Capacitance — parallel plate derivation, cylindrical, spherical

1.8.11 · Physics › Electromagnetism

Master recipe (KAISE koi bhi capacitor derive karen)

Neeche har derivation same chaar steps ki hai. Recipe seekho, teen alag formulas nahi.

1. Ek conductor pe $+Q$ charge rakho, doosre pe $-Q$.
2. Unke beech field $\vec{E}$ nikalo — almost always Gauss's law se.
3. Integrate karke voltage nikalo: $V=-{\int }_{-}^{+}\vec{E}\cdot d\vec{l}=\int \vec{E}\cdot d\vec{l}$ (+ plate se − plate ki taraf lene pe positive number aata hai).
4. Divide karo: $C=Q/V$. $Q$ cancel ho jaata hai — yeh proof hai ki yeh pure geometry hai.

Gauss's law (engine): $\oint \vec{E}\cdot d\vec{A}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$.

1. Parallel plate

Setup: do plates area $A$, separation $d$, charge $\pm Q$, surface charge density $\sigma =Q/A$.

Step 2 — field. Positive plate ke ird-gird ek pillbox Gaussian surface use karo. Flux sirf uss face se nikalta hai jo gap ke andar hai (conductor mein field ~0 hoti hai aur ideal capacitor ke bahar bhi): $E{A}_{\text{box}}=\frac{\sigma A_{\text{box}}}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{{\epsilon }_{0}A}.$ Yeh step KYUN? Pillbox charge $\sigma A_{\text{box}}$ enclose karta hai; area cancel ho jaata hai, aur ek uniform field milti hai jo position pe depend nahi karta — yeh infinite-plane field ki pehchaan hai.

Step 3 — voltage. Field uniform hai, toh integral bas $E\times d$ hai: $V={\int }_0^{d}Edx=Ed=\frac{Qd}{{\epsilon }_{0}A}.$

Step 4 — divide. $\boxed{C=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d}}$

2. Cylindrical (coaxial)

Setup: inner radius $a$, outer radius $b$, length $L$ (maano $L\gg b$ toh ends ignore hain). Inner pe $+Q$ charge, outer pe $-Q$; linear density $\lambda =Q/L$.

Step 2 — field. Coaxial Gaussian cylinder radius $r$ ($a<r<b$), length $L$. Symmetry se $E$ radial hai aur surface pe constant hai: $E(2\pi rL)=\frac{Q}{{\epsilon }_0}\Rightarrow E(r)=\frac{Q}{2\pi {\epsilon }_{0}Lr}=\frac{\lambda }{2\pi {\epsilon }_{0}r}.$ Yeh step KYUN? Sirf curved side pe flux hai; uska area $2\pi rL$ hai. Field $1/r$ ki tarah girta hai.

Step 3 — voltage. $a$ (high) se $b$ (low) tak integrate karo: $V={\int }_a^{b}Edr=\frac{Q}{2\pi {\epsilon }_{0}L}{\int }_a^b\frac{dr}{r}=\frac{Q}{2\pi {\epsilon }_{0}L}\ln \frac{b}{a}.$ Yeh step KYUN? $\int dr/r=\ln r$ — $1/r$ field exactly wahi hai jisse logarithm produce hota hai.

Step 4 — divide. $\boxed{C=\frac{2\pi {\epsilon }_{0}L}{\ln (b/a)}}$

3. Spherical

Setup: inner sphere radius $a$, concentric outer shell radius $b$, charge $\pm Q$.

Step 2 — field. Gaussian sphere radius $r$ ($a<r<b$): $E(4\pi r^2)=\frac{Q}{{\epsilon }_0}\Rightarrow E(r)=\frac{Q}{4\pi {\epsilon }_0{r}^2}.$ Yeh step KYUN? Shells ke beech pure point-charge field hai; area $4\pi r^2$ hai, field $1/r^2$ ki tarah girta hai.

Step 3 — voltage. $V={\int }_a^b\frac{Q}{4\pi {\epsilon }_0{r}^2}dr=\frac{Q}{4\pi {\epsilon }_0}{\left[-\frac{1}{r}\right]}_a^b=\frac{Q}{4\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{Q}{4\pi {\epsilon }_0}\frac{b-a}{ab}.$ Yeh step KYUN? $\int dr/r^2=-1/r$; evaluate karne pe $(1/a-1/b)$ combination milta hai.

Step 4 — divide. $\boxed{C=\frac{4\pi {\epsilon }_{0}ab}{b-a}}$

Flashcards

Connections

• Gauss's Law — teeno field calculations ka engine.
• Electric Potential — step 3 integration $V=\int \vec{E}\cdot d\vec{l}$.
• Dielectrics and Polarization — $\kappa$ factor.
• Energy Stored in a Capacitor — $U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}$.
• Capacitors in Series and Parallel — in geometries ko combine karna.
• Coaxial Cable Transmission Lines — real-world cylindrical case.

Concept Map