Dielectrics — polarization, dielectric constant, effect on capacitance: Worked examples

1.8.14 · D3 · Physics › Electromagnetism › Dielectrics — polarization, dielectric constant, effect on c

Yeh page ek drill hai jisme dielectric topic ke har possible case ko cover kiya gaya hai. Kuch bhi solve karne se pehle, hum ek matrix of case classes banate hain — yani jo alag-alag tarah ke questions exist karte hain — aur phir examples solve karte hain jab tak har cell cover na ho jaye. Agar tum yeh sab kar sako, toh koi bhi dielectric problem tumhe surprise nahi kar sakti.

Prerequisites jinpe hum depend karte hain: Capacitance and Capacitors, Electric Field of Parallel Plates, Energy Stored in a Capacitor, Electric Susceptibility, Gauss's Law and the D-field, aur parent Dielectrics topic note.

Definition Woh symbols jinpe sab kuch bana hai (

$C_{0}$ , $σ_{f}$ , $σ_{b}$ ) Neeche teen symbols baar baar aate hain — use karne se pehle har ek ko ek picture se jodo.

== $C_{0}$ = vacuum (empty-gap) capacitance== — woh capacitance jo inhi plates ( $A$ , $d$ ) ki hoti agar sirf vacuum (ya air) beech mein hoti. Yeh "pehle" ki value hai; dielectric daalne se yeh $C = κ C_{0}$ ban jaata hai. Toh " $C_{0} = 4 μ F$ empty" ka matlab bas yeh hai ki gadget ne koi slab daalne se pehle $4 μ F$ measure kiya.
== $σ_{f}$ = free-charge surface density== — charge per unit area jo battery ne physically metal plates par deliver kiya (woh "free" electrons jo wires se guzar sakte hain). $+$ plate par $σ_{f} > 0$ ; $-$ plate par $σ_{f} < 0$ . Dono plates par magnitude same, sign opposite.
== $σ_{b}$ = bound-charge surface density== — woh charge jo dielectric ki dono faces par appear hota hai kyunki dipoles line up ho jaate hain (yeh "bound" hai: yeh apna molecule nahi chod sakta). Yeh slab ki inner surfaces par rehta hai, bilkul har plate ke sath chipka hua.

Recall Teen master formulas + regime question (parent mein build ki gayi — yahan repeat kar di taaki scroll na karni pade)

Teen formulas hi saara kaam karti hain (symbols upar define hain):

Field girta hai: $E = \frac{E _{0}}{κ}$ , jahan $E_{0} = \frac{σ _{f}}{ε _{0}}$ akele free charge ka vacuum field hai.
Capacitance badhta hai: $C = κ C_{0} = \frac{κ ε _{0} A}{d}$ , jahan $C_{0}$ empty-gap capacitance hai.
Bound surface charge (magnitude): $σ_{b} = σ_{f} (1 - \frac{1}{κ})$ , woh free charge ke opposite sign mein baithta hai jisko yeh face karta hai.

Aur ek sawaal jo tumhe unhe apply karne se pehle poochna hai:

Kaunsi quantity fixed hai — $Q$ (battery off) ya $V$ (battery on)?

Scenario matrix

Har dielectric exam question in case classes mein se ek (ya combination) hota hai. Sabse right column mein us worked example ka naam hai jo use cover karta hai.

#	Case class	Kya test ho raha hai	Covered by
A	Plain $κ$ boost	$C = κ C_{0}$ , kuch subtle nahi	Ex 1
B	Battery disconnected ( $Q$ fixed)	$E, V, U$ sab $κ$ se girte hain	Ex 2
C	Battery connected ( $V$ fixed)	$E$ fixed, $Q, U$ badhte hain; charge kahan se aata hai	Ex 3
D	Bound charge & field numbers	$σ_{b}$ , $E_{0}$ , $E$ from $σ_{f}, κ$	Ex 4
E	Degenerate limits	$κ \to 1$ (vacuum) aur $κ \to \infty$ (conductor)	Ex 5
F	Partial fill (slab thinner than gap)	do regions ki series, geometry	Ex 6 (figure)
G	Do dielectrics side-by-side vs stacked	parallel vs series capacitors	Ex 7 (figure)
H	Real-world word problem	safety/size ke liye material choose karna	Ex 8
I	Exam-style twist (slab ko andar kheenchne wala force)	energy method, force ka sign	Ex 9

Cell E do limiting inputs cover karta hai; A–D roz ke signs aur regimes cover karte hain; F, G geometry degeneracies cover karte hain (ek slab jo gap ka sirf part fill kare, ya gap ko split kare). Chalo shuru karte hain.

Worked examples

Worked example Ex 2 — Case B · Battery DISCONNECTED (

$Q$ fixed) Ek $C_{0} = 2 μ F$ (empty-gap) capacitor ko $V_{0} = 100 V$ tak charge karo, phir battery unplug kar do. Ab $κ = 4$ insert karo. Naye $Q$ , $V$ , $E$ -ratio, aur energy $U$ nikalo.

Forecast: Battery jaane ke baad, kaunsi quantity frozen hai? Yeh jaannte ho toh baaki sab predict karo.

Battery gayi $\Rightarrow$ $Q$ badal nahi sakta (charge ko flow karne ki jagah nahi hai). $Q = C_{0} V_{0} = 2 \times 1 0^{- 6} \times 100 = 2 \times 1 0^{- 4} C = 200 μ C$ , aur yeh $200 μ C$ hi rehta hai. Yeh step kyun? Isolated plates charge conserve karti hain — yeh poore regime ka anchor hai.
Naya capacitance $C = κ C_{0} = 8 μ F$ . Yeh step kyun? Geometry unchanged, medium changed — Case A ka rule ab bhi apply hota hai.
Naya voltage $V = \frac{Q}{C} = \frac{200 μ C}{8 μ F} = 25 V$ . Yeh step kyun? $Q$ fixed, $C$ $4$ se badha $\Rightarrow$ $V$ $4$ se gira: $100/4 = 25$ ✔.
Field ratio $E / E_{0} = V / V_{0} = 25/100 = 1/4 = 1/ κ$ . Yeh step kyun? $E = V / d$ with $d$ fixed, toh $E$ , $V$ ke saath exactly track karta hai.
Energy $U = \frac{Q ^{2}}{2 C} = \frac{( 2 \times 1 0 ^{- 4} ) ^{2}}{2 \times 8 \times 1 0 ^{- 6}} = 2.5 \times 1 0^{- 3} J = 2.5 mJ$ (pehle $\frac{1}{2} C_{0} V_{0}^{2} = 10 mJ$ tha). Yeh step kyun? Energy ka $Q$ fixed form; $C$ denominator mein hai, toh energy giregi.

Verify: Energy $10 \to 2.5 mJ$ gayi. $7.5 mJ$ kahan gayi? Us work mein jo field ne kiya jab usne slab ko andar kheencha (Ex 9 mein yeh concrete hoga). Consistency: $U = \frac{1}{2} Q V = \frac{1}{2} (200 μ C) (25 V) = 2.5 mJ$ ✔.

Worked example Ex 3 — Case C · Battery CONNECTED (

$V$ fixed) Wahi $C_{0} = 2 μ F$ empty-gap capacitor, lekin $V = 100 V$ battery plugged in rehti hai. $κ = 3$ insert karo. $C$ , $Q$ , battery se aaya extra charge, field ratio, aur $U$ nikalo.

Forecast: Ab kaunsi quantity frozen hai? $Q$ aur $E$ ki direction guess karo.

Battery $V = 100 V$ fixed rakhti hai. Kyunki $E = V / d$ aur $d$ fixed hai, insertion se $E$ unchanged rehta hai. Yeh step kyun? Yeh trap-avoider hai: battery ke saath tum $E = E_{0} / κ$ use nahi karte.
$C = κ C_{0} = 3 \times 2 = 6 μ F$ .
$Q = C V = 6 \times 1 0^{- 6} \times 100 = 6 \times 1 0^{- 4} C = 600 μ C$ (pehle $C_{0} V = 200 μ C$ tha). Yeh step kyun? $V$ fixed, $C$ $3$ se badha $\Rightarrow$ $Q$ $3$ se badha.
Battery ka pump kiya extra charge $= 600 - 200 = 400 μ C$ . Yeh step kyun? Conservation — increase kisi na kisi jagah se aana chahiye; battery woh source hai.
$U = \frac{1}{2} C V^{2} = \frac{1}{2} (6 \times 1 0^{- 6}) (100)^{2} = 3 \times 1 0^{- 2} J = 30 mJ$ (pehle $10 mJ$ tha). Yeh step kyun? $V$ fixed form; $C$ numerator mein hai, toh energy badhti hai.

Verify: $E$ unchanged lekin $C, Q, U$ teeno $κ = 3$ se badhe ( $6/2 = 3$ , $600/200 = 3$ , $30/10 = 3$ ) ✔. Ex 2 se compare karo: same $κ$ -ish insertion, opposite energy trend — kyunki alag cheez fixed thi.

Worked example Ex 4 — Case D · Bound charge & fields scratch se

Plates par free-charge density $σ_{f} = 8.85 \times 1 0^{- 6} C/m^{2}$ (magnitude $+$ plate par; $-$ plate wahi magnitude, opposite sign carry karta hai), dielectric $κ = 4$ . $E_{0}$ , $E$ , aur bound surface charge $σ_{b}$ nikalo.

Forecast: $σ_{b}$ , $σ_{f}$ se bada hoga ya chhota? (Yeh bound charge hai jo sirf partially cancel karta hai — toh...?)

Vacuum field $E_{0} = \frac{σ _{f}}{ε _{0}} = \frac{8.85 \times 1 0 ^{- 6}}{8.85 \times 1 0 ^{- 12}} = 1.0 \times 1 0^{6} V/m$ . Yeh step kyun? Sirf free charge ka field, dielectric ke respond karne se pehle (Electric Field of Parallel Plates).
Net field $E = \frac{E _{0}}{κ} = \frac{1 0 ^{6}}{4} = 2.5 \times 1 0^{5} V/m$ . Yeh step kyun? $σ_{f}$ ke baare mein koi battery mention nahi — $σ_{f}$ hi given free charge hai, toh $E$ , $κ$ se shrink hota hai.
Bound charge $σ_{b} = σ_{f} (1 - \frac{1}{κ}) = 8.85 \times 1 0^{- 6} \times (1 - 0.25) = 6.64 \times 1 0^{- 6} C/m^{2}$ . Yeh step kyun? $E = \frac{σ _{f} - σ _{b}}{ε _{0}} = \frac{σ _{f}}{κ ε _{0}}$ se $σ_{b}$ solve karo. Hum $σ_{b}$ subtract karte hain kyunki $+$ plate ko hugging karne wali bound face opposite ( $-$ ) sign carry karti hai, use partially cancel karti hai (upar sign-convention callout dekho).

Verify: $σ_{b} < σ_{f}$ (partial cancellation, kabhi total nahi) ✔. Net check karo: $σ_{f} - σ_{b} = 8.85 - 6.64 = 2.21 \times 1 0^{- 6}$ , aur $E = \frac{2.21 \times 1 0 ^{- 6}}{8.85 \times 1 0 ^{- 12}} = 2.5 \times 1 0^{5} V/m$ — step 2 se match karta hai ✔.

Worked example Ex 5 — Case E · Do DEGENERATE limits

Ek $C_{0} = 10 μ F$ vacuum (empty-gap) capacitor. (a) "Vacuum" slide in karo ( $κ = 1$ ). (b) Ek perfect conductor slide in karo jo gap fill kar de — ise $κ \to \infty$ treat karo. $C$ , $E$ , aur reason discuss karo.

Forecast: (b) ke liye, ek conductor ke across voltage ka kya hoga? Woh $C$ ko kya force karta hai?

(a) $κ = 1 \Rightarrow C = 1 \cdot C_{0} = 10 μ F$ . Kuch nahi badla. Yeh step kyun? $κ = 1$ ka matlab zero susceptibility — "dielectric" bilkul polarize nahi hota, yani woh hai hi vacuum. Formula ka sanity floor.
(b) $κ \to \infty \Rightarrow C = κ C_{0} \to \infty$ . Yeh step kyun? Ek conductor apne andar internal field sustain nahi kar sakta, toh $E \to 0$ andar; $V = E d \to 0$ , aur $C = Q / V \to \infty$ finite $Q$ ke liye.
Field: (a) unchanged $E = E_{0}$ ; (b) $E = E_{0} / κ \to 0$ . Yeh step kyun? $κ = E_{0} / E$ ke consistent: infinite $κ$ matlab medium poora field cancel kar deta hai.

Verify: Dono limits $C = κ C_{0}$ ke saath monotone-consistent hain: jab $κ$ , $1 \to \infty$ badhta hai, $C$ , $C_{0} \to \infty$ badhta hai aur $E$ , $E_{0} \to 0$ girta hai. Formula kisi endpoint par break nahi karta ✔. (Thickness $t < d$ wala ek real conductor slab ek finite boost deta hai — woh partial-fill logic Ex 6 mein hai.)

Worked example Ex 6 — Case F · Partial fill (slab gap se thinner)

Plate separation $d = 3 mm$ , area $A$ jo $C_{0} = \frac{ε _{0} A}{d} = 6 μ F$ empty-gap capacitance deta hai. Ek dielectric slab $κ = 3$ , thickness $t = 1 mm$ , ek plate ke against baithy hai; bacha hua $2 mm$ air hai. Naya capacitance nikalo.

Figure — kya dikhata hai: do horizontal kaali bars plates hain ( $+ Q$ upar, $- Q$ neeche). Neeche wali plate se chipka red block $1 mm$ ka dielectric slab hai; uske upar white space (jis par "air" likha hai) bacha hua $2 mm$ hai. Right side ke double-headed arrows dono thicknesses $t = 1 mm$ aur $d - t = 2 mm$ mark karte hain; left side ka arrow poora gap $d = 3 mm$ mark karta hai. Picture yeh bata rahi hai ki gap do stacked regions mein split hai.

Forecast: Kya ek partial slab same $κ$ ke full slab se zyada effective hai ya kam? $6$ aur $18 μ F$ ke beech ek number predict karo.

Gap ab do regions series mein hai: ek dielectric layer (thickness $t = 1 mm$ ) aur ek air layer (thickness $d - t = 2 mm$ ). Jaise figure dikhata hai, red slab air region ke neeche baitha hai, ek ke upar ek stacked. Yeh step kyun? Wahi charge $Q$ dono layers se guzarta hai; voltages add hote hain. Yahi series ki definition hai (Capacitance and Capacitors).
Region capacitances (same area $A$ ): dielectric $C_{1} = \frac{κ ε _{0} A}{t}$ , air $C_{2} = \frac{ε _{0} A}{d - t}$ . Yeh step kyun? Har layer apni thickness ke saath apna parallel-plate capacitor hai.
Series combine: $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} = \frac{t}{κ ε _{0} A} + \frac{d - t}{ε _{0} A} = \frac{1}{ε _{0} A} (\frac{t}{κ} + d - t)$ . Yeh step kyun? Series capacitors reciprocals add karte hain (shared $Q$ ke liye voltages add hote hain).
Toh $C = \frac{ε _{0} A}{\frac{t}{κ} + d - t}$ . Plug in karo $ε_{0} A = C_{0} d = 6 μ F \times 3 mm = 18 μ F\cdotmm$ ke saath: $C = \frac{18 μ F\cdotmm}{\frac{1}{3} + ( 3 - 1 ) mm} = \frac{18}{\frac{1}{3} + 2} = \frac{18}{7/3} = \frac{54}{7} \approx 7.71 μ F .$ Yeh step kyun? $\frac{t}{κ} = \frac{1}{3} mm$ slab ki "electrical thickness" hai — field ko yeh $κ$ times thinner lagta hai.

Verify: Bounds check — full slab $κ C_{0} = 18 μ F$ deta, koi slab nahi toh $6 μ F$ ; hamara $7.71$ beech mein hai aur low end ke paas kyunki gap ka sirf $1/3$ fill hua hai ✔.

Worked example Ex 7 — Case G · Do dielectrics: side-by-side vs stacked

Wahi plates, $C_{0} = 6 μ F$ empty-gap capacitance. Do materials $κ_{1} = 2$ , $κ_{2} = 4$ , har ek aadhi jagah fill karta hai. Do tarike se karo: (a) side-by-side (har ek aadha area cover kare), (b) stacked (har ek aadhi thickness fill kare).

Figure — kya dikhata hai: do mini-capacitors side by side draw kiye hain. Left diagram ("side-by-side = PARALLEL"): gap ek left white column ( $κ_{1} = 2$ ) aur ek right red column ( $κ_{2} = 4$ ) mein split hai, har ek poori plate separation span karta hai lekin sirf aadha area. Right diagram ("stacked = SERIES"): gap ek lower white layer ( $κ_{1} = 2$ ) aur ek upper red layer ( $κ_{2} = 4$ ) mein split hai, har ek poora area span karta hai lekin sirf aadhi thickness. Left = dono halves mein same voltage (parallel); right = dono layers se same charge (series).

Forecast: Kaunsa arrangement zyada capacitance deta hai — area share karna, ya thickness share karna? Compute karne se pehle guess karo.

(a) Side-by-side = parallel. Har half ka area $A /2$ , full gap $d$ . $C_{a} = \frac{κ _{1} ε _{0} ( A /2 )}{d} + \frac{κ _{2} ε _{0} ( A /2 )}{d} = \frac{( κ _{1} + κ _{2} )}{2} C_{0}$ . Yeh step kyun? Dono halves mein same voltage, charges add hote hain — parallel. Yeh left diagram hai.
$C_{a} = \frac{2 + 4}{2} \times 6 = 3 \times 6 = 18 μ F$ . (Yeh $C_{0}$ ka average $κ = 3$ wala version hai.) Yeh step kyun? Equal areas ke saath parallel literally dono $κ$ 's average karta hai.
(b) Stacked = series. Har layer ki thickness $d /2$ , full area $A$ . $C_{i} = \frac{κ _{i} ε _{0} A}{d /2} = 2 κ_{i} C_{0}$ . Series: $\frac{1}{C _{b}} = \frac{1}{2 κ _{1} C _{0}} + \frac{1}{2 κ _{2} C _{0}}$ . Yeh step kyun? Dono stacked layers se same charge, voltages add hote hain — series. Yeh right diagram hai.
$\frac{1}{C _{b}} = \frac{1}{2 C _{0}} (\frac{1}{κ _{1}} + \frac{1}{κ _{2}})$ . Do-capacitor series ka clean closed form "product over sum" hai, toh $C_{b} = 2 C_{0} \cdot \frac{κ _{1} κ _{2}}{κ _{1} + κ _{2}} = 2 (6) \cdot \frac{2 \cdot 4}{2 + 4} = 12 \cdot \frac{8}{6} = 16 μ F .$ Yeh step kyun? Do capacitors ki series "product over sum" hai; effective $κ$ woh harmonic-type combination hai $\frac{2 κ _{1} κ _{2}}{κ _{1} + κ _{2}} = \frac{16}{6} \approx 2.67$ , aur $C_{b} = κ_{eff} C_{0} = 2.67 \times 6 = 16 μ F$ .

Verify: Side-by-side ( $C_{a} = 18 μ F$ ), stacked ( $C_{b} = 16 μ F$ ) se jeet jaata hai. General truth: parallel (arithmetic mean of $κ$ , yahan $3$ ) $\geq$ series (harmonic-type mean, yahan $2.67$ ) ✔. Dono $C_{0} = 6 μ F$ se zyada hain aur all- $κ_{2}$ value $4 \times 6 = 24 μ F$ se neeche hain ✔. Physical read: kamzor dielectric ko series mein daalne se poori stack us ki low value ki taraf khinch jaati hai, isliye stacked, side-by-side se haar jaata hai.

Worked example Ex 8 — Case H · Real-world word problem

Tumhe $C = 100 pF$ ka capacitor banana hai jo $V = 2000 V$ par spark kiye bina hold kare. Plate area fixed hai $A = 1.0 \times 1 0^{- 3} m^{2}$ . Air $3 \times 1 0^{6} V/m$ par break down karti hai; ek plastic $κ = 3$ ke saath $20 \times 1 0^{6} V/m$ par break down karti hai. Kya air yeh kar sakti hai? Plastic ke saath plate gap $d$ nikalo.

Forecast: Plastic ki zyada $κ$ bhi hai aur zyada breakdown strength bhi — yahan asal mein kaunsi cheez tumhe bachati hai?

Minimum gap taaki field air ki limit se neeche rahe: $d_{m i n} = \frac{V}{E _{max,air}} = \frac{2000}{3 \times 1 0 ^{6}} = 6.7 \times 1 0^{- 4} m$ . Yeh step kyun? $E = V / d$ ; $E$ ko breakdown se neeche rakhne ke liye tumhe kaafi bada $d$ chahiye.
Is gap par air ka capacitance: $C_{air} = \frac{ε _{0} A}{d _{m i n}} = \frac{8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 3}}{6.7 \times 1 0 ^{- 4}} = 1.33 \times 1 0^{- 11} F = 13.3 pF$ . Yeh step kyun? Safe gap par air sirf $13.3 pF$ tak pahunch sakti hai — $100 pF$ se bahut door. Toh plain air spec fail karti hai.
Plastic ke saath, breakdown ek chhota gap allow karta hai: $d = \frac{V}{E _{max,plastic}} = \frac{2000}{20 \times 1 0 ^{6}} = 1.0 \times 1 0^{- 4} m$ . Yeh step kyun? Zyada breakdown field field ko safely zyada intense hone deta hai, toh plates paas aa sakti hain — chhota $d$ matlab bada $C$ .
Resulting capacitance: $C = \frac{κ ε _{0} A}{d} = \frac{3 \times 8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 3}}{1.0 \times 1 0 ^{- 4}} = 2.66 \times 1 0^{- 10} F = 265 pF \geq 100 pF$ ✔. Yeh step kyun? Dono effects help karte hain — $κ$ , $C$ ko $3$ se multiply karta hai, aur chhota safe gap ise phir multiply karta hai — aur saath mein target cross ho jaata hai.

Verify: Air $13.3 pF$ par max out karti hai (fail); plastic $265 pF$ reach karti hai ( $100 pF$ target margin ke saath pass karta hai) ✔. Lesson: sirf $κ$ nahi tha — zyada breakdown field decisive property tha. Step 4 ki units: $\frac{(F/m) \cdot m ^{2}}{m} = F$ ✔.

Worked example Ex 9 — Case I · Exam twist: woh force jo slab ko andar kheenchti hai

Ex 2 ka disconnected capacitor ( $Q = 200 μ C$ , $C_{0} = 2 μ F$ , $κ = 4$ ). Humne paya tha $U$ , $10 mJ$ se $2.5 mJ$ tak gira. Dikhao ki yeh drop field ka kiya hua work ke barabar hai, aur isliye field slab ko andar kheenchta hai (bahar nahi).

Forecast: Systems lower energy ki taraf girte hain. Agar slab insert karna $U$ giraa deta hai, toh slab kis direction mein khicha jaata hai?

Full insertion par energy change: $Δ U = U_{final} - U_{initial} = 2.5 - 10 = - 7.5 mJ$ . Yeh step kyun? Dono ends Ex 2 mein already compute kiye hain; sign hi physics hai.
Work-energy: $Q$ fixed aur koi battery nahi ke saath, energy conservation deti hai $W_{by field on slab} = - Δ U = + 7.5 mJ > 0$ . Yeh step kyun? Jo stored energy gayi woh slab par mechanical work mein gayi.
Positive work by field along direction of motion ⇒ force inward act karta hai (slab ko gap mein kheenchta hai). Yeh step kyun? $W = \int F d x > 0$ inward motion par matlab $F$ inward point karta hai.

Verify: Direction check via energy landscape — zyada overlap matlab zyada $κ_{eff}$ , zyada $C$ , kam $U = Q^{2} /2 C$ ; system hamesha zyada overlap ki taraf slide karta hai ✔. Released energy ka magnitude $7.5 mJ = 10 - 2.5$ ✔, Ex 2 ke consistent.

Recall Matrix self-test (answers cover karo)

Battery ON, $κ$ insert karo: kya frozen hai? ::: Voltage $V$ (aur isliye $E = V / d$ ); $Q$ aur $U$ $κ$ se badhte hain. Battery OFF, $κ$ insert karo: kya frozen hai? ::: Charge $Q$ ; $V$ , $E$ , aur $U$ girte hain (last wala factor $κ$ se). Slab gap ka sirf part fill kare — series ya parallel? ::: Series (dielectric layer + air layer same $Q$ share karte hain). Do dielectrics area split karte hain — series ya parallel? ::: Parallel (same $V$ , charges add hote hain). Do dielectrics thickness split karte hain — series ya parallel? ::: Series (same $Q$ , voltages add hote hain). Limit $κ \to \infty$ kya model karta hai? ::: Gap fill karne wala conductor: $E \to 0$ , $C \to \infty$ . Disconnected slab insertion $U$ giraa deti hai — force direction? ::: Inward; field slab ko andar kheenchta hai.

Connections

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