Field Intensity on Equatorial Line of a Dipole

1. Statement of the Law / Concept Overview

The electric field intensity on the equatorial line of a dipole is the net electric field at any point lying on the perpendicular bisector of the dipole.
On this line, the fields due to +q and −q act in opposite directions, leading to partial cancellation.
Thus, the equatorial field is always opposite in direction to the dipole moment and decreases with distance as [\dfrac{1}{r^3}] (far field).

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2. Clear Explanation and Mathematical Derivation

Consider a dipole with:

• Charges: [+q] and [-q]
• Separation: [2a]
• Dipole moment: [\vec{p} = q(2a)\hat{l}]

Let point P lie on the equatorial line, at a distance [r] from the center.

Distances

The point P is equidistant from both charges:

[r_+ = r_-] [= \sqrt{a^2 + r^2}]

Electric Field Due to Each Charge

Electric field magnitude (same for both charges):

[E] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{a^2 + r^2}]

Each field is not collinear with the dipole moment. We break it into components:

• Horizontal components cancel.
• Vertical components add.

Vertical component:

[E_{\text{component}}] [= E\cos\theta]

Where:

[\cos\theta] [= \dfrac{a}{\sqrt{a^2+r^2}}]

Net Electric Field on Equatorial Line

[E_{\text{equatorial}}] [= 2E\cos\theta]

Substitute values:

[
E_{\text{equatorial}}
= 2 \cdot \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{a^2+r^2}
\cdot\dfrac{a}{\sqrt{a^2+r^2}}
]

[
E_{\text{equatorial}}
= \dfrac{1}{4\pi\varepsilon_0}
\cdot\dfrac{2qa}{(a^2+r^2)^{3/2}}
]

Using dipole moment [p = 2aq]:

[
E_{\text{equatorial}}
= \dfrac{1}{4\pi\varepsilon_0}
\cdot\dfrac{p}{(a^2+r^2)^{3/2}}
]

Direction

The field is opposite to dipole moment (\vec{p}).

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Far Field Approximation (when [r >> a])

[E_{\text{equatorial}}] [\approx \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{p}{r^3}]

Direction: opposite to [\vec{p}].

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3. Dimensions and Units

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4. Key Features

• Magnitude is half of axial field at far distances:
  [E_{\text{axial}}] [= 2E_{\text{equatorial}}]
• Field direction is opposite to dipole moment.
• Field decreases as [\dfrac{1}{r^3}].
• Horizontal components cancel; vertical components add.
• Works for any dipole in free space.
• Field is non-uniform.

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5. Important Formulas to Remember

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6. Conceptual Questions with Solutions (15)

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7. FAQ / Common Misconceptions (10)

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8. Practice Questions (With Step-by-Step Solutions)

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Q1.

Dipole moment [p = 2\times10^{-7}\ \text{C·m}]. Find equatorial field at [r = 20\ \text{cm}].

Solution:

[E] [= \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{p}{r^3}]
[E] [= 9\times10^9 \cdot \dfrac{2\times10^{-7}}{(0.20)^3}]
[= 9\times10^9 \cdot \dfrac{2\times10^{-7}}{0.008}]
[= 9\times10^9 \cdot 2.5\times10^{-5}]
[E] [= 2.25\times10^{5}\ \text{N/C}]

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Q2.

Find the equatorial field at a distance 10 cm for [p = 1\times10^{-8}\ \text{C·m}].

Solution:

[E] [= 9\times10^9 \cdot\dfrac{1\times10^{-8}}{(0.10)^3}]
[= 9\times10^9 \cdot\dfrac{10^{-8}}{10^{-3}}]
[
= 9\times10^9 \cdot10^{-5}
]
[
E = 9\times10^{4}\ \text{N/C}
]

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Q3.

For a dipole, compare equatorial fields at 5 cm and 10 cm.

Solution:

[
E\propto\dfrac{1}{r^3}
]
[\dfrac{E_1}{E_2}=\left(\dfrac{r_2}{r_1}\right)^3]
[
= \left(\dfrac{0.10}{0.05}\right)^3 = 8
]

Thus,
[
E_{5\text{cm}} = 8E_{10\text{cm}}
]

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Q4.

A dipole has [p = 5\times10^{-9}]. Find equatorial field at [r = 50\ \text{cm}].

Solution:

[E] [= 9\times10^9 \cdot\dfrac{5\times10^{-9}}{(0.50)^3}]
[
= 9\times10^9 \cdot\dfrac{5\times10^{-9}}{0.125}
]
[
= 9\times10^9 \cdot 4\times10^{-8}
]
[
= 3.6\times10^{2}\ \text{N/C}
]

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Q5.

For dipole with [p = 3\times10^{-7}], find equatorial field at 30 cm.

Solution:

[E] [= 9\times10^9 \cdot\dfrac{3\times10^{-7}}{(0.30)^3}]
[
= 9\times10^9 \cdot\dfrac{3\times10^{-7}}{0.027}
]
[
= 9\times10^9 \cdot 1.11\times10^{-5}
]
[
= 10^{5}\ \text{N/C (approx)}
]