Understand capacitance and the farad — Hinglish notes

1.1.11 · Hardware › Electricity & Charge Basics

Capacitance KYA hota hai?

Ye ratio constant kyun rehta hai? Ek fixed pair of conductors ke liye (fixed size, shape, spacing, material), charge double karne se voltage bhi double ho jaata hai. Ratio $Q / V$ sirf plates ki geometry aur beech ke insulator (dielectric) par depend karta hai — aapne kitna charge kiya iske upar nahi. Usi constant ratio ko hum "capacitance" kehte hain.

$C = Q / V$ ko first principles se KAISE derive karte hain?

Sabse simple capacitor lete hain: do flat parallel plates, area $A$ , separation $d$ , $+ Q$ aur $- Q$ charges liye hue.

Step 1 — Charge se Field. Har plate ka surface charge density $σ = Q / A$ hai. Gauss's law se, do opposite charged plates ke beech uniform field hai: $E = \frac{σ}{ε} = \frac{Q}{ε A}$ Ye step kyun? Gauss's law enclosed charge ko field se relate karta hai; infinite sheet ke liye field uniform hota hai aur charge per area se set hota hai.

Step 2 — Field se Voltage. Voltage = field × wo distance jitna charge push hota hai: $V = E d = \frac{Q d}{ε A}$ Ye step kyun? Potential difference $\int E d l$ hota hai; uniform $E$ ke saath ye bas $E \times d$ hai.

Step 3 — Ratio lo. Ab $Q / V$ banao: $C = \frac{Q}{V} = \frac{Q}{\frac{Q d}{ε A}} = \frac{ε A}{d}$ Ye step kyun? $Q$ cancel ho jaata hai — ye prove karta hai ki capacitance pure geometry + material hai, bilkul jaisa claim kiya tha.

Figure — Understand capacitance and the farad

Kitni ENERGY store hoti hai?

Charging free nahi hai — har extra charge ko already present voltage ke against push karna padta hai.

Derivation. Jab voltage $v = q / C$ ho tab charge $d q$ add karne par kaam $d W = v d q = \frac{q}{C} d q$ hoga. $0$ se $Q$ tak integrate karo: $W = \int_{0}^{Q} \frac{q}{C} d q = \frac{1}{C} \cdot \frac{Q ^{2}}{2} = \frac{Q ^{2}}{2 C}$ $Q = C V$ use karke: $E = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C}$

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Recall Feynman: 12-saal ke bachche ko samjhao

Ek paani ki tank imagine karo. Voltage hai kitna uncha paani push kiya; charge hai andar kitna paani hai. Badi tank kam paani level par bhi bahut saara paani hold karti hai — "kitna paani per unit height" hi uski capacitance hai. Farad ek ridiculously huge tank hai: 1-farad tank ko sirf 1 volt tak bharne ke liye 1 coulomb chahiye (electrons ka pahad!). Zyada store karne ke liye ya to tank wider banao (badi plates) ya uski walls aapas mein karib karo taaki paani asaani se pile up ho.

Flashcards

What is the definition of capacitance?

Store hue charge per unit voltage,

C = Q / V

What is the unit of capacitance and its base definition?

Farad (F);

1 F = 1 coulomb per volt

Why is capacitance independent of the charge placed on it?

Kyunki voltage charge ke proportional hota hai, isliye unka ratio

Q / V

sirf geometry aur dielectric par depend karta hai.

Formula for a parallel-plate capacitor?

C = ε_{0} ε_{r} A / d

Two physical ways to increase capacitance?

Plate area

A

badhao ya gap

d

ghataao (aur higher

ε_{r}

bhi).

Energy stored in a capacitor?

E = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = Q^{2} / (2 C)

Why the factor ½ in the energy formula?

Charge ko average voltage

V /2

ke against push kiya jaata hai jab ye 0 se

V

tak build hota hai.

Charge on a 2 µF cap at 3 V?

Q = C V = 6 μ C

Why is 1 farad considered a huge unit?

Ye 1 coulomb per volt store karta hai; typical caps pF–mF hoti hain.

Value of

ε_{0}

8.85 \times 1 0^{- 12} F/m

Connections

Electric Charge and the Coulomb — $C = Q / V$ mein $Q$ coulombs mein measure hota hai.
Voltage and Potential Difference — $V$ wo "pressure" hai jo bucket bharta hai.
Gauss's Law and Electric Fields — parallel-plate field derive karne mein use hua.
Dielectrics and Permittivity — $ε_{r}$ capacitance boost karta hai.
Capacitors in Series and Parallel — buckets combine karna.
Energy Storage in Circuits — $\frac{1}{2} C V^{2}$ flashes, filters, backups power karta hai.