion billion se zyada electrons hain. $n$ aur $n+1$ electrons ke beech "step size" $6.25\times 10^{18}$ mein se $1$ brick hai — bilkul negligible. Isliye macroscopic charge continuous lagta hai, haalaanki woh actually discrete bricks se bana hai (jaise baalu door se smooth lagti hai). Answer: $\approx 6.25\times 10^{18}$ electrons.

Step 1 — pair production bookkeeping: photon ka charge $0$ hota hai. Yeh ek electron ($-e$) aur positron ($+e$) banata hai. Net charge change $=(-e)+(+e)=0$. Charge conserved hai — equal-and-opposite pair mein create hota hai, kabhi akela nahi. Step 2 — sphere system mein aane wala charge collect karo: P laata hai $+6e$, electron laata hai $-e$, Q laata hai $0$. $Q_{\text{total}}=(+6e)+(-e)+0=+5e.$ Step 3 — do identical spheres P aur Q ke beech share karo (electron ka charge contact par P mein merge ho jaata hai, isliye hum $+5e$ ko do spheres par split karte hain): $q_P=q_Q=\frac{+5e}{2}=+2.5e.$ Phir se ideal average $+2.5e$ hai; physically whole-electron constraint $+3e$ ek par aur $+2e$ doosri par deta hai. Step 4 — global conservation check: event se pehle, isolated system mein tha photon ($0$) + P ($+6e$) + Q ($0$) = $+6e$. Baad mein: positron ($+e$) ud gaya, aur do spheres total $+5e$ carry karti hain. Sum $=(+e)+(+5e)=+6e$. ✓ Total unchanged. Answer: $+2.5e$ average per sphere (physically $+3e$ aur $+2e$); total system charge poore time $+6e$ rehta hai.

KYUN net electrons track karein: electrons kho dena positive charge add karta hai; electrons gain karna negative charge add karta hai. Net electron change: $\Delta n=-(5\times 10^{13})+(2\times 10^{13})=-3\times 10^{13}\text{electrons (net lost)}.$ Net lost electrons → net positive charge. Magnitude: $|q|=(3\times 10^{13})(1.6\times 10^{-19})=4.8\times 10^{-6}\text{C}=+4.8\mu \text{C}.$ Answer: $+4.8\mu \text{C}$.

KYUN yeh kaam karta hai: agar har charge $q=ne$ hai integer $n$ ke saath, toh $e$ ek common "brick size" hona chahiye jo saare measured values ko whole numbers mein divide kare. Hum sabse bada number dhundh rahe hain jo fit ho. Har ek ko $1.6\times 10^{-19}$ ke multiple ke roop mein likho:

• $3.2=2\times 1.6$
• $6.4=4\times 1.6$
• $8.0=5\times 1.6$
• $4.8=3\times 1.6$
• $11.2=7\times 1.6$

Har measurement $1.6\times 10^{-19}\text{C}$ ka whole-number multiple hai, aur koi bada common divisor kaam nahi karta (jaise $3.2$ sirf $2$ bricks hai, isliye brick $1.6$ se bada nahi ho sakta). Isliye $e=1.6\times 10^{-19}\text{C}.$ Answer: $e=1.6\times 10^{-19}\text{C}$ — sirf data ki discreteness se recover kiya gaya. Yahi Millikan ke real experiment ki logic hai.

$+\frac{1}{3}e$ ka charge exactly kuch quarks ka charge hai. Lekin quarks confined hote hain — woh kabhi akele nahi milte. Kisi bhi free, isolated, observable particle ka charge $q=ne$ hona chahiye jahan $n$ ek whole number ho. Kyunki $\frac{1}{3}$ whole number nahi hai, yeh claim ek free particle ke liye impossible hai. Free charge ka quantum $e$ hai, $e/3$ nahi.

$n=|q|/e$ se bricks count karo; charges ko unke signs ke saath add karo; isolated total ko constant rakho; aur hamesha sanity-check karo ki koi bhi final charge $e$ ka whole multiple hai.