Consider a thin infinitely long straight line charge of linear charge density λ.
Let P be the point at a distance ‘a’ from the line. To find electric field at point P, draw a cylindrical surface of radius ‘a’ and length l.
If E is the magnitude of electric field at point P, then electric flux through the Gaussian surface is given by,
Φ = E × Area of the curved surface of a cylinder of radius r and length l
Because electric lines of force are parallel to end faces (circular caps) of the cylinder, there is no component of field along the normal to the end faces.
Φ = E × 2πal … (i)
According to Gauss theorem, we have
From equations (i) and (ii), we obtain
Electric Field Due To An Infinite Plane Sheet Of Charge
Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σon both sides of the sheet. Let P be the point at a distance ‘a’ from the sheet at which electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.
The electric flux crossing through the Gaussian surface is given by,
Φ = E × Area of the circular caps of the cylinder
Since electric lines of force are parallel to the curved surface of the cylinder, the flux due to electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.
Φ = E × 2A … (i)
According to Gauss theorem, we have
Here, the charge enclosed by the Gaussian surface,
q = σA
From equations (i) and (ii), we obtain
Electric Field Due To A Uniformly Charged Thin Spherical Shell
When point P lies outside the spherical shell
Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let be the electric field at point P. Then, the electric flux through area element is given by,
Since is also along normal to the surface,
dΦ = Eds
∴ Total electric flux through the Gaussian surface is given by,
Now,
Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,
From equations (i) and (ii), we obtain
When point P lies inside the spherical shell
In such a case, the Gaussian surface encloses no charge.
The electric flux, through a surface, held inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal to the surface.
Electric flux is a scalar quantity and is denoted by Φ.
SI unit − Nm2 C−1
Gauss Theorem
It states that the total electric flux through a closed surface enclosing a charge is equal to times the magnitude of the charge enclosed.
However,
∴Gauss theorem may be expressed as
Proof
Consider that a point electric charge q is situated at the centre of a sphere of radius ‘a’.
According to Coulomb’s law,
Where, is unit vector along the line OP
The electric flux through area element is given by,
Therefore, electric flux through the closed surface of the sphere,
Electric Dipole & Dipole in a Uniform External Field
Electric Dipole − System of two equal and opposite charges separated by a certain small distance.
Electric Dipole Moment − It is a vector quantity, with magnitude equal to the product of either of the charges and the length of the electric dipole
Its direction is from the negative charge to the positive charge.
Electric Field on Axial Line of an Electric Dipole
Let P be at distance r from the centre of the dipole on the side of charge q. Then,
Where, is the unit vector along the dipole axis (from − q to q). Also,
The total field at P is
For r >> a
Electric Field for Points on the Equatorial Plane
The magnitudes of the electric field due to the two charges +q and −q are given by,
The directions of E+q and E−q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
∴ Total electric field
[Negative sign shows that field is opposite to]
At large distances (r >> a), this reduces to
Dipole in a Uniform External Field
Consider an electric dipole consisting of charges −q and +q and of length 2a placed in a uniform electric field making an angle θ with electric field.
Force on charge −q at (opposite to)
Force on charge +q at (along)
Electric dipole is under the action of two equal and unlike parallel forces, which give rise to a torque on the dipole.
τ = Force × Perpendicular distance between the two forces
be the electric field at point P. Then, the electric flux through area element 
is given by,
is also along normal to the surface,
times the magnitude of the charge enclosed.
is unit vector along the line OP
is given by,
is the unit vector along the dipole axis (from − q to q). Also,
[Negative sign shows that field is opposite to
making an angle θ with electric field.
(opposite to
(along