Capacitance — parallel plate derivation, cylindrical, spherical

The master recipe (HOW to derive any capacitor)

Every derivation below is the same four steps. Learn the recipe, not three formulas.

1. Put charge $+Q$ on one conductor, $-Q$ on the other.
2. Find the field $\vec E$ between them — almost always via Gauss's law.
3. Integrate to get the voltage: $V = -\int_{-}^{+}\vec E\cdot d\vec l = \int \vec E\cdot d\vec l$ (taken from + plate to − plate gives a positive number).
4. Divide: $C = Q/V$. The $Q$ cancels — proof it's pure geometry.

Gauss's law (the engine): $\displaystyle \oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0}$.

1. Parallel plate

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

Step 2 — field. Use a pillbox Gaussian surface straddling the positive plate. Flux only exits through the face inside the gap (field is ~0 in the conductor and outside an ideal capacitor): $E\,A_{\text{box}} = \frac{\sigma A_{\text{box}}}{\varepsilon_0} \;\Rightarrow\; E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}.$ Why this step? The pillbox encloses charge $\sigma A_{\text{box}}$; the area cancels, giving a uniform field independent of position — that's the signature of an infinite-plane field.

Step 3 — voltage. Field is uniform, so the integral is just $E\times d$: $V = \int_0^d E\,dx = E d = \frac{Q d}{\varepsilon_0 A}.$

Step 4 — divide. $\boxed{C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}}$

2. Cylindrical (coaxial)

Setup: inner radius $a$, outer radius $b$, length $L$ (assume $L \gg b$ so ends ignored). Charge $+Q$ on inner, $-Q$ on outer; linear density $\lambda = Q/L$.

Step 2 — field. Coaxial Gaussian cylinder radius $r$ ($a<r<b$), length $L$. By symmetry $E$ is radial and constant on the surface: $E(2\pi r L) = \frac{Q}{\varepsilon_0}\;\Rightarrow\; E(r) = \frac{Q}{2\pi\varepsilon_0 L\, r} = \frac{\lambda}{2\pi\varepsilon_0 r}.$ Why this step? Only the curved side has flux; its area is $2\pi r L$. The field falls as $1/r$.

Step 3 — voltage. Integrate from $a$ (high) to $b$ (low): $V = \int_a^b E\,dr = \frac{Q}{2\pi\varepsilon_0 L}\int_a^b \frac{dr}{r} = \frac{Q}{2\pi\varepsilon_0 L}\ln\!\frac{b}{a}.$ Why this step? $\int dr/r = \ln r$ — the $1/r$ field is exactly what produces a logarithm.

Step 4 — divide. $\boxed{C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}}$

3. Spherical

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

Step 2 — field. Gaussian sphere radius $r$ ($a<r<b$): $E(4\pi r^2) = \frac{Q}{\varepsilon_0}\;\Rightarrow\; E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}.$ Why this step? Pure point-charge field between the shells; area is $4\pi r^2$, field falls as $1/r^2$.

Step 3 — voltage. $V = \int_a^b \frac{Q}{4\pi\varepsilon_0 r^2}\,dr = \frac{Q}{4\pi\varepsilon_0}\left[-\frac1r\right]_a^b = \frac{Q}{4\pi\varepsilon_0}\left(\frac1a-\frac1b\right) = \frac{Q}{4\pi\varepsilon_0}\,\frac{b-a}{ab}.$ Why this step? $\int dr/r^2 = -1/r$; evaluating gives the $(1/a-1/b)$ combination.

Step 4 — divide. $\boxed{C = \frac{4\pi\varepsilon_0\, ab}{b-a}}$

Flashcards

Connections

• Gauss's Law — the engine for all three field calculations.
• Electric Potential — step 3 integration $V=\int\vec E\cdot d\vec l$.
• Dielectrics and Polarization — the $\kappa$ factor.
• Energy Stored in a Capacitor — $U=\tfrac12 CV^2=\tfrac{Q^2}{2C}$.
• Capacitors in Series and Parallel — combining these geometries.
• Coaxial Cable Transmission Lines — real-world cylindrical case.

Concept Map

Hinglish (regional understanding)