Electric Flux , Gauss Law , Applications of Gauss Law

Flux of electric field :– Flux of electric field E through any area A is defined as

$\displaystyle \phi = \vec{E}.\vec{S} = E S cos\theta$

(Here the direction of area vector is outward normal to the area S)

In case of non uniform electric field or curved area
$\displaystyle \phi = \int \vec{E}.\vec{dS}$

A cylinder of radius R and length l  is placed in uniform electric field E such that its axis is perpendicular to E.

Flux linked with flat surfaces , $\displaystyle \phi = \vec{E}.\vec{S} = 0$ (As θ = 90° )

Flux linked with curved surfaces , $\displaystyle \phi = \vec{E}.\vec{S} = E S$ (As θ= 0° )

Gauss’s Theroem:

The net electric flux emerging out of a closed surface is equal to the net charge enclosed by it divided by εo

$\displaystyle \phi = \frac{\Sigma q_{in}}{\epsilon_0}$

$\displaystyle \oint \vec{E}.\vec{dS} = \frac{\Sigma q_{in}}{\epsilon_0}$

This law is true for any closed surface, but is useful only in case of symmetry, because in other cases the integral on the left hand side becomes very difficult to evaluate .

Applications of Gauss’s Law

(A) Electric field due to a uniformly charged spherical shell:

From symmetry the electric field at any point on the Gaussian surface is same and is directed radially outward So ,

$\displaystyle \oint \vec{E}.\vec{dS} = E (4 \pi r^2 )$ (Since cos θ = 1)

From Gauss’s law

$\displaystyle E (4 \pi r^2 ) = \frac{Q}{\epsilon_0}$ (if r > R )

$\displaystyle E = \frac{Q}{4 \pi \epsilon_0 r^2}$

or,  E (4 π r2) = 0 ( if r < R )

⇒  E = 0

Thus, electric field inside the shell is zero and outside it is similar to that of a point charge. The same result is true for a charged solid conducting sphere because the charge given to a conductor resides on its outer surface.

Similarly for a uniformly charged solid dielectric sphere carrying a charge Q and having radius R it can be shown that :

$\displaystyle E = \frac{Q}{4 \pi \epsilon_0 r^2}$ (If r > R)

$\displaystyle E = \frac{Q r}{4 \pi \epsilon_0 R^3}$ (If r < R)

(B) For an infinite line charge using a cylindrical Gaussian surface, it can be shown that

$\displaystyle E = \frac{\lambda}{2 \pi \epsilon_0 r}$

where λ is the charge per unit length and r is the distance of the point from the line charge

(C) Electric field close to an infinite plane sheet having uniform surface charge density (charge per unit area) σ is

$\displaystyle E = \frac{\sigma}{2 \epsilon_0 }$

which is directed perpendicular to the plane sheet.

Note: that this value is independent of the distance of the point from the sheet. Thus it is a uniform electric field.

(D) Electric field close to an infinite plane-conducting surface of charge density σ is given as –

$\displaystyle E = \frac{\sigma}{ \epsilon_0 }$