5.7 Electric dipoles (Page 2/6)

University physics volume 2 Page 2 / 6

A dipole, consisting of a negative charge on the left and a positive charge on the right is in a uniform electric field pointing to the right. The dipole moment, p, points to the right. The field lines of the net electric field are the sum of the dipole field and the uniform external field, horizontal far from the dipole and similar to the dipole field near the dipole. — The net electric field is the vector sum of the field of the dipole plus the external field.

Recall that we found the electric field of a dipole in [link] . If we rewrite it in terms of the dipole moment we get:

\vec{E} (z) = \frac{1}{4 π ε_{0}} \frac{\vec{p}}{z^{3}} .

The form of this field is shown in [link] . Notice that along the plane perpendicular to the axis of the dipole and midway between the charges, the direction of the electric field is opposite that of the dipole and gets weaker the further from the axis one goes. Similarly, on the axis of the dipole (but outside it), the field points in the same direction as the dipole, again getting weaker the further one gets from the charges.

Summary

If a permanent dipole is placed in an external electric field, it results in a torque that aligns it with the external field.
If a nonpolar atom (or molecule) is placed in an external field, it gains an induced dipole that is aligned with the external field.
The net field is the vector sum of the external field plus the field of the dipole (physical or induced).
The strength of the polarization is described by the dipole moment of the dipole, $\vec{p} = q \vec{d}$ .

Key equations

Coulomb’s law	${\vec{F}}_{12} (r) = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{r}}_{12}$
Superposition of electric forces	$\vec{F} (r) = \frac{1}{4 π ε_{0}} Q \sum_{i = 1}^{N} \frac{q_{i}}{r_{i}^{2}} {\hat{r}}_{i}$
Electric force due to an electric field	$\vec{F} = Q \vec{E}$
Electric field at point P	$\vec{E} (P) \equiv \frac{1}{4 π ε_{0}} \sum_{i = 1}^{N} \frac{q_{i}}{r_{i}^{2}} {\hat{r}}_{i}$
Field of an infinite wire	$\vec{E} (z) = \frac{1}{4 π ε_{0}} \frac{2 λ}{z} \hat{k}$
Field of an infinite plane	$\vec{E} = \frac{σ}{2 ε_{0}} \hat{k}$
Dipole moment	$\vec{p} \equiv q \vec{d}$
Torque on dipole in external E-field	$\vec{τ} = \vec{p} \times \vec{E}$

Conceptual questions

What are the stable orientation(s) for a dipole in an external electric field? What happens if the dipole is slightly perturbed from these orientations?

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Problems

Consider the equal and opposite charges shown below. (a) Show that at all points on the x -axis for which $| x | ≫ a, E \approx {Q a / 2 π ε}_{0} x^{3} .$ (b) Show that at all points on the y -axis for which $| y | ≫ a, E \approx {Q a / π ε}_{0} y^{3} .$

Two charges are shown on the y axis of an x y coordinate system. Charge +Q is a distance a above the origin, and charge −Q is a distance a below the origin.

$E_{x} = 0,$
$E_{y} = \frac{1}{4 π ε_{0}} [\frac{2 q}{(x^{2} + a^{2})} \frac{a}{\sqrt{(x^{2} + a^{2})}}]$
$\Rightarrow x ≫ a \Rightarrow \frac{1}{2 π ε_{0}} \frac{q a}{x^{3}}$ ,
$E_{y} = \frac{q}{4 π ε_{0}} [\frac{2 y a + 2 y a}{{(y - a)}^{2} {(y + a)}^{2}}]$
$\Rightarrow y ≫ a \Rightarrow \frac{1}{π ε_{0}} \frac{q a}{y^{3}}$

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(a) What is the dipole moment of the configuration shown above? If $Q = 4.0 μ C$ , (b) what is the torque on this dipole with an electric field of $4.0 \times 10^{5} N/C \hat{i}$ ? (c) What is the torque on this dipole with an electric field of $−4.0 \times 10^{5} N/C \hat{i}$ ? (d) What is the torque on this dipole with an electric field of $\pm 4.0 \times 10^{5} N/C \hat{j}$ ?

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A water molecule consists of two hydrogen atoms bonded with one oxygen atom. The bond angle between the two hydrogen atoms is $104 °$ (see below). Calculate the net dipole moment of a water molecule that is placed in a uniform, horizontal electric field of magnitude $2.3 \times 10^{−8} N/C .$ (You are missing some information for solving this problem; you will need to determine what information you need, and look it up.)

A schematic representation of the outer electron cloud of a neutral water molecule is shown. Three atoms are at the vertices of a triangle. The hydrogen atom has positive q charge and the oxygen atom has minus two q charge, and the angle between the line joining each hydrogen atom with the oxygen atom is one hundred and four degrees. The cloud density is shown as being greater at the oxygen atom.

The net dipole moment of the molecule is the vector sum of the individual dipole moments between the two O-H. The separation O-H is 0.9578 angstroms:
$\vec{p} = 1.889 \times 10^{−29} Cm \hat{i}$

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Additional problems

Point charges $q_{1} = 2.0 μ C$ and $q_{1} = 4.0 μ C$ are located at $r_{1} = (4.0 \hat{i} - 2.0 \hat{j} + 2.0 \hat{k}) m$ and $r_{2} = (8.0 \hat{i} + 5.0 \hat{j} - 9.0 \hat{k}) m$ . What is the force of $q_{2} on q_{1} ?$

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What is the force on the $5.0 - μ C$ charges shown below?

The following charges are shown on an x y coordinate system: Minus 3.0 micro Coulomb on the x axis, 3.0 meters to the left of the origin. Positive 5.0 micro Coulomb at the origin. Positive 9.0 micro Coulomb on the x axis, 3.0 meters to the right of the origin. Positive 6.0 micro Coulomb on the y axis, 3.0 meters above the origin.

${\vec{F}}_{net} = [−8.99 \times 10^{9} \frac{3.0 \times 10^{−6} (5.0 \times 10^{−6})}{{(3.0 m)}^{2}} - 8.99 \times 10^{9} \frac{9.0 \times 10^{−6} (5.0 \times 10^{−6})}{{(3.0 m)}^{2}}] \hat{i}$ ,
$−8.99 \times 10^{9} \frac{6.0 \times 10^{−6} (5.0 \times 10^{−6})}{{(3.0 m)}^{2}} \hat{j} = −0.06 N \hat{i} - 0.03 N \hat{j}$

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Source: OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3

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