Electrostatics

Charge

Quantized property, incrementing in order $q_e=1.602\times 10^{-19}C$ which is always conserved in a closed system. Net charge of matter is EXACTLY zero. It does NOT depend on relativistic speed.

$${\vec{F}}_{1{\to }_2}=-{\vec{F}}_{2{\to }_1}=k\frac{Q_1{Q}_2}{r^2}{u}_{r_1}=\vec{E}q$$ $$\vec{E}=\frac{kQ}{r^2}{\vec{u}}_r$$ $${\vec{E}}_{TOT}=\Sigma {\vec{E}}_i$$

“Electrostatic” means that charges are stationary with respect to a reference frame.

Electrostatic Field Properties

• Central Force
• Conservative Force
• Gauss Law holds
• NOT an acceleration field, since charge does not depend on mass
• Has a scalar potential associated
• Force can be attractive or repulsive
• Radial
• Measured in $N{C}^{-1}$
• A charge is needed to measure it

If the path is closed, then:

$$\begin{array}{rl}& \vec{A}=\vec{B}\\ & ⟹\oint \vec{E}\cdot d\vec{s}=0\end{array}$$

Since electrostatic force is conservative.

Hollow Charged Sphere

If we have a hollow sphere of radius $R$ with uniform charge on the surface (total charge is $Q$, $Q_{\text{inner}}=0$), at a distance $r$ from the centre of the sphere, we get two scenarios:

When $r\le R$:

$$\begin{array}{rl}& E=0\\ & V=k\frac{Q}{R}\end{array}$$

And when $r>R$ the sphere acts like a point charge at its centre, so laws below hold.

Electrostatic Potential

$V_{es}$ with unit $V$. Can be used to find forces and field.

$$\begin{array}{rl}& \vec{E}=\frac{F}{q}\\ & \vec{F}=-\nabla U\\ & U=-\int \vec{F}\cdot d\vec{s}\\ & V_{es}=\frac{U}{q}\\ & \vec{E}=-\nabla V_{es}\\ & V_{es}=-\int \vec{E}\cdot d\vec{s}\\ & V_{TOT}=\Sigma V_i\end{array}$$

Potential difference:

$$\Delta V_{AB}=-{\int }_A^B\vec{E}\cdot d\vec{s}=V_A-V_B$$

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Gauss’ Law

DEFINITION

For a charge INSIDE a CLOSED surface generating an electrostatic field, we have flux:

$${\Phi }_{\vec{E}}={\oint }_S\vec{E}\cdot d\vec{S}=\frac{Q}{{\epsilon }_0}$$

However, if the charge is outside the closed surface, we get: ${\Phi }_{\vec{E}}=0$

Infinite Wire

Assume an infinite cylinder with charge density $\lambda =\frac{dq}{dL}$. We then have an IMAGINARY Gaussian surface as a cylinder with radius $r$ and length $h$ with central axis aligned with the wire.

$$⟹Q=\lambda h$$

On the top and bottom lids, ${\Phi }_{\vec{E}}=0$, while on the side walls:

$$\begin{array}{rl}& {\Phi }_{\vec{E}}=2E\pi rh=4k\lambda h\pi \\ & V_{es}=\text{const}-2j\lambda \ln r\end{array}$$

Infinite Plane

Assume an infinite plane with charge density $\sigma$ ($[\sigma ]=C{m}^{-2}$). We then have a Gaussian surface being a cylinder cutting perpendicularly through the plane.

On the side walls, ${\Phi }_{\vec{E}}=0$, while on the taps:

$${\Phi }_{\vec{E}}=2ES=\frac{\sigma S}{{\epsilon }_0}=\frac{Q}{{\epsilon }_0}$$

Inside the cylinder,

$$\begin{array}{rl}& \vec{E}=\frac{\sigma }{2{\epsilon }_0}{\vec{u}}_{\perp }\\ & ⟹V_{es}=0\end{array}$$

Summary

Geometry	Field Formula
Point Charge / Sphere (Outside)	$E=\frac{kQ}{r^2}$
Sphere (Inside)	$E=0$
Infinite wire	$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$
Infinite plane	$E=\frac{\sigma }{2{\epsilon }_0}$