CBSE · Class 12 · Physics · Electric Charges and FieldsWhat is an electric dipole? Define electric dipole moment and state its SI unit and dimensional formula. Derive an expression for the electric field intensity at a point situated on the axial line (end-on position) of an electric dipole.
1. Definitions and Fundamental Concepts
- Electric Dipole: A system consisting of two equal and opposite point charges separated by a small distance is called an electric dipole. Example: Molecules like $\text{HCl}$, $\text{H}_2\text{O}$, $\text{NH}_3$.
- Electric Dipole Moment ($\vec{p}$): The product of the magnitude of either charge ($q$) and the distance between them ($2l$) is called the electric dipole moment.
$$\vec{p} = q \cdot (2\vec{l})$$
- Nature: Vector quantity.
- Direction: Directed from the negative charge ($-q$) to the positive charge ($+q$) along the axis.
- SI Unit: Coulomb-metre ($\text{C}\cdot\text{m}$).
- Dimensional Formula: $[\text{M}^0 \text{L}^1 \text{T}^1 \text{A}^1]$.
2. Derivation of Electric Field Intensity on the Axial Line
A. System Setup
- Let $AB$ be an electric dipole consisting of two point charges $-q$ at point $A$ and $+q$ at point $B$, separated by distance $2l$.
- Let $O$ be the midpoint of the dipole.
- Let $P$ be a point on the axial line at a distance $r$ from the midpoint $O$.
- Therefore, distance $AP = r + l$ and distance $BP = r - l$.
B. Electric Field Intensity due to individual charges
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Electric field at point $P$ due to charge $+q$ (at $B$): $$E_1 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{(BP)^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{(r - l)^2} \quad \text{(directed along } \vec{BP}\text{, away from dipole)}$$
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Electric field at point $P$ due to charge $-q$ (at $A$): $$E_2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{(AP)^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{(r + l)^2} \quad \text{(directed along } \vec{PA}\text{, towards dipole)}$$
C. Net Electric Field Intensity ($E$)\nSince $E_1$ and $E_2$ act along the same line in opposite directions and $E_1 > E_2$ (as $r - l < r + l$):
$$E = E_1 - E_2$$ $$E = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{(r - l)^2} - \frac{q}{(r + l)^2} \right]$$ $$E = \frac{q}{4\pi \varepsilon_0} \left[ \frac{(r + l)^2 - (r - l)^2}{(r - l)^2 (r + l)^2} \right]$$ \nUsing algebra $(r+l)^2 - (r-l)^2 = 4rl$: $$E = \frac{q}{4\pi \varepsilon_0} \left[ \frac{4rl}{(r^2 - l^2)^2} \right]$$ $$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{(q \cdot 2l) \cdot 2r}{(r^2 - l^2)^2}$$ \nSince $p = q \cdot 2l$ (Electric Dipole Moment): $$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2pr}{(r^2 - l^2)^2}$$
D. Special Case (Short Dipole: $r \gg l$)\nIf the dipole is very short compared to distance $r$ ($l^2 \ll r^2$), then $l^2$ can be neglected in comparison to $r^2$:
$$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2pr}{(r^2)^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2pr}{r^4}$$ $$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2p}{r^3}$$