Class 12th Physics Chapter-2 Notes on Potential due to an Electric Dipole

Potential due to an Electric Dipole (Derivation)

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1. Physical setup

• An electric dipole consists of two equal and opposite charges:

  $$+q\text{and}-q$$

• Separation between the charges = $2a$

• Dipole moment:

  $$\vec{p}=q(2a)\widehat{p}$$

• The origin $O$ is taken at the centre of the dipole

• Point $P$ is at a distance $r$ from the centre

• Angle between $\vec{p}$ and $\vec{r}$ is $\theta$

• Distances of point $P$ from the charges:

  • From $+q$: $r_1$

  • From $-q$: $r_2$

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2. Potential due to a system of charges

Electric potential is a scalar and obeys the superposition principle.

So, potential at $P$ due to the dipole is:

$$V=\frac{1}{4\pi {\epsilon }_0}(\frac{q}{r_1}-\frac{q}{r_2})$$

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3. Geometrical relations

From geometry:

$$r_1^2=r^2+a^2-2ar\cos \theta$$
$$r_2^2=r^2+a^2+2ar\cos \theta$$

For a short dipole:

$$r\gg a$$

So we keep only first-order terms in ${\displaystyle \frac{a}{r}}$.

Hence,

$$r_1\approx r-a\cos \theta$$

$$r_2\approx r+a\cos \theta$$

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4. Apply binomial approximation

$$\frac{1}{r_1}=\frac{1}{r-a\cos \theta }=\frac{1}{r}{(1-\frac{a\cos \theta }{r})}^{-1}$$

Using

$$(1+x{)}^{-1}\approx 1-x(\mathrm{\mid }x\mathrm{\mid }\ll 1)$$

$$\frac{1}{r_1}\approx \frac{1}{r}(1+\frac{a\cos \theta }{r})$$

Similarly,

$$\frac{1}{r_2}\approx \frac{1}{r}(1-\frac{a\cos \theta }{r})$$

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5. Substitute into the potential expression

$$V=\frac{q}{4\pi {\epsilon }_0}[\frac{1}{r}(1+\frac{a\cos \theta }{r})-\frac{1}{r}(1-\frac{a\cos \theta }{r})]$$

Simplifying:

$$V=\frac{q}{4\pi {\epsilon }_0}\cdot \frac{2a\cos \theta }{r^2}$$

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6. Use dipole moment $p=2aq$

$$\boxed{{\displaystyle V=\frac{1}{4\pi {\epsilon }_0}\frac{p\cos \theta }{r^2}}}$$

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7. Vector form (most compact result )

Since $p\cos \theta =\vec{p}\cdot \widehat{r}$,

$$\boxed{{\displaystyle V(r,\theta )=\frac{1}{4\pi {\epsilon }_0}\frac{\vec{p}\cdot \widehat{r}}{r^2}(r\gg a)}}$$

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8. Special cases (often asked)

(a) Axial line ($\theta =0^{\circ }$)

$$V=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^2}$$

(b) Equatorial line ($\theta ={90}^{\circ }$)

$$V=0$$

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9. Important points to mention in exams 📝

• Assumption: $r\gg a$ (short dipole)

• Potential depends on distance and angle

• Potential of a dipole varies as $1\mathrm{/}{r}^2$ (not $1\mathrm{/}r$)

• Potential is zero on the equatorial plane