Electric Charges and Fields — Part 2: Dipoles, flux and Gauss’s law: visual lesson, CBSE Class 12 Physics

Lesson

Dipoles, flux and Gauss’s law

Part 2 — a pair of opposite charges, charge spread over lines and surfaces, and a powerful law that turns hard integrals into one-line results.

In this lesson you will

Find the field of a dipole on its axis and on its equatorial line
Calculate the torque on a dipole in a uniform field
Describe continuous charge distributions with λ, σ and ρ
Define electric flux, state Gauss’s law and use it to find the field of a wire, a sheet and a shell

Concept

The electric dipole

Two equal and opposite charges $\pm q$ separated by $2a$ form an electric dipole. Its dipole moment is $\vec p = q \times 2a\,\hat p, \qquad \text{unit C m},$ directed from $-q$ to $+q$ . The total charge is zero, but the field is not.

Polar molecules (H₂O, HCl) have permanent dipole moments.
Non-polar molecules get an induced dipole moment in an external field.
A point dipole: $2a \to 0$ , $q \to \infty$ with $p$ fixed.

Compare

Field of a dipole: axis and equatorial line

Let the point be at distance $r$ from the centre of the dipole.

On the axis (P) $E_\text{axis} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{2pr}{(r^2 - a^2)^2} \approx \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{2p}{r^3}$ for $r \gg a$ ; direction along $\vec p$ .

The nearer charge ( $+q$ for P beyond $+q$ ) gives the larger field.

On the equatorial line (Q) $E_\text{eq} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{p}{(r^2 + a^2)^{3/2}} \approx \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{p}{r^3}$ for $r \gg a$ ; direction opposite to $\vec p$ .

The components along the bisector cancel; those parallel to the axis add.

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