Applications — sphere, cylinder, infinite plane

0. The master tool

1. The Sphere

Consider total charge $Q$ on a sphere of radius $R$ (either a conducting shell with charge on the surface, OR a uniformly charged insulator with volume density $\rho$).

Derivation (outside, $r>R$): $\oint \vec E\cdot d\vec A = E\oint dA = E\,(4\pi r^2)$ Why this step? $E$ is constant on the surface so it factors out; $\oint dA$ is just the sphere's area.

Set equal to $Q_{enc}/\varepsilon_0 = Q/\varepsilon_0$: $E(4\pi r^2)=\frac{Q}{\varepsilon_0}\;\Rightarrow\; \boxed{E_{out}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}}$

Inside a conducting shell ($r<R$): $Q_{enc}=0$, so $\boxed{E_{in}=0}$.

Inside a uniformly charged insulator ($r<R$): only the charge within radius $r$ counts. $Q_{enc}=\rho\cdot\tfrac{4}{3}\pi r^3,\qquad \rho=\frac{Q}{\tfrac43\pi R^3}$ $E(4\pi r^2)=\frac{\rho\,\tfrac43\pi r^3}{\varepsilon_0}\;\Rightarrow\;\boxed{E_{in}=\frac{\rho\,r}{3\varepsilon_0}=\frac{Q\,r}{4\pi\varepsilon_0 R^3}}$ Why this step? Inside a solid ball, enclosed charge grows like $r^3$ while area grows like $r^2$, leaving $E\propto r$ — it grows linearly from the center.

2. The Cylinder / Infinite Line

A line charge with linear density $\lambda$ (C/m), or a long charged cylinder.

Derivation (line charge): Gaussian cylinder radius $s$, length $L$. $\oint\vec E\cdot d\vec A = \underbrace{E(2\pi s L)}_{\text{curved side}} + \underbrace{0}_{\text{two caps}}$ Why this step? Curved area $=2\pi s L$; end caps give nothing because field is parallel to them.

Enclosed charge $Q_{enc}=\lambda L$: $E(2\pi sL)=\frac{\lambda L}{\varepsilon_0}\;\Rightarrow\;\boxed{E=\frac{\lambda}{2\pi\varepsilon_0 s}}$ Why this step? The $L$ cancels — the field doesn't care how long your box is, which proves the choice was legitimate.

3. The Infinite Plane

A flat sheet with surface charge density $\sigma$ (C/m²).

Derivation: pillbox with face area $A$. $\oint\vec E\cdot d\vec A = E\,A + E\,A = 2EA$ Why this step? Field exits both faces (two contributions $EA$); side walls have $\vec E\perp d\vec A$ ⇒ zero.

Enclosed charge $=\sigma A$: $2EA=\frac{\sigma A}{\varepsilon_0}\;\Rightarrow\;\boxed{E=\frac{\sigma}{2\varepsilon_0}}$ Why this step? $A$ cancels — confirming the field is uniform, independent of distance from the sheet!

Common mistakes

Flashcards

Connections

• Gauss's Law — the master equation applied here
• Electric Flux — $\oint\vec E\cdot d\vec A$ definition
• Coulomb's Law — sphere result recovers point-charge field
• Conductors in Electrostatics — why $E=0$ inside, $\sigma/\varepsilon_0$ outside
• Parallel Plate Capacitor — uses the two-sheet superposition
• Electric Potential — integrate these $E$ fields to get $V$
• Symmetry in Physics — the deciding principle for surface choice

Concept Map

Hinglish (regional understanding)