MP Board · Class 12 · Physics · Electric Charges and FieldsDefine electric dipole and electric dipole moment. Derive an expression for the electric field intensity at any point on the axial line of an electric dipole.

Definition of Electric Dipole and Dipole Moment

• Electric Dipole: An electric dipole is a system of two equal and opposite point charges separated by a very small distance. Examples include molecules like $\text{HCl}$, $\text{H}_2\text{O}$, etc.
• Electric Dipole Moment: It is defined as the product of the magnitude of either charge and the distance between the two charges. It is denoted by $\vec{p}$: $$\vec{p} = q \cdot 2\vec{a}$$ where $2\vec{a}$ is the vector representing the separation distance from $-q$ to $+q$. Its SI unit is Coulomb-meter ($"),\text{C}\cdot\text{m}$). It is a vector quantity directed from the negative charge to the positive charge.

Derivation of Electric Field Intensity on the Axial Line

1. Setup and Geometry:

   • Consider an electric dipole consisting of charges $-q$ and $+q$ separated by a distance $2a$.
   • Let $O$ be the center of the dipole. The distance between $-q$ and $O$ is $a$, and between $+q$ and $O$ is $a$.
   • Let $P$ be a point on the axial line (extended line of the dipole) at a distance $r$ from the center $O$ ($r > a$).

2. Electric Field due to Individual Charges:

   • Distance from $+q$ to $P$: $r - a$

   • Distance from $-q$ to $P$: $r + a$

   • The electric field at point $P$ due to the positive charge $+q$ is given by: $$\vec{E}_1 = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r - a)^2} \hat{p}$| (directed away from $+q$, i.e., outwards along the axis)

   • The electric field at point $P$ due to the negative charge $-q$ is given by: $$\vec{E}_2 = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r + a)^2} \hat{p}$| (directed towards $-q$, i.e., inwards along the axis)

3. Net Electric Field: Since $\vec{E}_1$ and $\vec{E}_2$ act along the same straight line in opposite directions, the magnitude of the net electric field $\vec{E}$ at point $P$ is: $$E = E_1 - E_2$$ $$E = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right]$|

4. Simplification: Taking a common denominator: $$E = \frac{q}{4\pi \varepsilon_0} \left[ \frac{(r + a)^2 - (r - a)^2}{(r^2 - a^2)^2} \right]$| Expanding the numerator: $(r+a)^2 - (r-a)^2 = (r^2 + 2ra + a^2) - (r^2 - 2ra + a^2) = 4ra$. $$\therefore E = \frac{q}{4\pi \varepsilon_0} \frac{4ra}{(r^2 - a^2)^2}$|

   Rearranging the terms by substituting $p = q \cdot 2a$: $$E = \frac{1}{4\pi \varepsilon_0} \frac{2 \cdot (q \cdot 2a) \cdot r}{(r^2 - a^2)^2}$| $$E = \frac{1}{4\pi \varepsilon_0} \frac{2pr}{(r^2 - a^2)^2}$|

5. Special Case ($r \gg a$): If the point $P$ is very far away compared to the dipole length ($r \ge a$), then $a^2$ can be neglected in comparison to $r^2$. $$E \approx \frac{1}{4\pi \varepsilon_0} \frac{2pr}{(r^2)^2} = \frac{1}{4\pi \varepsilon_0} \frac{2pr}{r^4}$| $$E = \frac{1}{4\pi \varepsilon_0} \frac{2p}{r^3}$|

   In vector form: $$\vec{E} = \frac{1}{4\pi \varepsilon_0} \frac{2\vec{p}}{r^3}$$