Dielectrics — polarization, dielectric constant, effect on capacitance: Question bank

1.8.14 · D5 · Physics › Electromagnetism › Dielectrics — polarization, dielectric constant, effect on c

Quick symbol reminder taaki kuch bhi unexplained na rahe:

$σ_{f}$ = free surface charge (wo jo battery metal plates par daalta hai).
$σ_{b}$ = bound surface charge (dielectric ke faces par exposed, uncancelled charge).
$E_{0} = σ_{f} / ε_{0}$ = sirf free charge se field (vacuum value).
$E$ = dielectric ke andar actual net field; $κ = E_{0} / E \geq 1$ dielectric constant hai.
" $Q$ fixed" ⇔ battery disconnected; " $V$ fixed" ⇔ battery abhi bhi connected.

True or false — justify karo

Dielectric face par bound charge hamesha paas wali plate ke free charge se opposite sign ka hota hai.

True — plate ka field har molecule ke opposite-sign end ko apni taraf kheenchta hai, isliye

+

plate ki taraf wala face

-

charge expose karta hai. Ye opposition hi exactly woh reason hai jis se net field kam hota hai.

Ek perfect vacuum ka

κ = 1

hota hai.

True — koi matter nahi toh koi dipole nahi, isliye

χ_{e} = 0

aur

κ = 1 + χ_{e} = 1

. Ye floor value hai; koi bhi ordinary cheez isse neeche nahi jaati.

Dielectric insert karne par

κ < 1

ho sakta hai agar material polarize hone ka resist kare.

False — polarization resist karna sirf matlab hai chhota

χ_{e}

, aur

χ_{e} \geq 0

, isliye

κ = 1 + χ_{e} \geq 1

. "Resistance"

κ

1

ki taraf dhakelta hai, kabhi neeche nahi.

Battery disconnected hone par, dielectric insert karne se stored energy kam hoti hai.

True —

Q

fixed hai aur

U = Q^{2} /2 C

; kyunki

C

κ

se badhta hai,

U

κ

se kam hota hai. Missing energy woh kaam ban gayi jo field ne slab ko andar kheenchte waqt kiya.

Battery connected hone par, dielectric insert karne se stored energy kam hoti hai.

False —

V

fixed hai aur

U = \frac{1}{2} C V^{2}

;

C

κ

se badhta hai isliye

U

bhi

κ

se badhta hai. Battery extra energy supply karti hai (aur kuch jo heat/work mein jaata hai).

Jab battery connected rahe toh dielectric ke andar field unchanged rehta hai.

True — battery

V

pin karti hai, aur

E = V / d

mein

d

fixed hai, isliye

E

change nahi ho sakta; balki

Q

κ Q_{0}

tak badh jaata hai.

Bound charge Gauss's law for

E

mein contribute karta hai.

True —

E

ke liye Gauss's law sabhi charge count karta hai, free aur bound dono.

D = ε_{0} κ E

ki convenience ye hai ki

D

sirf free charge ko respond karta hai, jisse tum

σ_{b}

skip kar sakte ho — dekho Gauss's Law and the D-field.

Polarization

P

ke units surface charge density ke same hote hain.

True — dipole moment per volume hai

(C\cdotm) / m^{3} = C / m^{2}

, isliye aligned dipoles literally dikhte hain jaise har face par charge ki sheet

σ_{b} = P

ho.

Water jaisi ek polar molecule stretching se polarize hoti hai, bilkul non-polar ki tarah.

False — water mein pehle se permanent dipole hai; field mainly use partially align karne ke liye rotate karta hai (torque, dekho Electric Dipoles and Torque), jabki non-polar molecules charge-cloud stretching se induced dipole gain karte hain.

Error dhundo

"Maine dielectric add kiya, toh matter add kiya, toh andar ka field zyada strong hona chahiye."

Added matter bound charge contribute karta hai jo free charge ka opposition karta hai, isliye net field

κ

se kam hota hai. Yahan zyada matter matlab zyada opposition, zyada field nahi.

"Battery connected hone par maine

E = E_{0} / κ

use karke naya field nikala."

Galat regime —

E_{0} / κ

rule assume karta hai ki

Q

(matlab

σ_{f}

) fixed hai. Battery ke saath,

V

aur isliye

E = V / d

pin hote hain;

Q

(aur

σ_{f}

)

κ

se badhta hai.

"Bound charge surface par real charge hai, isliye main ise wire karke drain kar sakta hoon."

Bound charge flow nahi kar sakta — ye locked-in molecular dipoles ka exposed end hai, free carriers nahi. Tum ise conduct karke bahar nahi nikaal sakte; field hatao aur ye khatam ho jaata hai.

σ_{b} = σ_{f}

kyunki bound charge free charge ko cancel karta hai."

Ye sirf partially cancel karta hai:

σ_{b} = σ_{f} (1 - 1/ κ)

, jo hamesha

σ_{f}

se kam hai. Agar dono equal hote toh net field zero hota, jo sirf

κ \to \infty

(conductor) limit mein hota hai.

"Jab maine slab insert kiya toh energy kam ho gayi, isliye energy destroy ho gayi."

Energy conserved hai — drop

Δ U

us mechanical work ke barabar hai jo field ne slab ko kheenchte waqt kiya. Stored field energy kinetic/mechanical energy mein convert hui, khatam nahi hui.

κ = 1 + χ_{e}

, isliye bada field bada

κ

deta hai."

Ek linear dielectric ke liye

χ_{e}

(aur isliye

κ

) ek material property hai jo field strength se independent hai.

P

E

ke saath scale karta hai, lekin unka ratio

χ_{e}

fixed rehta hai.

"Main capacitor ka field total charge

σ_{f} + σ_{b}

aur

ε_{0}

use karke compute karunga."

Net field hai

E = (σ_{f} - σ_{b}) / ε_{0}

— dono faces par bound aur free charge ke opposite signs hain, isliye ye subtract hote hain, add nahi. Unhe add karna double-counting hai aur physics ulti ho jaati hai.

Why questions

Dielectric insert karne par capacitor same voltage par zyada charge kyun store kar sakta hai?

Bound charge field ka part cancel karta hai, isliye given

V = E d

ke liye tum ab

κ

times zyada free charge daalo tab field (aur voltage) same value tak pahunche — woh extra packing hi higher capacitance hai, per Capacitance and Capacitors.

Relation

σ_{b} = P \cdot \overset{n}{^}

likhte hain instead of sirf

σ_{b} = P

— kyun?

Exposed charge depend karta hai ki polarization face ke through kitna pokes karta hai; sirf outward normal

\overset{n}{^}

ke direction mein component usi face par charge chhodta hai. Ek slab mein jiske faces

P

ke perpendicular hain, dot product

\pm P

ho jaata hai.

Net field exactly

κ

factor se kyun girta hai aur koi aur number se nahi?

Kyunki

P = ε_{0} χ_{e} E

opposition ko field ke proportional banata hai;

E = E_{0} - χ_{e} E

solve karne par

E (1 + χ_{e}) = E_{0}

milta hai, aur

1 + χ_{e}

ko hi

κ

define karte hain. "Field polarization cause karta hai jo field ko reduce karta hai" ki self-consistency mein single factor bak jaata hai.

Aligned dipoles ke interior ko electrically neutral kyun treat kar sakte hain?

Slab ke andar, har molecule ka

+

end apne neighbour ke

-

end ke saath baithta hai, isliye interior charges pairwise cancel ho jaate hain. Sirf do outer faces uncancelled charge ke saath reh jaate hain.

Parallel-plate formula mein

κ

kyun aata hai lekin

A

aur

d

same rehte hain?

Geometry (

A

d

) gap fill karne se untouched rehti hai; sirf medium badla, aur medium sirf

κ

ke through enter karta hai, deta hai

C = κ ε_{0} A / d

(vacuum case se compare karo Electric Field of Parallel Plates mein).

Water jaisa strong dielectric (

κ \approx 80

) andar almost koi field kyun nahi chodta?

Bada

χ_{e}

matlab dipoles strongly align hote hain aur almost utna hi bound charge produce karte hain jitna free charge hai, isliye

E = E_{0} / κ

tiny hai — free-charge field ka zyaadatar hissa neutralise ho jaata hai, per Electric Susceptibility.

Edge cases

Jab material perfect conductor banne ki taraf jaaye (

χ_{e} \to \infty

) toh

κ

aur field ka kya hoga?

κ \to \infty

, isliye

E = E_{0} / κ \to 0

— bound charge free charge ko poori tarah cancel kar deta hai, bilkul conductor ke zero interior field ki tarah. Conductor "infinitely polarizable" dielectric ki limiting case hai.

Agar external field zero ho toh water jaisi polar dielectric ki polarization kya hogi?

Average par zero — permanent dipoles random directions mein point karte hain aur cancel ho jaate hain, isliye

P = 0

hai chahe har molecule dipole carry kare. Net alignment ke liye field ka torque chahiye.

Ek dielectric gap ko sirf half bharta hai (thickness

t < d

ka slab). Kya gap mein poori jagah

E = E_{0} / κ

hai?

Nahi — reduced field

E_{0} / κ

sirf slab ke andar hota hai; baaki vacuum mein field abhi bhi

E_{0}

hai. Dono regions series mein hain aur alag handle karne padte hain.

Trivial case

κ = 1

(vacuum ya air) mein

σ_{b}

kya hai?

σ_{b} = σ_{f} (1 - 1/ κ) = σ_{f} (1 - 1) = 0

— koi polarizable matter nahi matlab koi bound charge nahi aur koi field reduction nahi, consistent hai

E = E_{0}

ke saath.

Battery disconnected: tum dielectric ko wapas bahar kheenchte ho. Stored energy badhti hai ya kam hoti hai?

Badhti hai — slab hatane se

C

kam hota hai, aur

Q

fixed hone par

U = Q^{2} /2 C

badhta hai. Tum ne field ki inward pull ke against kaam kiya usse nikaalte waqt, aur woh kaam stored energy ban gaya.

Do identical capacitors, ek

κ = 2

se bhara, dono same

Q

par charged. Andar ka field kiska zyada hai?

Empty wala —

Q

(matlab

σ_{f}

) fixed hone par, filled capacitor ka field uske bound charge se half ho jaata hai

E_{0} /2

. Same charge, weaker field, kyunki dielectric oppose karta hai.

Agar ek dielectric ka

χ_{e}

negative ho (ek hypothetical "diamagnetic-like" dielectric), toh

E

ka kya hoga?

κ = 1 + χ_{e} < 1

dega

E > E_{0}

— field increase ho jaayega. Ordinary dielectrics aisa kabhi nahi karte kyunki dipoles induce karna hamesha ek tarah se energy cost karta hai jo field ko reinforce nahi, oppose karta hai.

Recall Jaane se pehle ek-line self-test

Wo single question jo zyaadatar traps suljha deta hai ::: "Kya fixed hai — charge $Q$ (battery off) ya voltage $V$ (battery on)?" Baaki sab ussi se follow karta hai.

Connections

Hinglish parent
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