Gauss Law

Gauss's Law is a key concept in electromagnetism that connects the electric flux passing through a closed surface to the total charge inside it. This law is especially helpful for finding electric fields in situations with symmetrical charge distributions, like spheres, cylinders, or flat surfaces. Beyond electrostatics, Gauss's Law also plays a role in deriving the magnetic field. This leads to Gauss's Law for magnetism, which states that the total magnetic flux through any closed surface is always zero, indicating that magnetic monopoles do not exist.

1.0Electric Flux

• Electric flux through a surface in an electric field shows the total number of electric field lines crossing it and is a property of the field.
• The SI unit of electric flux is $N{m}^2{C}^{-1}$(Gauss) or $Jm{C}^1$. Electric flux is a scalar quantity. It can be positive, negative or zero.

$d\varphi =Eds\cos \theta$

$d\varphi =(E\cos \theta )ds$

$d\varphi =E_{n}ds$

$E_n$ is the component of electric field in the direction of $d\vec{s}$

The electric flux over the whole area =$\varphi ={\int }_S\vec{E}\cdot d\vec{s}={\int }_S{E}_{n}ds$

If the electric field is uniform over that area then $\varphi =\vec{E}\cdot \vec{S}$

Special Cases:

Case 1:If the electric field is normal to the surface, then angle of electric field E

 with normal will be zero.

$\varphi =ES\cos 0^{\circ }\Rightarrow \varphi =ES$

Case 2:If electric field is parallel of the surface (grazing), then angle made by $\vec{E}\text{with normal}=90°.$

$\varphi =ES\cos 90^{\circ }\Rightarrow \varphi =0$

Case 3: Curved surface in uniform electric field

Consider a circular surface of radius R placed in a uniform electric field, as depicted.

Flux passing through the surface $\varphi =E(\pi R^2)$

Now suppose, a hemispherical surface is placed in the electric field. Flux through

hemispherical surface:

$\varphi =\int Eds\cos \theta =E\int ds\cos \theta$

$\int ds\cos \theta$ is projection of the spherical surface Area on base.

$\int ds\cos \theta =\pi R^2$

$\varphi =E(\pi R^2)$

Note: If two surfaces have the same number of electric field lines passing through them, the flux will be the same, regardless of shape.

${\varphi }_1={\varphi }_2={\varphi }_3=E(\pi R^2)$

Case 4: Flux through a closed surface 

Flux through the spherical surface

$\varphi =\int \vec{E}\cdot d\vec{s}=\int Eds(\text{as }\vec{E}\text{ is along }d\vec{s}\text{ (normal)})$

$\varphi =\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R^2}\int dS\left(\int ds=4\pi R^2\right)$

$\varphi =\left(\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R^2}\right)(4\pi R^2)\Rightarrow \varphi =\frac{Q}{{\epsilon }_0}$

Even if the charge Q is enclosed by a different closed surface, the same number of lines of force will pass through that surface.

$\varphi =\frac{Q}{{\epsilon }_0}$

2.0 Gauss Theorem

• Gauss's law, formulated by Carl F. Gauss, relates the electric field on a closed surface to the net enclosed charge. The Gaussian surface, a hypothetical closed surface, is validated through experiments and is used to calculate electric fields for symmetric charge distributions.
• The surface integral of the electric field intensity over any closed hypothetical surface (Gaussian surface) in free space equals 1/ε₀ times the total charge enclosed within the surface, where ε₀ is the permittivity of free space.
• If S is the Gaussian surface and $su{m}_{i=1}^n{q}_i$ is the total charge enclosed by the Gaussian surface, then according to Gauss's law,

$\varphi =\oint \vec{E}\cdot d\vec{s}=\frac{1}{{\epsilon }_0}{\sum }_{i=1}^n{q}_i$

Note:

(1) Flux through the Gaussian surface is independent of its shape.

(2) The flux through a Gaussian surface depends solely on the total charge enclosed within the surface.

(3) Flux through the Gaussian surface is independent of the position of charges inside the Gaussian surface.

(4) The electric field intensity at a Gaussian surface is influenced by all the charges both inside and outside the surface.

(5) In a closed surface incoming flux is taken negative, while outgoing flux is taken positive, because $\hat{n}$ is taken positive in an outward direction.

(6) A flux of zero through a Gaussian surface (Φ = 0) does not necessarily mean that the electric field (E) is zero at every point on the surface. However, if the electric field is zero at every point on the surface (E = 0), then the flux must also be zero (Φ = 0)

3.0Electric Field Due to a Spherical Shell

Electric field outside the Sphere

$\varphi =\int \vec{E}\cdot d\vec{s}={\varphi }_{\text{net}}=\frac{q_{\text{in}}}{{\epsilon }_0}=\frac{q}{{\epsilon }_0}$

$\int \vec{E}\cdot d\vec{s}=\int Eds\cos 0^{\circ }=E\int ds=E\left(4\pi R^2\right)$

$E\left(4\pi R^2\right)=\frac{q_{\text{in}}}{{\epsilon }_0}$

$E_{\text{out}}=\frac{q}{4\pi {\epsilon }_0{r}^2}$

Electric field inside a sphere

$\varphi =\int \vec{E}\cdot d\vec{s}={\varphi }_{\text{net}}=\frac{q_{\text{in}}}{{\epsilon }_0}=0$

$\int \vec{E}\cdot d\vec{s}=\int Eds\cos 0^{\circ }=E\int ds=E\left(4\pi R^2\right)$

$E\left(4\pi R^2\right)=0$

$E_{\text{in}}=0$

4.0Electric field due to uniformly charged solid sphere

For a solid sphere having uniformly distributed charge Q  and radius R.

Electric field outside the sphere

$\varphi =\int \vec{E}\cdot d\vec{s}={\varphi }_{\text{net}}=\frac{q_{\text{in}}}{{\epsilon }_0}=\frac{Q}{{\epsilon }_0}$

$\int \vec{E}\cdot d\vec{s}=\int Eds\cos 0^{\circ }=E\int ds=E\left(4\pi r^2\right)$

$E\left(4\pi r^2\right)=\frac{Q}{{\epsilon }_0}$

$E_{\text{out}}=\frac{Q}{4\pi {\epsilon }_0{r}^2}$

Electric field inside a solid sphere

$\varphi =\int \vec{E}\cdot d\vec{s}={\varphi }_{\text{net}}=\frac{q_{\text{in}}}{{\epsilon }_0}=\frac{\frac{Q}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi r^3}{{\epsilon }_0}=\frac{Q{r}^3}{{\epsilon }_0{R}^3}$

$\int \vec{E}\cdot d\vec{s}=\int Eds\cos 0^{\circ }=E\int ds=E\left(4\pi r^2\right)$

$E\left(4\pi r^2\right)=\frac{Q{r}^3}{{\epsilon }_0{R}^3}$

$E=\frac{Qr}{4\pi {\epsilon }_0{R}^3}$

$E_{\text{in}}=\frac{KQ}{R^3}r$

$\vec{E}=\frac{\rho \vec{r}}{3{\epsilon }_0}(\rho =\text{Volume Charge Density})$

5.0Electric field due to infinite line charge

${\varphi }_3=\int \vec{E}\cdot d\vec{s}=\frac{q_{\text{in}}}{{\epsilon }_0}$

$\int Eds=\frac{\lambda l}{{\epsilon }_0}$

$E\int ds=\frac{\lambda l}{{\epsilon }_o}$

$E(2\pi rl)=\frac{\lambda l}{{\epsilon }_o}$

$E=\frac{\lambda }{2\pi {\epsilon }_{o}r}=\frac{2k\lambda }{r}$

6.0Electric field due to infinitely long charged Tube

E outside the tube

${\varphi }_{\text{net}}=\frac{q_{\text{in}}}{{\epsilon }_o}=\frac{\sigma 2\pi Rl}{{\epsilon }_o}$

$\Rightarrow E_{\text{out}}\times 2\pi rl=\frac{\sigma 2\pi Rl}{{\epsilon }_o}$

$E=\frac{\sigma R}{r{\epsilon }_o}$

E inside the tube 

${\varphi }_{\text{net}}=\frac{q_{\text{in}}}{{\epsilon }_o}=0$

So $E_{\text{in}}=0$

7.0Electric Field due to infinitely long solid cylinder of radius R 

E at outside point:

$E\times 2\pi rl=\frac{q_{\text{in}}}{{\epsilon }_o}=\frac{\rho \times \pi R^{2}l}{{\epsilon }_o}$

$E_{\text{out}}=\frac{\rho R^2}{2r{\epsilon }_o}$

E at inside point:

$E\times 2\pi rl=\frac{q_{\text{in}}}{{\epsilon }_o}=\frac{\rho \times \pi r^{2}l}{{\epsilon }_o}$

$\Rightarrow E_{\text{in}}=\frac{\rho r}{2{\epsilon }_o}$