Field Intensity on Axial Line of Dipole

1. Concept Overview

The electric field intensity on the axial line of an electric dipole is the net electric field produced by the two equal and opposite charges of the dipole at any point lying along its axis.
It is a strong field region and decreases with distance as [\dfrac{1}{r^3}] for distant points.

---

2. Clear Explanation and Mathematical Derivation

Consider a dipole consisting of:

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

Let point (P) lie on the axial line of the dipole at a distance [r] from the center.

Electric Field at P

Distance from charges to point P:

[r_+ = r – a],[\qquad] [r_- = r + a]

Electric field due to:

• Positive charge (+q) at distance [r – a]:
  [E_+] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2}]
• Negative charge (-q) at distance [r + a]:
  [E_-] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2}]

Both fields are in the same direction (from +q to -q).

Net Electric Field on Axial Line

[E_{\text{axial}}] $= \dfrac{1}{4\pi\varepsilon_0}\left[ \dfrac{q}{(r-a)^2} – \dfrac{q}{(r+a)^2} \right]$

Simplifying:
[E_{\text{axial}}] [= \dfrac{1}{4\pi\varepsilon_0}\cdot \dfrac{4qar}{(r^2-a^2)^2}]

Substitute (p = 2aq):
[E_{\text{axial}}] [= \dfrac{1}{4\pi\varepsilon_0}\cdot \dfrac{2pr}{(r^2-a^2)^2}]

At Large Distance (r >> a)

If the observation point is far:
[E_{\text{axial}}] [\approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}]

This is the commonly used form.

---

3. Dimensions and Units

---

4. Key Features

• Field is strongest along the axial line.
• Decreases as [\dfrac{1}{r^3}] for distant points.
• Direction of field: from +q to -q along the dipole axis.
• Both charges contribute fields in the same direction on the axial line.
• Depends on dipole moment (p).
• The axial field is twice the equatorial field in magnitude (at far distances).
• The field is non-uniform.

---

5. Important Formulas to Remember

---

6. Conceptual Questions with Solutions

---

7. FAQ / Common Misconceptions

---

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

---

Q1.

A dipole has charges [q = 5\ \text{μC}] separated by [2a = 6\ \text{cm}].
Find the axial field at a point 20 cm from center.
[\varepsilon_0] [= 8.85\times10^{-12}\ \text{F/m}]

Solution:

1. Dipole moment:
   [p = q(2a)] [= 5\times10^{-6} \times 0.06] [= 3\times10^{-7}\ \text{C·m}]
2. Far-field approximation (r >> a):
   [E] [= \dfrac{1}{4\pi\varepsilon_0}\cdot \dfrac{2p}{r^3}]
3. Substitute:
   [E] [= 9\times10^9 \cdot\dfrac{2(3\times10^{-7})}{(0.20)^3}]
   [E] [= 9\times10^9 \cdot \dfrac{6\times10^{-7}}{0.008}]
   [E] [= 9\times10^9 \cdot 7.5\times10^{-5}]
   [E] [= 6.75\times10^{5}\ \text{N/C}]

Answer:
[
E = 6.75\times10^{5}\ \text{N/C}
]

---

Q2.

For a dipole having [p = 4\times10^{-8}\ \text{C·m}], find the axial field at 10 cm.

Solution:
[E] [= \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2p}{r^3}]
[E] [= 9\times10^9 \cdot \dfrac{2(4\times10^{-8})}{(0.10)^3}]
[E] [= 9\times10^9 \cdot \dfrac{8\times10^{-8}}{10^{-3}}]
[E] [= 9\times10^9 \cdot 8\times10^{-5}]
[E] [= 7.2\times10^{5}\ \text{N/C}]

Answer:
[
E = 7.2\times10^{5}\ \text{N/C}
]

---

Q3.

A dipole has [p = 6\times10^{-9}\ \text{C·m}]. Find the ratio of axial field at distances 5 cm and 10 cm.

Solution:
[
E \propto \dfrac{1}{r^3}
]
[\dfrac{E_1}{E_2}] $= \left(\dfrac{r_2}{r_1}\right)^3$
$= \left(\dfrac{0.10}{0.05}\right)^3 = 2^3 = 8$

Answer:
[
E_{5\text{cm}} = 8E_{10\text{cm}}
]

---

Q4.

Find axial field at 50 cm for dipole moment [p = 1\times10^{-6}\ \text{C·m}].

Solution:
[E] [= 9\times10^9 \cdot \dfrac{2(10^{-6})}{(0.50)^3}]
[= 9\times10^9 \cdot \dfrac{2\times10^{-6}}{0.125}]
[= 9\times10^9 \cdot 1.6\times10^{-5}]
[= 1.44\times10^{5}\ \text{N/C}]

---

Q5.

A point is at 30 cm on axial line. Dipole moment: [p = 3\times10^{-7}\ \text{C·m}].
Find axial field.

Solution:
[E] [= 9\times10^9 \cdot \dfrac{2(3\times10^{-7})}{(0.30)^3}]
[= 9\times10^9 \cdot \dfrac{6\times10^{-7}}{0.027}]
[= 9\times10^9 \cdot 2.22\times10^{-5}]
[= 2\times10^{5}\ \text{N/C}]