Electric Field on the Axial Line of an Electric Dipole.
Consider an electric dipole consisting of charges \( -q \) and \( +q \), separated by a distance \( 2a \), placed in free space.
The electric field \( \mathbf{E} \) at point \( P \) is the resultant of the electric fields \( \mathbf{E_A} \) (due to \( -q \) at \( A \)) and \( \mathbf{E_B} \) (due to \( +q \) at \( B \)):
\[
\mathbf{E} = \mathbf{E_B} - \mathbf{E_A}
\]
Using Coulomb’s law:
\[
E_B = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r - a)^2}, \quad E_A = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r + a)^2}
\]
So,
\[
E = \frac{1}{4\pi\varepsilon_0} q \left( \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right)
\]
Using the identity:
\[
\frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} = \frac{4ra}{(r^2 - a^2)^2}
\]
we obtain:
\[
E = \frac{1}{4\pi\varepsilon_0} \frac{q(4ra)}{(r^2 - a^2)^2}
\]
Since the dipole moment is:
\[
P = q(2a)
\]
we rewrite:
\[
E = \frac{1}{4\pi\varepsilon_0} \frac{2Pr}{(r^2 - a^2)^2}
\]
For a small dipole (\( a \ll r \)):
\[
\mathbf{E} = \frac{1}{4\pi\varepsilon_0} \frac{2\mathbf{P} r}{r^3}
\]
Derivation 2).
Electric Field on the Equatorial Line of a Dipole
For a dipole consisting of charges \( -q \) and \( +q \),
the resultant electric field at point \( P \) on the perpendicular bisector is:
\[
E = 2 E_A \cos\theta
\]
Using:
\[
E_A = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r^2 + a^2)}
\]
\[
\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}
\]
we get:
\[
E = \frac{1}{4\pi\varepsilon_0} \frac{q(2a)}{(r^2 + a^2)^{3/2}}
\]
In terms of the dipole moment \( P \):
\[
E = \frac{1}{4\pi\varepsilon_0} \frac{P}{(r^2 + a^2)^{3/2}}
\]
For \( a \ll r \), we approximate:
\[
\mathbf{E} = -\frac{1}{4\pi\varepsilon_0} \frac{\mathbf{P}}{r^3}
\]
Derivation 3).
Electric Dipole in a Uniform Electric Field
Consider an electric dipole consisting of charges \( -q \) and \( +q \) and of length \( 2a \), placed in a uniform electric field \( \mathbf{E} \) making an angle \( \theta \) with the direction of the field.
Force on charge \( -q \) at \( A \):
\[
\mathbf{F_A} = -q\mathbf{E}
\]
Force on charge \( +q \) at \( B \):
\[
\mathbf{F_B} = q\mathbf{E}
\]
Since both forces are equal and opposite, they produce a torque:
\[
\tau = \text{force} \times \text{perpendicular distance}
\]
\[
\tau = qE (2a \sin\theta) = (q(2a)) E \sin\theta
\]
Since \( P = q(2a) \), we get:
\[
\tau = P E \sin\theta
\]
In vector form:
\[
\boldsymbol{\tau} = \mathbf{P} \times \mathbf{E}
\]
Derivation 4.
Electric Field Due to a Line Charge
Consider an infinitely long thin line charge having uniform linear charge density \( \lambda \) placed along the \( YY' \) axis. The Gaussian surface is a cylinder with radius \( r \) and length \( l \).
From Gauss’s theorem:
\[
\oint \mathbf{E} \cdot d\mathbf{S} = \frac{q}{\varepsilon_0}
\]
Since the flux passes through the curved surface:
\[
\Phi = E \cdot 2\pi r l
\]
Charge enclosed by the Gaussian surface:
\[
q = \lambda l
\]
From Gauss’s law:
\[
E \cdot 2\pi r l = \frac{\lambda l}{\varepsilon_0}
\]
Simplifying:
\[
E = \frac{\lambda}{2\pi\varepsilon_0 r}
\]
Derivattion 5.
Electric Field Due to an Infinite Charged Plane Sheet
Consider an infinite thin plane sheet of charge having uniform surface charge density \( \sigma \). The Gaussian surface is a cylinder with area \( A \).
Applying Gauss’s law:
\[
\Phi = E \cdot 2A
\]
Charge enclosed:
\[
q = \sigma A
\]
Using Gauss’s theorem:
\[
E \cdot 2A = \frac{\sigma A}{\varepsilon_0}
\]
Solving for \( E \):
\[
E = \frac{\sigma}{2\varepsilon_0}
\]
This shows that the electric field due to an infinite plane sheet is independent of distance from the sheet.
Great! Here’s the next set of derivations in MathJax format:
Derivation 6).
Electric Field Due to a Charged Spherical Shell
Consider a thin spherical shell of radius \( R \) with charge \( +q \).
The Gaussian surface is a sphere in three cases:
(a) When the Point Lies Outside the Shell (\( r > R \))
By Gauss’s law:
\[
\oint \mathbf{E} \cdot d\mathbf{S} = \frac{q}{\varepsilon_0}
\]
Since the Gaussian surface is spherical:
\[
E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}
\]
Solving for \( E \):
\[
E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}
\]
For a medium with dielectric constant \( K \):
\[
E = \frac{1}{4\pi\varepsilon_0 K} \frac{q}{r^2}
\]
(b) When the Point Lies on the Surface (\( r = R \))
Applying Gauss’s law:
\[
E = \frac{1}{4\pi\varepsilon_0} \frac{q}{R^2}
\]
(c) When the Point Lies Inside the Shell (\( r < R \))
Since there is no charge enclosed within the Gaussian surface:
\[
E = 0
\]
Thus, the electric field inside a charged spherical shell is always zero.
