Gauss's Law

Electric Flux $\left({\varphi }_{\mathrm{E}}\right)$

Electric flux over an area is equal to the total number of electrical field lines crossing this area.

Electric flux through a small area element dS is given by

${\varphi }_E=E.dS$

Where E = electric field intensity

And dS = area vector.

Its SI unit is N-m²C^–1.

Gauss' Theorem

The electric flux over any closed surface is $\frac{1}{{\epsilon }_0}$ times the total charge enclosed by that surface, i.e.

${\varphi }_E={\oint }_{s}E.dS=\frac{1}{{\epsilon }_0}\sum q$

If a charge q is placed at the centre of a cube, then

Total electric flux linked with the whole cube $=\frac{q}{{\epsilon }_0}$

Electric flux linked with one face of the cube $=\frac{q}{6{\epsilon }_0}$

(i) Electric Field at Any Point on the Axis of a Uniformly Charged Ring A ring-shaped conductor with radius a carries a total charge Q uniformly distributed around it. Let us calculate the electric field at a point P that lies on the axis of the ring at a distance x from its centre.

$E_x=\frac{1}{4\pi {\epsilon }_0}.\frac{xQ}{{\left(x^2+a^2\right)}^{\frac{3}{2}}}$

The maximum value of electric field

$E=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q}{3\sqrt{3}{R}^2}\right)$

(ii) Electric Field due to a Charged Spherical Shell

(a) At an extreme point (r>R)

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

(b) At the surface of a shell (r=R)

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2}$

(c) At an internal point (r<R)

E = 0

(iii) Electric Field Intensity due to an Infinite Line Charge

$E=\frac{1}{2\pi {\epsilon }_0}\frac{\lambda }{r}$

Where $\lambda$ is linear charge density and r is distance from the line charge.

(iv) Electrical Field Near an Infinite Plane Sheet of Charge

$E=\frac{\sigma }{2{\epsilon }_0}$

Where $\sigma =$ surface charge density.

If infinite plane sheet has uniform thickness, then

$E=\frac{\sigma }{{\epsilon }_0}$

(v) Electric Potential due to a Charge Conducting Spherical Shell

(a) At an extreme point, (r > R)

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

(b) At the surface of a shell, (r = R)

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

(c) At an internal point (r < R)

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

Therefore potential inside a charged conducting spherical shell is equal to the potential at its surface.

(vi) Electric field and Potential due to a Charged Non-Conducting Sphere

At an extreme point, (r > R)

(a) Electric field intensity $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

(b) Electric potential, (r = R) $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

On the suface, (r =R )

(a) Electric field intensity $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2}$

(b) Electrical potential $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

Inside the sphere, (r < R)

(a) Electric field intensity $E=\frac{1}{4\pi {\epsilon }_0}\frac{qr}{R^3}$

(b) Electric potential $V=\frac{1}{4\pi {\epsilon }_0}\frac{q\left(3R^2-r^2\right)}{2R^3}$

At the centre of the sphere, (r = 0)

(a) Electric field intensity E = 0

(b) Electric potential $V=\frac{3}{2}\times \frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$