Potential due to an Electric Dipole

Introduction

In the previous section, we saw that the potential of a single point charge falls off as $1/r$. An electric dipole consists of two equal and opposite charges ($+q$ and $-q$) separated by a small distance $2a$. Because the charges are opposite, their potentials tend to cancel out, leading to a potential that decreases much faster with distance than that of a single charge.

Detailed Derivation: Potential at a General Point

1. The Setup:

Consider an electric dipole consisting of charges $-q$ and $+q$ placed at points $A$ and $B$ respectively. The center of the dipole is at the origin $O$. The dipole moment $\vec{p}$ has magnitude $p = q \times 2a$ and is directed from $-q$ to $+q$.

We wish to find the potential $V$ at a point $P$ defined by the position vector $\vec{r}$ relative to the center $O$. Let the angle between $\vec{r}$ and the dipole moment $\vec{p}$ be $\theta$.

2. Superposition Principle:

The potential at $P$ is the algebraic sum of the potentials due to individual charges $+q$ and $-q$: $$V = V_{+q} + V_{-q} = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_1} - \frac{q}{r_2} \right)$$ $$V = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \quad \text{--- (Eq. 1)}$$ Where $r_1$ is the distance from $+q$ to $P$, and $r_2$ is the distance from $-q$ to $P$.

3. Geometrical Approximation (for $r \gg a$):

Using the law of cosines in triangles $OPB$ and $OPA$: $$r_1^2 = r^2 + a^2 - 2ra \cos \theta$$ $$r_2^2 = r^2 + a^2 + 2ra \cos \theta$$

Since $r \gg a$, we can factor out $r^2$ and ignore the $(a/r)^2$ term: $$r_1^2 = r^2 \left( 1 + \frac{a^2}{r^2} - \frac{2a}{r} \cos \theta \right) \approx r^2 \left( 1 - \frac{2a}{r} \cos \theta \right)$$ Taking the square root and inverting (using Binomial expansion $(1-x)^{-1/2} \approx 1 + x/2$): $$\frac{1}{r_1} \approx \frac{1}{r} \left( 1 - \frac{2a}{r} \cos \theta \right)^{-1/2} \approx \frac{1}{r} \left( 1 + \frac{a}{r} \cos \theta \right)$$ Similarly: $$\frac{1}{r_2} \approx \frac{1}{r} \left( 1 - \frac{a}{r} \cos \theta \right)$$

4. Final Potential Calculation:

Substitute these into Eq. 1: $$V = \frac{q}{4\pi\epsilon_0 r} \left[ \left( 1 + \frac{a}{r} \cos \theta \right) - \left( 1 - \frac{a}{r} \cos \theta \right) \right]$$ $$V = \frac{q}{4\pi\epsilon_0 r} \left( \frac{2a \cos \theta}{r} \right)$$ $$V = \frac{1}{4\pi\epsilon_0} \frac{(q \times 2a) \cos \theta}{r^2}$$ Since $p = q \times 2a$, we get: $$V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2}$$

Special Cases

Case 1: At a point on the Axial Line

For points on the axis near the positive charge, $\theta = 0^{\circ}$, so $\cos 0^{\circ} = 1$. $$V_{axial} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$$ For points near the negative charge, $\theta = 180^{\circ}$, so $V_{axial} = -\frac{k p}{r^2}$.

Case 2: At a point on the Equatorial Line

For points on the perpendicular bisector, $\theta = 90^{\circ}$, so $\cos 90^{\circ} = 0$. $$V_{equatorial} = 0$$ Critical Fact: No work is done in moving a charge along the equatorial plane of a dipole because the potential is zero everywhere on that plane.

Dipole Potential vs. Point Charge Potential

Feature	Point Charge	Electric Dipole
Symmetry	Spherical symmetry	Axial symmetry about $\vec{p}$
Distance Dependence	$V \propto 1/r$	$V \propto 1/r^2$
Angular Dependence	Independent of $\theta$	Depends on $\theta$ (angle with $\vec{p}$)
Potential Sign	Depends on sign of charge	Depends on $\theta$

Memory Capsule

• The Master Formula: $V = \frac{kp \cos \theta}{r^2}$.
• Power of r: For a dipole, potential drops much faster ($1/r^2$) than for a point charge ($1/r$).
• The Zero Zone: The equatorial plane is an equipotential surface with $V = 0$.
• Vector Relation: $V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}$.

Example 1: Potential on the Axial Line

A dipole has a dipole moment of $4 \times 10^{-9}$ C m. Calculate the potential at a point 0.3 m away from the center on the axial line near the positive charge.

Solution:

1. Formula for axial point: $$V = \frac{k p}{r^2}$$
2. Given: $$p = 4 \times 10^{-9}\ \text{C m}, \quad r = 0.3\ \text{m}, \quad k = 9 \times 10^9$$
3. Substitute: $$V = \frac{(9 \times 10^9)(4 \times 10^{-9})}{(0.3)^2}$$
4. Calculate: $$V = \frac{36}{0.09} = 400\ \text{V}$$
5. Final Answer: $$V = 400\ \text{V}$$

Example 2: Potential at a General Point

Find the potential at a point 1 m away from the center of a dipole of dipole moment $2 \times 10^{-12}$ C m, if the point makes an angle of $60^{\circ}$ with the dipole axis.

Solution:

1. General formula: $$V = \frac{k p \cos\theta}{r^2}$$
2. Given: $$p = 2 \times 10^{-12}\ \text{C m}, \quad r = 1\ \text{m}, \quad \theta = 60^{\circ}$$
3. Since $$\cos 60^{\circ} = \frac{1}{2}$$
4. Substitute: $$V = \frac{(9 \times 10^9)(2 \times 10^{-12})(1/2)}{1^2}$$
5. Calculate: $$V = 9 \times 10^{-3}\ \text{V}$$
6. Final Answer: $$V = 9\ \text{mV}$$

Example 3: Potential on the Equatorial Line

What is the potential at a point on the equatorial line of a dipole at a distance 0.5 m from the center?

Solution:

1. On the equatorial line, $$\theta = 90^{\circ}$$
2. Therefore, $$\cos 90^{\circ} = 0$$
3. Using the general formula, $$V = \frac{k p \cos\theta}{r^2} = \frac{k p \cdot 0}{r^2} = 0$$
4. Final Answer: $$V = 0$$

Concept Check: The result is independent of the actual value of $p$ and $r$ as long as the point lies on the equatorial line.

Example 4: Potential on the Negative Axial Side

A dipole has dipole moment $6 \times 10^{-8}$ C m. Find the potential at a point 0.6 m from the center on the axial line near the negative charge.

Solution:

1. On the axial line near the negative charge, $$\theta = 180^{\circ}$$
2. Hence, $$\cos 180^{\circ} = -1$$
3. Formula becomes: $$V = -\frac{k p}{r^2}$$
4. Substitute values: $$V = -\frac{(9 \times 10^9)(6 \times 10^{-8})}{(0.6)^2}$$
5. Calculate: $$V = -\frac{540}{0.36} = -1500\ \text{V}$$
6. Final Answer: $$V = -1.5 \times 10^3\ \text{V}$$

Example 5: Finding Dipole Moment from Potential

At a point on the axial line 0.2 m from the center, the potential due to a dipole is 1800 V. Find the dipole moment.

Solution:

1. For an axial point, $$V = \frac{k p}{r^2}$$
2. Rearranging: $$p = \frac{V r^2}{k}$$
3. Substitute the values: $$p = \frac{1800 \times (0.2)^2}{9 \times 10^9}$$
4. Calculate: $$p = \frac{1800 \times 0.04}{9 \times 10^9} = \frac{72}{9 \times 10^9} = 8 \times 10^{-9}\ \text{C m}$$
5. Final Answer: $$p = 8 \times 10^{-9}\ \text{C m}$$

Example 6: Finding Distance from Given Potential

The potential on the axial line of a dipole of moment $9 \times 10^{-9}$ C m is 900 V. Find the distance of the point from the center.

Solution:

1. Formula on axial line: $$V = \frac{k p}{r^2}$$
2. Rearranging for distance: $$r^2 = \frac{k p}{V}$$ $$r = \sqrt{\frac{k p}{V}}$$
3. Substitute values: $$r = \sqrt{\frac{(9 \times 10^9)(9 \times 10^{-9})}{900}}$$
4. Calculate inside the square root: $$r = \sqrt{\frac{81}{900}} = \sqrt{0.09}$$
5. Therefore, $$r = 0.3\ \text{m}$$
6. Final Answer: $$r = 0.3\ \text{m}$$

Example 7: Comparing Potentials at Two Angles

For a fixed dipole moment and fixed distance, compare the potentials at $\theta = 60^{\circ}$ and $\theta = 120^{\circ}$.

Solution:

1. General relation: $$V = \frac{k p \cos\theta}{r^2}$$
2. At $60^{\circ}$: $$V_1 = \frac{k p \cos 60^{\circ}}{r^2} = \frac{k p}{2r^2}$$
3. At $120^{\circ}$: $$V_2 = \frac{k p \cos 120^{\circ}}{r^2} = \frac{k p(-1/2)}{r^2} = -\frac{k p}{2r^2}$$
4. Hence, $$V_1 = -V_2$$
5. Final Answer: The two potentials are equal in magnitude but opposite in sign.

Example 8: Potential Becomes One-Fourth

At a distance $r$ from a dipole, the potential at a certain angle is $V$. At what distance will the potential become $V/4$ if the angle remains the same?

Solution:

1. For fixed $p$ and fixed $\theta$, $$V \propto \frac{1}{r^2}$$
2. Therefore, $$\frac{V_1}{V_2} = \frac{r_2^2}{r_1^2}$$
3. Given: $$V_2 = \frac{V_1}{4}$$ so, $$\frac{V_1}{V_1/4} = \frac{r_2^2}{r_1^2}$$
4. Hence, $$4 = \frac{r_2^2}{r_1^2}$$ $$r_2 = 2r_1$$
5. Final Answer: The distance must be doubled.

Example 9: Work Done Along the Equatorial Line

How much work is done in moving a $5 \mu C$ charge from one point to another on the equatorial line of a dipole?

Solution:

1. On the equatorial line of a dipole, $$V = 0$$ at every point.
2. Therefore the potential difference between any two such points is: $$\Delta V = 0$$
3. Work done is: $$W = q\Delta V$$
4. Substitute: $$W = (5 \times 10^{-6})(0) = 0$$
5. Final Answer: $$W = 0$$

Reason: No work is needed because the entire equatorial line is an equipotential region.

Example 10: Potential at the Center of a Dipole

Find the potential at the center of an electric dipole whose charges are $+q$ and $-q$.

Solution:

1. The center is equidistant from both charges.
2. Let the distance of each charge from the center be $a$.
3. Potential due to $+q$ at the center: $$V_{+} = \frac{kq}{a}$$
4. Potential due to $-q$ at the center: $$V_{-} = -\frac{kq}{a}$$
5. Net potential: $$V = V_{+} + V_{-} = \frac{kq}{a} - \frac{kq}{a} = 0$$
6. Final Answer: $$V = 0$$

Example 11: Finding Angle for Half the Axial Potential

At a fixed distance from a dipole, the potential is half of the axial potential on the positive side. Find the angle $\theta$.

Solution:

1. Potential at a general point: $$V = \frac{k p \cos\theta}{r^2}$$
2. Axial potential on positive side: $$V_{axial} = \frac{k p}{r^2}$$
3. Given: $$V = \frac{1}{2}V_{axial}$$
4. Therefore, $$\frac{k p \cos\theta}{r^2} = \frac{1}{2}\cdot\frac{k p}{r^2}$$
5. Cancel common terms: $$\cos\theta = \frac{1}{2}$$
6. Hence, $$\theta = 60^{\circ}$$
7. Final Answer: $$\theta = 60^{\circ}$$

Example 12: Ratio of Potentials at Two Distances

A point lies on the axial line of a dipole. If its distances from the center are 2 m and 4 m in two cases, find the ratio of potentials.

Solution:

1. On the axial line, $$V \propto \frac{1}{r^2}$$
2. Therefore, $$\frac{V_1}{V_2} = \frac{r_2^2}{r_1^2}$$
3. Substitute: $$\frac{V_1}{V_2} = \frac{4^2}{2^2} = \frac{16}{4} = 4$$
4. Final Answer: $$V_1 : V_2 = 4 : 1$$