Dielectrics — polarization, dielectric constant, effect on capacitance: Visual walkthrough

1.8.14 · D2 · Physics › Electromagnetism › Dielectrics — polarization, dielectric constant, effect on c

Hum bas itna assume karte hain jo pehle se hai: ek field $E$ har point par ek arrow hai jo kehta hai "kis taraf, aur kitni force se, ek $+ 1$ charge ko dhakka lagega" (dekho Electric Field of Parallel Plates), aur ek capacitor charge $Q$ ko voltage $V$ par store karta hai jahan $C = Q / V$ (dekho Capacitance and Capacitors). Baaki sab kuch, hum scratch se banate hain.

Definition Symbol glossary (is page par use hone wale sab symbols)

Ek baar skim karo, zaroorat par wapas dekho. Symbols wahan phir define kiye jaate hain jahan pehli baar kaam aate hain.

$Q$ — ek plate par charge (coulombs, C).
$A$ — ek plate ka area (m²).
$d$ — do plates ke beech ka gap (m). Jab slab poora gap fill kar le, to uski thickness $t$ equals $d$ — hum poore walkthrough mein yahi full-fill case maante hain (Step 9 partial fill handle karta hai).
$σ_{f}$ — plates par free surface charge density, $= Q / A$ (C/m²).
$σ_{b}$ — slab ke faces par bound surface charge density (C/m²).
$ε_{0}$ — permittivity of free space, $8.85 \times 1 0^{- 12}$ C²/(N·m²) — charge-per-area se field ka "exchange rate."
$E_{0}$ — bina slab ke vacuum field; $V_{0} = E_{0} d$ vacuum voltage hai; $C_{0} = Q / V_{0} = ε_{0} A / d$ vacuum capacitance hai.
$E, V, C$ — wahi quantities slab ke saath.
$p, P$ — ek molecule ka dipole moment, aur dipole moment per volume.
$χ_{e}$ — electric susceptibility (dimensionless); $κ = 1 + χ_{e}$ — dielectric constant (dimensionless).

Step 1 — Do bare plates: woh field jahan se hum shuru karte hain

KYA. Left plate par $+ Q$ aur right plate par $- Q$ daalo, dono ka area $A$ , aur gap $d$ se alag. Phail kar, charge ek surface charge density hai: $σ_{f} = \frac{Q}{A} .$

$σ$ (sigma) = charge packed per square metre, units $C/m^{2}$ .
Subscript $f$ ka matlab free charge hai — woh charge jo battery ne actually metal par deliver kiya. Hum ise "free" isliye kehte hain kyunki yeh wire se hokar move kar sakta hai; yeh word yaad rakhna, ek aur tarah ka charge aane wala hai.
$d$ = woh gap jise field cross karta hai; Step 8 mein ise field ko voltage mein convert karne ke liye use karenge.

YUN YAHAN SE SHURU KARTE HAIN. Koi insulator exist karne se pehle bas yahi tha. Iska field hamara reference hai — woh cheez jise dielectric baad mein fight karega.

PICTURE. Uniform blue arrows seedhe $+$ plate se $-$ plate ki taraf march kar rahe hain. Ek charged sheet ka field $σ /2 ε_{0}$ hota hai; do opposite sheets plates ke beech add ho jaate hain (dekho Electric Field of Parallel Plates), jo deta hai: $E_{0} = \frac{σ _{f}}{ε _{0}} .$

$E_{0}$ = vacuum field, subscript $0$ = "empty space, kuch insert nahi."
$ε_{0}$ (epsilon-nought) = nature ka ek fixed constant, SI units mein $8.85 \times 1 0^{- 12}$ . Yeh "charge per area" ko "field strength" mein convert karta hai. Ise charge-crowding aur push ke beech exchange rate samjho.

Figure — Dielectrics — polarization, dielectric constant, effect on capacitance

Step 2 — Neutral matter ka ek slab andar slide karo

KYA. Hum ek insulating slab (glass, plastic, ek bag mein paani) poori tarah plates ke beech push karte hain, taaki uski thickness $t$ gap $d$ ke equal ho. Yeh har jagah electrically neutral hai — andar utni hi $+$ aur $-$ amount hai, isliye pehle yeh kuch nahi badalta.

KYUN. Hume jaanna hai ki bahari field $E_{0}$ matter ke saath kya karta hai. Metal mein kuch nahi badlata; ab poori kahaani slab ke andar hone wali cheez ke baare mein hai.

PICTURE. Slab ko ek translucent block ke roop mein draw karo. Uske andar, molecules ko chhote neutral circles ki tarah sketch karo: ek $+$ core aur ek $-$ cloud bilkul ek doosre ke upar baithe hain, isliye har chhoti unit ka net displacement zero hai — abhi koi arrows nahi.

Step 3 — Field har molecule ko stretch karke dipole banata hai

KYA. Blue field $E_{0}$ har $+$ core ko thoda right ki taraf dhakelta hai aur har $-$ cloud ko thoda left ki taraf kheenchta hai. Har molecule ek dipole ban jaata hai: ek $+$ aur $-$ ek chhoti si distance $δ$ se alag.

YEH IDEA KYUN, AUR "DIPOLE" KYUN? Field opposite charges ko opposite taraf push karta hai — yahi field ki definition hai. Jab tum $+ q$ aur $- q$ ko $δ$ se alag karte ho, physicists us cheez ko ek number mein band kar lete hain, dipole moment: $p = q δ,$ kyunki pair ka poora bahari effect sirf is product par depend karta hai, $q$ aur $δ$ alag-alag par nahi. (Yahi object hai jo Electric Dipoles and Torque mein torque experience karta hai.)

$q$ = woh charge jo shift hua, $δ$ = kitna aur kis taraf $+$ gaya $-$ ke relative ( $δ$ use karte hain, $d$ nahi, taaki plate gap se alag rahe).
$p$ $-$ se $+$ ki taraf point karta hai, yaani yahan field ke saath saath.

PICTURE. Step 2 wale hi molecules, ab stretched: red $+$ thoda right, blue $-$ thoda left, har ek par ek chhota green arrow $p$ right ki taraf point karta hua.

Step 4 — Dipoles ko add karo: polarization $P$

KYA. Aisi billions chhoti $p$ vectors, sab same taraf point kar rahi hain, ek ek karke track karna mushkil hai. Isliye hum unhe add karte hain aur jo volume they fill karte hain us se divide karte hain. Woh density hai polarization: $P = \frac{\sum p _{i}}{Volume} .$

VOLUME SE KYUN DIVIDE KARTE HAIN? Kyunki ek moti slab mein ek patli slab se zyada molecules hote hain lekin material ke har bit ke liye woh "zyada polarized" nahi hoti. Volume se divide karne par ek intensive quantity milti hai — material-in-this-field ki property, chahe tumne kitna bhi slab liya ho uski parwah nahi.

Units check (yeh crucial bridge hai). $p$ ke units $C \cdot m$ hain; volume $m^{3}$ hai; isliye: $[P] = \frac{C \cdot m}{m ^{3}} = \frac{C}{m ^{2}} .$ Yeh ek charge per area hai — $σ$ ke same units. Yeh coincidence nahi hai; Step 5 batata hai kyun.

PICTURE. Zoom out karo: individual arrows blur hokar ek smooth green field $P$ bann jaate hain jo slab mein fill ho jaata hai, right ki taraf point karta hai, corner mein $C/m^{2}$ units-badge dikhta hai.

Step 5 — Woh bada cancellation: bound surface charge

KYA. Sab stretched molecules ko ek row mein line karo. Ek molecule ka $+$ end apne neighbour ke $-$ end ke bilkul paas baithta hai — woh cancel ho jaate hain. Andar har jagah, plus minus se milta hai aur net charge zero hai. Sirf do faces par koi neighbour cancel karne wala nahi hota: left face par exposed $-$ reh jaata hai, right face par exposed $+$ . Us leftover ko bound surface charge $σ_{b}$ bolte hain.

Bound charge $σ_{b}$ : real charge hai, lekin molecules se chipka hua — yeh wire se hokar nahi ja sakta. "Free" ke opposite.

Sign vs magnitude — ek clean rule. Poora statement vector wala hai: $σ_{b} = P \cdot \overset{n}{^},$ jahan $\overset{n}{^}$ ek face par outward-pointing unit arrow hai. Dot product sign carry karta hai: right face par $P$ bahar ki taraf point karta hai ( $P \cdot \overset{n}{^} = + P$ , positive charge); left face par $P$ andar ki taraf point karta hai ( $P \cdot \overset{n}{^} = - P$ , negative charge). Jab hum baad mein sirf " $σ_{b} = P$ " likhte hain to hum ek aisi face par magnitude ki baat kar rahe hain jahan $P$ perpendicular hai; sign yeh decide karta hai ki kaun si face hai. Vector form ko sach maano aur $σ_{b} = P$ ko hamare do flat faces par uski magnitude.

Interior neutral kyun rehta hai. Volume bound charge ke liye general rule hai $ρ_{b} = - \nabla \cdot P$ (kitna $P$ ek point par "spread out ya pile up" hota hai). Haara $P$ slab ke andar uniform hai, isliye $\nabla \cdot P = 0$ aur $ρ_{b} = 0$ — koi bulk charge nahi, bilkul wahi pairwise cancellation jo humne draw ki. Saara leftover charge do faces par squeeze ho jaata hai.

$σ_{b}$ magnitude mein $P$ ke equal kyun hai. Slab ke total dipole moment ke do honest tarike agree karne chahiye. Tarika 1 (density $\times$ volume): $P \cdot (A t)$ , jahan $A$ face area hai, $t$ thickness hai. Tarika 2 (charge $\times$ lever arm): har face par exposed charge $Q_{b} = σ_{b} A$ , slab thickness $t$ se separated, khud ek giant dipole hai: $Q_{b} \cdot t = σ_{b} A t$ . Dono ko equal karo: $P (A t) = σ_{b} A t ⟹ σ_{b} = P .$ $A$ aur $t$ cancel ho jaate hain — geometry saara kaam karti hai.

PICTURE. Ek rectangle ke andar dipoles ki ek row; har interior $\pm$ pair grey mein struck through (cancelled); sabse left wala $-$ aur sabse right wala $+$ red/blue mein circled — wahi $σ_{b}$ hain.

Step 6 — Bound charge fight back karta hai: net field shrink ho jaata hai

KYA. Ab slab ke faces par charge ki do sheets hain: bound $- σ_{b}$ $+$ plate se chipki hui, bound $+ σ_{b}$ $-$ plate se chipki hui. Har sheet apna field $E_{b} = σ_{b} / ε_{0}$ banati hai jo $E_{0}$ ke opposite point karta hai (bound $+$ se bound $-$ ki taraf, yaani leftward). Andar ka real field tug-of-war winner hai: $E = E_{0} - E_{b} = \frac{σ _{f} - σ _{b}}{ε _{0}} .$

$E_{0}$ = original push (right), $E_{b}$ = dielectric ka counter-push (left).
Addition nahi, subtraction — kyunki dono opposite taraf point karte hain — wahi ek minus sign "dielectrics field ko kamzor karte hain" ki poori kahaani hai.

BOUND SHEET OPPOSE KYUN KARTA HAI. $+$ plate ne har molecule ka $-$ end attract kiya, isliye $+$ plate ke paas wala face negative hai. Ek negative sheet ek positive plate ke paas ek aisa field banati hai jo wapas plate ki taraf point karta hai — $E_{0}$ ke against. Yeh oppose karna zaroori hai; yeh geometry hai, luck nahi.

PICTURE. Strong blue $E_{0}$ arrows, ek chhota red $E_{b}$ arrow har jagah opposite direction mein, aur surviving green net arrow $E$ — $E_{0}$ se visibly chhota.

Step 7 — Loop close karo: $E$ exactly $κ$ se shrink hota hai

KYA. Hamare linear dielectric ke liye (assumption box dekho), jitna zyada push karo, utna zyada polarize hoga — proportionally: $P = ε_{0} χ_{e} E .$

$χ_{e}$ (chi, "kai") = electric susceptibility — material ki polarize hone ki eagerness (dekho Electric Susceptibility). Yeh dimensionless hai ( $ε_{0} E$ ke units pehle hi $P$ ke C/m² se match karte hain, isliye $χ_{e}$ koi units nahi carry karta). Bada $χ_{e}$ = bahut stretch hota hai; $χ_{e} = 0$ = vacuum. Yeh zyada common permittivity language se is tarah juda hai: $ε = ε_{0} (1 + χ_{e}) = ε_{0} κ$ , jahan $ε$ material ki permittivity hai aur $ε_{r} = ε / ε_{0} = κ$ "relative permittivity" hai.

$σ_{b} = P = ε_{0} χ_{e} E$ ko Step 6 mein feed karo: $E = \frac{σ _{f} - ε _{0} χ _{e} E}{ε _{0}} = E_{0} - χ_{e} E .$ Ab $E$ ke liye solve karo. AGLA LINE KYUN: $E$ dono sides par appear karta hai, isliye hum dono sides mein $χ_{e} E$ add karte hain taaki saara $E$ left par aa jaaye — "unknown ko akela karo" ka standard move: $E + χ_{e} E = E_{0} \Rightarrow E (1 + χ_{e}) = E_{0} \Rightarrow E = \frac{E _{0}}{1 + χ _{e}} = \frac{E _{0}}{κ} .$

YEH FINISH LINE KYUN HAI. Hum define karte hain: $κ = 1 + χ_{e},$ isliye messy susceptibility ek dimensionless number mein collapse ho jaati hai: woh factor jitna field shrink hua. $κ = 1$ matlab "koi eagerness nahi, vacuum"; bada $κ$ (paani $\approx 80$ ) matlab "field almost wiped out."

PICTURE. Side-by-side: left panel tall blue arrow $E_{0}$ ; right panel wahi arrow $E_{0} / κ$ tak scale down, ek $\div κ$ label dono ko bridge karta hua.

Step 8 — Kamzor field se badi capacitance

KYA. Plates ke beech voltage field $\times$ gap hota hai: $V = E d$ , jahan $d$ plate separation hai (from Electric Field of Parallel Plates). Slab se pehle, vacuum voltage hai: $V_{0} = E_{0} d,$ aur vacuum capacitance hai: $C_{0} = \frac{Q}{V _{0}} = \frac{Q}{E _{0} d} = \frac{Q}{( σ _{f} / ε _{0} ) d} = \frac{ε _{0} A}{d} .$ Ab slab daalo aur charge $Q$ fixed rakho (battery unplugged). Step 7 se $E = E_{0} / κ$ use karke: $V = E d = \frac{E _{0}}{κ} d = \frac{E _{0} d}{κ} = \frac{V _{0}}{κ} .$ Same charge, lekin iske across voltage $κ$ times chhota hai. Capacitance charge-per-volt hai: $C = \frac{Q}{V} = \frac{Q}{V _{0} / κ} = κ \frac{Q}{V _{0}} = κ C_{0} .$

YEH KYUN BADHTA HAI. $C$ measure karta hai "kitna charge tum per volt of strain store kar sakte ho." Dielectric strain relieve karta hai ( $V$ kam karta hai) free mein, isliye ab har volt $κ$ times zyada charge hold karta hai.

PICTURE. Do bar-meters: $V$ bar $1/ κ$ height tak girta hai; $C$ bar $κ$ height tak badhta hai. Same $Q$ label dono par.

Step 9 — Woh edge cases jo tumhe skip nahi karni chahiye

Har woh scenario jo reader ko face ho sakta hai, dikhaaya gaya taaki koi surprise na ho.

Vacuum / air ( $χ_{e} = 0$ ): koi molecule stretch nahi hota, $σ_{b} = 0$ , $E = E_{0}$ , $κ = 1$ , $C = C_{0}$ . Poori derivation quietly bare-plate case mein reduce ho jaati hai — acha hai, karna hi chahiye.
Battery kept connected ( $V$ fixed): ab $V$ , hence $E = V / d$ , battery se force constant hai. Field drop nahi karta. Balki $Q = C V = κ C_{0} V$ $κ$ se badhta hai; battery extra free charge pump karti hai naye-hungry capacitor ko feed karne ke liye (energy story in Energy Stored in a Capacitor).
Kya $κ < 1$ ho sakta hai? Nahi. Ordinary matter ke liye $χ_{e} \geq 0$ hota hai (dipoles field ke saath align hote hain, kabhi against nahi), isliye $κ = 1 + χ_{e} \geq 1$ . Field sirf kamzor ho sakta hai, kabhi strong nahi.
Slab gap se patla (partial fill, $t < d$ ): sirf filled fraction ko $E_{0} / κ$ milta hai; khali hissa $E_{0}$ rakhta hai. Capacitor series mein do capacitors ki tarah behave karta hai — is page se aage ki baat, lekin note karo ki yeh sirf $κ C_{0}$ nahi hai. Isliye humne poori derivation mein full-fill case $t = d$ assume kiya.

Ek-picture summary

Poori causal chain, left to right: free charge $E_{0}$ banata hai → $E_{0}$ molecules ko $p$ mein stretch karta hai → sum karne par woh $P$ hai → $P$ faces par bound $σ_{b}$ chhodta hai → $σ_{b}$ $E_{0}$ ko oppose karta hai, $E$ ko $κ$ se shrink karta hai → chhota $E$ matlab chhota $V$ matlab bada $C$ .

Recall Poore walkthrough ki Feynman-style retelling

Do charged plates ek gap $d$ apart se shuru karo: arrows seedhe across shoot karte hain (Step 1). Ab ek neutral block slide karo jo gap ko poora fill kar le (Step 2). Arrows har molecule ko grab karte hain aur stretch karte hain — plus end is taraf, minus end us taraf — laakhon chhote see-saws banate hain sab same direction mein tip kiye hue (Step 3). Un chhote tips ko blur karo aur ek smooth "amount-of-tipping-per-cubic-metre" field milta hai jise $P$ kehte hain (Step 4). Block ke andar, har plus apne neighbour ke minus ko touch karta hai aur woh cancel ho jaate hain; sirf do bahari faces par naked charge reh jaata hai (Step 5). Woh naked face-charges har plate ke paas galat sign ke saath baithte hain, isliye woh wapas push karte hain aur original arrows ko partly cancel karte hain (Step 6). Jitna zyada push karo, utna zyada tip hote hain aur push back karte hain, aur jab tum us tug-of-war ko solve karte ho to arrows exactly $κ$ times chhote nikalte hain (Step 7). Chhote arrows matlab same gap mein kam voltage strain, isliye ab har volt $κ$ times zyada charge hold karta hai — capacitance $C = Q / V$ $κ$ times badi hai (Step 8). Eagerness band karo ( $κ = 1$ ) aur sab kuch bare plates par snap back; battery plugged in rakho aur field wahi rehta hai jabki extra charge flood in ho jaata hai (Step 9). Glass ka ek block, ek number $κ$ , aur sab kuch follow karta hai.

Recall

$E = E_{0} - E_{b}$ mein minus sign kahan se aata hai? ::: Bound charge sheet har plate ke paas opposite sign ke saath baithti hai, isliye uska field $E_{0}$ ke against point karta hai. $σ_{b} = P$ (magnitude mein) exactly kyun hota hai? ::: "Dipole density × volume" ko "exposed face charge × slab thickness" ke equal karne par — area aur thickness cancel ho jaate hain. Bulk mein bound charge kyun nahi hota? ::: $ρ_{b} = - \nabla \cdot P = 0$ kyunki $P$ slab ke andar uniform hai. Insulator add karne ke baad bhi $C$ kyun badhti hai? ::: Dielectric same charge ke liye voltage strain kam karta hai, isliye har volt $κ$ times zyada charge store karta hai. Battery connected ho — $E$ ki jagah kya change hota hai? ::: $Q$ (aur stored energy) $κ$ se badhti hai; $E$ aur $V$ battery se pinned hain. Parallel plate ke liye $C_{0}$ kya hai? ::: $C_{0} = ε_{0} A / d$ .

Connections

Parent topic
Capacitance and Capacitors
Electric Field of Parallel Plates
Gauss's Law and the D-field
Electric Dipoles and Torque
Energy Stored in a Capacitor
Electric Susceptibility