Electric Dipole, Dipole Moment & Electric Field Due to a Dipole

(With FULL DERIVATIONS – Class 12 NCERT, One-Go Complete Notes)

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1️⃣ Electric Dipole (Revision)

An electric dipole consists of:

• Two equal and opposite charges $+q$and $-q$

• Separated by a small distance $2a$

📌 Condition:

$$a\ll r(\text{distance of observation point from dipole})$$

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2️⃣ Electric Dipole Moment (𝒑)

Definition:

Electric dipole moment is the product of one charge and separation vector.

$$\vec{p}=q\cdot 2\vec{a}$$

• SI unit: C m

• Vector quantity

• Direction: from –q to +q

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ELECTRIC FIELD DUE TO AN ELECTRIC DIPOLE (DERIVATIONS)

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🔵 Case I: Electric Field on the Axial Line

🔹 Geometry:

• Point $P$ lies on the axis of dipole

• Distance of point from centre $O$ = $r$

• Distance from +q = $r-a$

• Distance from –q = $r+a$

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🔹 Step 1: Electric field due to +q at point P

$$E_{+}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r-a{)}^2}$$

(Direction: away from +q)

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🔹 Step 2: Electric field due to –q at point P

$$E_{-}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r+a{)}^2}$$

(Direction: towards –q)

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🔹 Step 3: Net electric field

Both fields act along the axis, but in opposite directions.

$$E=E_{+}-E_{-}$$

$$E=\frac{1}{4\pi {\epsilon }_0}[\frac{q}{(r-a{)}^2}-\frac{q}{(r+a{)}^2}]$$

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🔹 Step 4: Simplification

$$E=\frac{1}{4\pi {\epsilon }_0}\frac{4qar}{(r^2-a^2{)}^2}$$

For short dipole $(r\gg a)$:

$$(r^2-a^2{)}^2\approx r^4$$

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✅ Final Result (Axial Line):

$$\boxed{{\displaystyle E_{\text{axial}}=\frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}}}$$

🔹 Direction:

👉 Along the dipole moment

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🟢 Case II: Electric Field on the Equatorial Line

🔹 Geometry:

• Point $P$ lies on the perpendicular bisector

• Distance of point from centre = $r$

• Distance from each charge:

$$\sqrt{r^2+a^2}$$

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🔹 Step 1: Field due to +q

$$E_{+}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2+a^2}$$

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🔹 Step 2: Field due to –q

$$E_{-}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2+a^2}$$

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🔹 Step 3: Resolution of Fields

• Horizontal components cancel

• Vertical components add

$$E=2E_{+}\cos \theta$$

$$\cos \theta =\frac{a}{\sqrt{r^2+a^2}}$$

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🔹 Step 4: Net Field

$$E=\frac{1}{4\pi {\epsilon }_0}\frac{2qa}{(r^2+a^2{)}^{3\mathrm{/}2}}$$

For short dipole $(r\gg a)$:

$$(r^2+a^2{)}^{3\mathrm{/}2}\approx r^3$$

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✅ Final Result (Equatorial Line):

$$\boxed{{\displaystyle E_{\text{equatorial}}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^3}}}$$

🔹 Direction:

👉 Opposite to dipole moment

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4️⃣ Axial vs Equatorial (Very Important Table 🔥)

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5️⃣ One-Line Exam Conclusions ⭐

• Electric field due to dipole decreases rapidly with distance.

• Axial field is twice the equatorial field.

• Dipole moment controls strength and direction.

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6️⃣ Memory Shortcut 🧠

• Axial → 2p

• Equatorial → p

• Both → $1\mathrm{/}{r}^3$