Electric Dipoles and Fields

Electric Dipole ($ \vec{p} = q \cdot 2\vec{a} $)

An electric dipole is a system formed by two equal and opposite point charges, $+q$ and $-q$, separated by a small distance $2a$. The behaviour of an electric dipole is often described by its electric dipole moment.

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Definition of Electric Dipole Moment ($\vec{p}$)

The electric dipole moment ($\vec{p}$) is a vector quantity defined as the product of the magnitude of one of the charges ($q$) and the distance ($2a$) between the charges. The direction of the dipole moment vector is conventionally taken from the negative charge $(-q)$ to the positive charge ($+q$).

If the negative charge is at position $\vec{r}_1$ and the positive charge is at position $\vec{r}_2$, the vector $\vec{d} = \vec{r}_2 - \vec{r}_1$ is the displacement vector from $-q$ to $+q$, and its magnitude is $2a = |\vec{r}_2 - \vec{r}_1|$. The electric dipole moment is:

$ \vec{p} = q \cdot \vec{d} = q \cdot (\vec{r}_2 - \vec{r}_1) $

If the origin is placed at the midpoint of the dipole, with $-q$ at $-\vec{a}$ and $+q$ at $+\vec{a}$, where $2a = |2\vec{a}|$, then the dipole moment is $\vec{p} = q (+\vec{a} - (-\vec{a})) = q(2\vec{a})$. Using $2a$ for magnitude and defining $\vec{2a}$ as the vector from $-q$ to $+q$, the definition is simply $\vec{p} = q \cdot \vec{2a}$. The magnitude of the dipole moment is $p = q(2a)$.

The SI unit of electric dipole moment is Coulomb-metre (C·m).

Dipoles are important because many molecules (polar molecules like water, HCl) behave as electric dipoles due to uneven distribution of charge. Also, when an insulator is placed in an electric field, its molecules can become induced dipoles.

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The Field Of An Electric Dipole (Axial and Equatorial)

An electric dipole creates an electric field in the space around it. The electric field of a dipole is different from that of a single point charge and decreases faster with distance.

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Electric Field on the Axial Line

Consider an electric dipole with charges $+q$ at $(a, 0)$ and $-q$ at $(-a, 0)$ on the x-axis, so the dipole moment $\vec{p}$ is along the positive x-axis. We want to find the electric field at a point P on the x-axis (axial line) at a distance $r$ from the midpoint of the dipole (origin), where $r \gg a$. Let P be at $(r, 0)$.

(Image Placeholder: A dipole with -q at -a and +q at +a on the x-axis. The midpoint is origin O. Point P is on the x-axis at r from O, with r > a. Show field vector E+ from +q away from +q, and E- from -q away from -q (towards P if P > -a). Indicate resultant E field.)

The distance of P from $+q$ is $r_1 = r-a$. The distance of P from $-q$ is $r_2 = r+a$.

Electric field at P due to $+q$: $\vec{E}_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)^2} \hat{i}$ (direction is away from $+q$, along +x axis).

Electric field at P due to $-q$: $\vec{E}_- = \frac{1}{4\pi\epsilon_0} \frac{-q}{(r+a)^2} \hat{i}$ (direction is towards $-q$, along -x axis).

The net electric field at P is the vector sum $\vec{E} = \vec{E}_+ + \vec{E}_-$.

$ \vec{E} = \frac{q}{4\pi\epsilon_0} \left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right] \hat{i} $

$ \vec{E} = \frac{q}{4\pi\epsilon_0} \left[\frac{(r+a)^2 - (r-a)^2}{(r-a)^2 (r+a)^2}\right] \hat{i} $

$ (r+a)^2 - (r-a)^2 = (r^2 + 2ra + a^2) - (r^2 - 2ra + a^2) = 4ra $

$ (r-a)^2 (r+a)^2 = [(r-a)(r+a)]^2 = (r^2 - a^2)^2 $

$ \vec{E} = \frac{q}{4\pi\epsilon_0} \frac{4ra}{(r^2 - a^2)^2} \hat{i} $

$ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q(2a)(2r)}{(r^2 - a^2)^2} \hat{i} $

Using the magnitude of the dipole moment $p = q(2a)$:

$ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - a^2)^2} \hat{i} $

For points far from the dipole, $r \gg a$. In the denominator, $r^2 - a^2 \approx r^2$, so $(r^2 - a^2)^2 \approx (r^2)^2 = r^4$.

$ \vec{E}_{axial} \approx \frac{1}{4\pi\epsilon_0} \frac{2pr}{r^4} \hat{i} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \hat{i} $ (for $r \gg a$)

The electric field on the axial line of a dipole falls off as $1/r^3$ for large distances, which is faster than the $1/r^2$ falloff for a single point charge. The direction of the field is along the dipole moment vector ($\hat{i}$).

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Electric Field on the Equatorial Line (Perpendicular Bisector)

Consider a point P on the y-axis (equatorial line) at a distance $r$ from the midpoint of the dipole (origin). The charges are $+q$ at $(a, 0)$ and $-q$ at $(-a, 0)$. P is at $(0, r)$.

(Image Placeholder: A dipole with -q at -a and +q at +a on the x-axis. Midpoint origin O. Point P on the y-axis at distance r from O. Show field vector E+ from +q towards P, and E- from -q towards P. Show components and resultant E field.)

The distance of P from both $+q$ and $-q$ is $R = \sqrt{r^2 + a^2}$.

Magnitude of electric field at P due to $+q$: $E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2}$. The direction is away from $+q$, along the line from $+q$ to P.

Magnitude of electric field at P due to $-q$: $E_- = \frac{1}{4\pi\epsilon_0} \frac{|-q|}{R^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2}$. The direction is towards $-q$, along the line from P to $-q$.

Magnitudes are equal, $E_+ = E_-$. The resultant field $\vec{E}$ is the vector sum. By symmetry, the vertical components cancel out, and the horizontal components add up. The horizontal component of $\vec{E}_+$ is $E_+ \cos\theta$, where $\theta$ is the angle between the line joining $+q$ and P, and the negative x-axis. $\cos\theta = a/R = a/\sqrt{r^2 + a^2}$. The horizontal component of $\vec{E}_-$ is $E_- \cos\theta$, where $\theta$ is the angle between the line joining $-q$ and P, and the positive x-axis. $\cos\theta = a/R$. Both horizontal components are directed towards the negative x-axis.

$ E_{net, x} = E_+ \cos\theta + E_- \cos\theta = 2 E_+ \cos\theta $

$ E_{net, x} = 2 \left(\frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2}\right) \left(\frac{a}{\sqrt{r^2 + a^2}}\right) $

$ E_{net, x} = \frac{1}{4\pi\epsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} $

Using $p = q(2a)$:

$ E_{net, x} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} $

The direction is towards the negative x-axis, opposite to the direction of the dipole moment vector ($\vec{p}$ is along +x axis).

$ \vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \hat{i} $

For points far from the dipole, $r \gg a$. In the denominator, $r^2 + a^2 \approx r^2$, so $(r^2 + a^2)^{3/2} \approx (r^2)^{3/2} = r^3$.

$ \vec{E}_{equatorial} \approx -\frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \hat{i} $ (for $r \gg a$)

The electric field on the equatorial line also falls off as $1/r^3$ for large distances, but its magnitude is half that on the axial line at the same large distance, and its direction is opposite to the dipole moment vector.

In general, the electric field of a dipole falls off as $1/r^3$ for large distances in all directions, which is a faster decay than the $1/r^2$ field of a monopole (single charge). This is because a dipole is electrically neutral overall, and its field arises from the slight separation of positive and negative charges, whose effects partially cancel at large distances.

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Physical Significance Of Dipoles

Electric dipoles and the concept of dipole moment are important in physics and chemistry for several reasons:

• Molecular Properties: Many molecules, like water ($H_2O$), ammonia ($NH_3$), and hydrogen chloride (HCl), have a permanent electric dipole moment due to the uneven distribution of charge resulting from the electronegativity differences between atoms and the molecular geometry. These are called polar molecules. The dipole moment influences the physical and chemical properties of these substances, such as their boiling points, solubility, and behaviour in electric fields. Non-polar molecules (like $O_2, N_2, CO_2$ - linear) have zero net dipole moment due to symmetry.
• Behavior of Materials in Electric Fields: When an electric field is applied to a material, especially an insulator (dielectric), it causes a rearrangement of charges.
  • In polar dielectrics, the permanent molecular dipoles tend to align with the external electric field.
  • In non-polar dielectrics, the electric field induces a separation of charge within each molecule, creating induced dipoles that are aligned with the field.
  This phenomenon, called polarization of the dielectric material, reduces the net electric field inside the material. This is why the force between charges is reduced when placed in a dielectric medium ($\epsilon_r > 1$).
• Intermolecular Forces: Dipole moments contribute to intermolecular forces, such as dipole-dipole interactions between polar molecules or dipole-induced dipole interactions.
• Electromagnetic Waves: Oscillating electric dipoles radiate electromagnetic waves. For example, antennas transmit radio waves by creating oscillating electric dipoles.

The concept of an electric dipole provides a useful model for understanding the behavior of atoms and molecules in electric fields and the macroscopic properties of dielectric materials.

Dipole In A Uniform External Field (Torque $ \vec{\tau} = \vec{p} \times \vec{E} $)

When an electric dipole is placed in an external electric field, the positive and negative charges of the dipole experience forces. If the electric field is uniform, the net force on the dipole is zero, but it experiences a net torque that tends to align the dipole with the field.

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Force on a Dipole in a Uniform Field

Consider an electric dipole with charge $+q$ at position $\vec{r}_+$ and charge $-q$ at position $\vec{r}_-$, placed in a uniform external electric field $\vec{E}$.

The force on the positive charge is $\vec{F}_+ = q\vec{E}$.

The force on the negative charge is $\vec{F}_- = -q\vec{E}$.

The net force on the dipole is $\vec{F}_{net} = \vec{F}_+ + \vec{F}_- = q\vec{E} + (-q\vec{E}) = 0$.

Thus, a dipole in a uniform electric field experiences zero net force. Its centre of mass will not accelerate.

However, even though the net force is zero, if the forces $\vec{F}_+$ and $\vec{F}_-$ are not collinear, they form a couple, which exerts a torque on the dipole.

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Torque on a Dipole in a Uniform Field

Let the dipole be oriented such that the vector from $-q$ to $+q$ is $\vec{2a}$. The dipole moment is $\vec{p} = q(2\vec{a})$. Let the uniform electric field be $\vec{E}$. The force $+q\vec{E}$ acts on $+q$, and $-q\vec{E}$ acts on $-q$. The lines of action of these forces are parallel but generally separated.

The torque on the dipole can be calculated about any point. Let's calculate it about the position of the negative charge, $-q$. The force on $-q$ is at the origin, so its torque about the origin is 0. The torque is due to the force on $+q$ acting at position $\vec{r}_+ = \vec{2a}$ relative to $-q$.

$ \vec{\tau}_- = \vec{r}_- \times \vec{F}_- = \vec{0} \times (-q\vec{E}) = \vec{0} $ (if calculating about $-q$'s position)

$ \vec{\tau}_+ = \vec{r}_+ \times \vec{F}_+ = \vec{2a} \times (q\vec{E}) = q (\vec{2a} \times \vec{E}) $

Since $\vec{p} = q(2\vec{a})$, the net torque about the position of $-q$ is:

$ \vec{\tau} = \vec{\tau}_+ + \vec{\tau}_- = \vec{p} \times \vec{E} + \vec{0} = \vec{p} \times \vec{E} $

If we calculate the torque about the midpoint of the dipole (origin), where $+q$ is at $\vec{a}$ and $-q$ is at $-\vec{a}$: Torque on $+q$: $\vec{\tau}_+ = \vec{a} \times (q\vec{E})$ Torque on $-q$: $\vec{\tau}_- = (-\vec{a}) \times (-q\vec{E}) = (-\vec{a}) \times (q\vec{E}) \times (-1) = \vec{a} \times (q\vec{E})$ Net torque: $\vec{\tau} = \vec{\tau}_+ + \vec{\tau}_- = (\vec{a} \times q\vec{E}) + (\vec{a} \times q\vec{E})$. This result $\vec{\tau} = 2(\vec{a} \times q\vec{E})$ doesn't look right. The formula $\vec{\tau} = \vec{p} \times \vec{E}$ is standard. Let's re-verify the torque calculation about the midpoint.

Torque of $\vec{F}_+$ about origin: $\vec{r}_+ = \vec{a}$, $\vec{\tau}_+ = \vec{a} \times \vec{F}_+ = \vec{a} \times (q\vec{E})$. Torque of $\vec{F}_-$ about origin: $\vec{r}_- = -\vec{a}$, $\vec{\tau}_- = \vec{r}_- \times \vec{F}_- = (-\vec{a}) \times (-q\vec{E}) = \vec{a} \times (q\vec{E})$. Net torque: $\vec{\tau} = \vec{\tau}_+ + \vec{\tau}_- = (\vec{a} \times q\vec{E}) + (\vec{a} \times q\vec{E}) = 2 (\vec{a} \times q\vec{E})$. Wait, $\vec{p} = q(2\vec{a})$. So $\vec{a} = \vec{p}/(2q)$. Net torque $ = 2 (\vec{p}/(2q) \times q\vec{E}) = 2 (\vec{p}/2 \times \vec{E}) = \vec{p} \times \vec{E}$. Yes, the calculation is correct. It is $\vec{p} \times \vec{E}$ regardless of the pivot point when the net force is zero.

The net torque on an electric dipole in a uniform electric field is given by the vector product of the dipole moment vector and the electric field vector:

$ \vec{\tau} = \vec{p} \times \vec{E} $

The magnitude of the torque is $\tau = pE \sin\theta$, where $\theta$ is the angle between the direction of the dipole moment $\vec{p}$ and the direction of the electric field $\vec{E}$.

The direction of the torque is perpendicular to the plane containing $\vec{p}$ and $\vec{E}$, determined by the right-hand rule for vector products. The torque tends to rotate the dipole such that its dipole moment $\vec{p}$ aligns with the external electric field $\vec{E}$ (towards the minimum potential energy configuration).

The torque is maximum when the dipole is perpendicular to the field ($\theta = 90^\circ$), and it is zero when the dipole is aligned parallel or anti-parallel to the field ($\theta = 0^\circ$ or $180^\circ$). The equilibrium position is when $\vec{p}$ is parallel to $\vec{E}$ ($\theta = 0^\circ$, stable equilibrium) or anti-parallel ($\theta = 180^\circ$, unstable equilibrium).

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Potential Energy of a Dipole in a Uniform Field

Work is done by the electric field when the dipole rotates. This work changes the potential energy of the dipole in the field. The potential energy ($U$) of an electric dipole in a uniform electric field is defined as the work done by an external agent to rotate the dipole from a reference orientation (usually perpendicular to the field, $\theta = 90^\circ$, where $U=0$) to a specific orientation $\theta$.

$ U(\theta) - U(90^\circ) = -\int_{90^\circ}^{\theta} \tau_{field} \, d\theta' $

The torque exerted by the field is $\tau_{field} = -pE \sin\theta'$ (negative sign because torque tends to decrease $\theta$ towards 0). The torque by external agent is $\tau_{ext} = pE \sin\theta'$.

$ U(\theta) - 0 = \int_{90^\circ}^{\theta} pE \sin\theta' \, d\theta' $

$ U(\theta) = pE [-\cos\theta']_{90^\circ}^{\theta} = pE [-\cos\theta - (-\cos 90^\circ)] = pE [-\cos\theta - 0] $

$ U(\theta) = -pE \cos\theta $

In vector form, this is the dot product of the dipole moment and electric field vectors:

$ U = -\vec{p} \cdot \vec{E} $

The potential energy is minimum when $\vec{p}$ is parallel to $\vec{E}$ ($\theta = 0^\circ, \cos 0^\circ = 1$, $U = -pE$, stable equilibrium) and maximum when $\vec{p}$ is anti-parallel to $\vec{E}$ ($\theta = 180^\circ, \cos 180^\circ = -1$, $U = +pE$, unstable equilibrium).

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