Question

Derive the expression for Electric Field Intensity on the axial and equatorial line of a dipole.

Hint:

Key dipole results to remember: - Axial field: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{2p}{r^3} \) - Equatorial field: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{p}{r^3} \) - Both decrease as \( 1/r^3 \), faster than a point charge field.

Answer 1

Concept: An electric dipole consists of two equal and opposite charges \( +q \) and \( -q \) separated by a small distance \( 2a \). Dipole moment: \[ \vec{p} = q \cdot (2a) \quad \text{directed from } -q \text{ to } +q \] The electric field due to a dipole is obtained using the principle of superposition by adding fields due to both charges.
Part 1: Electric Field on the Axial Line of a Dipole The axial line is the line passing through both charges. Let:

• Point \( P \) be at distance \( r \) from the center of dipole
• Charges located at \( -a \) and \( +a \)

Step 1: Electric field due to each charge Field due to \( +q \): \[ E_+ = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r-a)^2} \] Field due to \( -q \): \[ E_- = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r+a)^2} \] Both fields act along the same line but in opposite directions.
Step 2: Net electric field \[ E_{\text{axial}} = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{(r-a)^2} - \frac{q}{(r+a)^2} \right] \] Taking common denominator and simplifying: \[ E_{\text{axial}} = \frac{1}{4\pi \varepsilon_0} \frac{4qar}{(r^2 - a^2)^2} \] Using dipole moment \( p = 2aq \): \[ E_{\text{axial}} = \frac{1}{4\pi \varepsilon_0} \frac{2pr}{(r^2 - a^2)^2} \]
For a short dipole (\( r \gg a \)) Neglect \( a^2 \): \[ \boxed{E_{\text{axial}} = \frac{1}{4\pi \varepsilon_0} \frac{2p}{r^3}} \] Direction: Along dipole moment.
Part 2: Electric Field on the Equatorial Line of a Dipole The equatorial line is perpendicular to the dipole axis and passes through the center. Let point \( P \) be at distance \( r \) from the center.
Step 1: Distance from charges Distance from each charge: \[ \sqrt{r^2 + a^2} \] Field due to each charge: \[ E = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2 + a^2} \]
Step 2: Resolve components Horizontal components cancel due to symmetry. Vertical components add. Vertical component of each field: \[ E \cos \theta = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2 + a^2} \cdot \frac{a}{\sqrt{r^2 + a^2}} \]
Step 3: Net electric field \[ E_{\text{equatorial}} = 2 \times \frac{1}{4\pi \varepsilon_0} \frac{qa}{(r^2 + a^2)^{3/2}} \] Using \( p = 2aq \): \[ E_{\text{equatorial}} = \frac{1}{4\pi \varepsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \]
For a short dipole (\( r \gg a \)) \[ \boxed{E_{\text{equatorial}} = \frac{1}{4\pi \varepsilon_0} \frac{p}{r^3}} \] Direction: Opposite to dipole moment.
Comparison: Axial vs Equatorial Field

• Axial field is twice the equatorial field at same distance.
• Both vary inversely as \( r^3 \).

\[ E_{\text{axial}} = 2E_{\text{equatorial}} \]