Gauss's Law (Concise Notes)

Prerequisite

Area vector, \(\vec A\) or \(\vec S\)
- vector quantity,
- magnitude = area of surface, S
- direction along positive normal (outward normal), \(\hat S\)

Electric Flux

measure of electric field passing through a given surface
dot product of \(\vec E\) and \(\vec S\)
mathematically, \(d\Phi = {\vec E} \cdot {d\vec S}\)
Total electric flux through a closed surface, \(\Phi_E = \int \vec{E} \cdot d\vec{A}\)
scalar quantity
SI unit \(Nm^2C^{-1}\)

Statement

For, total electric flux passing through a closed surface, \(\oint \vec E \cdot d\vec S\)
and net charge enclosed by the surface, \(q_{enc}\)
total electric flux, \(\Phi_E\)=\(\oint \vec E \cdot d\vec S\) = \(\frac{q_{enc}}{\epsilon_0}\)
where,
- \(\vec E\) is the net electric field due to charges existing in the space, inside or outside of the closed surface
  - for a charge outside, the amount of electric field that enters and exits a closed surface are equal
- \(q_{enc}\) is the charge enclosed by the surface only

Significance of Gauss’s Law

It helps in calculation of electric field due to a system of charges, especially symmetrical one

Derivation of Gauss’s Law

For a single charge, q
spherical gaussian surface at a distance r
electric field, \(\vec E = \frac{q}{4\pi \epsilon_0 r^2} \hat r\), \(E\) is constant
Area element, \(\vec S = S \hat r\)
elemental electric flux, \(\vec E \cdot d \vec S\)
Total flux through closed surface
- \(\Phi_E = \oint \vec E \cdot d \vec S\)
- \(\Phi_E = \oint E \cdot dS\) , \(\vec E\) and \(d\vec S\) are parallel
- \(\Phi_E = E \oint dS\)
- \(\Phi_E = E \cdot 4 \pi r^2\)
- \(\Phi_E = \frac{q}{\epsilon_0}\), q is enclosed by the gaussian surface

Gaussian Surface

imaginary closed surface on which Gauss’s law is applied
could be of any shape or size but
- symmetrical shape help easy calculation of electric field
- preferred shape – part of surface is either parallel or perpendicular to \(\vec E\)
should not pass through discrete charge
- \(\vec E\) is not defined at the location of the charge
may pass through continuous charge distribution

Steps to find \(\vec E\) using Gauss’s Law

Choose suitable symmetrical Gaussian surface
Calculate charge enclosed by the surface, \(q_{enc}\)
Find electric flux, \(\Phi = \oint \vec E \cdot d \vec S\) through the gaussian surface
Equate \(\Phi\) with \(\frac{q_{enc}}{\epsilon_0}\) and solve for \(\vec E\)

Infinitely long straight wire of charge density, \(\lambda\)
Choice of a point, P, where \(\vec E\) need to be found
- at a distance \(R\) from the wire
- parallel components of \(\vec E\) cancel each other
- net direction of \(\vec E\) is radially outwards
Choice of a cylindrical gaussian surface, radius \(R\) and length \(l\)
- curved surface passing through P
- plane surfaces, perpendicular to the wire
Electric flux, \(\Phi = \Phi_C + \Phi_P\)
- \(\Phi_C = \oint E \cdot dS = E\int dS = E \cdot 2\pi R l\) as \(\theta\ = 0^o\)
- \(Phi_P = 0\) as \(\theta = 90^o\)
Apply Gauss’s Law
- \(\Phi = \frac{q_{enc}}{\epsilon_0}\)
- \(E \cdot 2\pi R l = \frac{q_{enc}}{\epsilon_0}\)
- \(E \cdot 2\pi R l = \frac{\lambda l}{\epsilon_0}\)
- \(E = \frac{\lambda }{2\pi \epsilon_0 R}\)