Electric Dipoles - Physics 2

Two point charges $+2q$ and $-q$ are separated by a distance $d$ . An observer at a very large distance $r \gg d$ measures the electric field. Which of the following best describes how the field falls off with distance $r$ at leading order?
Answer: The field falls off as $r^{-2}$ at leading order because the net charge is $+q \neq 0$ , and at large distances the system appears as a point charge regardless of the dipole contribution. When analyzing the electric field from a charge distribution at large distances, the key tool is the multipole expansion. This framework decomposes the field into contributions ordered by how fast they decay: monopole ( $\sim r^{-2}$ ), dipole ( $\sim r^{-3}$ ), quadrupole ( $\sim r^{-4}$ ), and so on. The dominant term at large distances is always the lowest-order nonzero term. For the system $+2q$ and $-q$ , first check the net charge: $Q_{net} = +2q + (-q) = +q \neq 0$ . A nonzero net charge means the monopole term survives. Since the monopole field falls off as $r^{-2}$ and the dipole falls off as $r^{-3}$ , the monopole term completely dominates at large $r$ . The field therefore falls off as $r^{-2}$ , making A correct. Choice B claims the net charge is "zero when properly decomposed," which is simply false — decomposing into a dipole doesn't erase the physical net charge of $+q$ . Choice C makes the same error while adding a further misconception: a two-charge system is only a pure dipole when the charges are equal and opposite (e.g., $+q$ and $-q$ ). Here, the charges are unequal, so there is no cancellation of the monopole. Choice D describes a scenario involving interference between monopole and dipole terms producing $r^{-4}$ decay — this doesn't reflect how multipole expansions work; terms don't "interfere" to create faster decay, they simply add independently. Study tip: Always check $Q_{net}$ first. If it's nonzero, the answer is $r^{-2}$ — no further analysis needed.
Far from a dipole, on its perpendicular bisector, the field is
Answer: opposite to $p$ ; $r^{-3}$ At any point on the perpendicular bisector, the two charges' horizontal field components cancel, leaving a resultant pointing opposite to the dipole moment p. The far-field dipole strength falls as 1/ $r^3$ , not 1/ $r^2$ . The tempting mistake is thinking the field points parallel to p or that it follows the point-charge 1/ $r^2$ law; both miss the dipole nature.
Dipole $\mathbf{p}=p\hat{x}$ in $\mathbf{E}=E\hat{y}$ . Torque direction?
Answer: $+\hat{z}$ Torque is p x E. With p along +x and E along +y, x cross y is +z, so the torque points toward +z. The tempting wrong pick is -z, which comes from reversing the cross-product order or flipping the right-hand rule; dipole torque is defined as p x E, not E x p.
A dipole in uniform $E$ is slightly displaced and returns to its original orientation. That orientation was
Answer: parallel to field A dipole's potential energy is lowest when its moment is parallel to the field, so that is the stable equilibrium. If you displace it slightly, the torque pushes it back to parallel. The tempting wrong answer is antiparallel, but that is unstable: the slightest displacement makes it rotate farther away, not back.
Which of the following correctly describes the kinetic energy of the dipole when it reaches $\theta = 0°$ (aligned with the field), assuming no energy dissipation?
Answer: $K = pE$ , because the potential energy decreases from $U(90°) = 0$ to $U(0°) = -pE$ , and energy conservation requires this decrease to appear entirely as kinetic energy. When a dipole rotates in an electric field, treat it like any other energy conservation problem: the work done on the system equals the change in kinetic energy, and potential energy "lost" becomes kinetic energy "gained." The potential energy of a dipole in a uniform field is $U = -pE\cos\theta$ . At $\theta_0 = 90°$ , $U = -pE\cos(90°) = 0$ . At $\theta = 0°$ , $U = -pE\cos(0°) = -pE$ . The potential energy decreases by $pE$ , so by conservation of energy, the kinetic energy gained is exactly $K = pE$ . That makes D correct. A confuses the torque at the final position with the work done during the entire journey. Yes, torque is zero at $\theta = 0°$ , but that just means the dipole stops accelerating there — it doesn't mean the dipole is at rest. The dipole has been accelerating the entire way from $90°$ to $0°$ , accumulating kinetic energy. B uses $U(90°) = +pE$ , which is wrong. Plugging $\theta = 90°$ into $U = -pE\cos\theta$ gives zero, not $+pE$ . This error doubles the actual energy change, inflating the answer to $2pE$ . C invents a shortcut that doesn't exist. Averaging torque over angle is not a valid method for calculating work in rotational systems. The correct approach is $W = \int \tau\, d\theta$ , which is equivalent to using $\Delta U$ — and that integral yields $pE$ , not $\tfrac{1}{2}pE$ . Study tip: Always write out $U = -pE\cos\theta$ explicitly and plug in your angles carefully. The negative sign is the most common source of errors on dipole energy questions.
A rigid electric dipole $\vec{p} = p\hat{z}$ is placed at the origin in an external electric field $\vec{E} = E_0(1 + \alpha x)\hat{z}$ , where $\alpha$ is a small positive constant and $E_0 > 0$ . The dipole is oriented along $\hat{z}$ and is free to translate but not rotate. Which statement best describes the translational motion of the dipole?
Answer: The dipole accelerates in the $+\hat{x}$ direction because the field gradient $\partial E_z/\partial x = E_0\alpha$ produces a net force $F_x = p\alpha E_0$ via $\vec{F} = \nabla(\vec{p}\cdot\vec{E})$ . When a permanent dipole sits in a non-uniform electric field, the key formula to reach for is $\vec{F} = \nabla(\vec{p} \cdot \vec{E})$ . This tells you the net force comes from how the field varies in space — not just its direction relative to $\vec{p}$ . Here, $\vec{p} = p\hat{z}$ and $\vec{E} = E_0(1 + \alpha x)\hat{z}$ , so the dot product is $\vec{p} \cdot \vec{E} = pE_0(1 + \alpha x)$ . Taking the gradient gives $\vec{F} = \nabla[pE_0(1+\alpha x)] = pE_0\alpha\,\hat{x}$ . The dipole feels a net force in the $+\hat{x}$ direction — toward the region of stronger field — confirming D is correct. A is wrong on two counts: dipoles in external fields are attracted toward stronger field regions (not repelled), and the energy-density argument here misapplies thermodynamic intuition to a mechanical force problem. B contains a subtle but critical error: it conflates "field varies along $\hat{x}$ " with "no gradient relevant to $\vec{p}$ ." In fact, $\partial E_z/\partial x \neq 0$ is exactly the gradient that drives the force — the gradient doesn't need to be along $\hat{z}$ to matter. C sounds plausible but incorrectly imagines the $+q$ and $-q$ charges displaced along $\hat{z}$ , where the field is uniform (no $z$ -dependence in $E$ ), so those forces cancel perfectly. A useful rule of thumb: **the net force on a dipole points along the gradient of the field component parallel to $\vec{p}$ **, regardless of which spatial direction that gradient points. Always check $\nabla(\vec{p}\cdot\vec{E})$ explicitly rather than relying on geometric intuition.
An electric dipole consists of charges $+q$ and $-q$ separated by distance $d$ , with the dipole moment $\vec{p}$ pointing from $-q$ to $+q$ . A point P lies on the perpendicular bisector of the dipole at distance $r \gg d$ from the center. Which of the following correctly describes the direction of the electric field at P due to the dipole?
Answer: The field at P points antiparallel to $\vec{p}$ , because both charges contribute field components along the bisector that partially cancel, leaving a net component opposing $\vec{p}$ . When analyzing the electric field of a dipole at a point on its perpendicular bisector, you need to carefully track the vector contributions from each charge separately before combining them. Place the dipole vertically, with $+q$ on top and $-q$ on bottom, so $\vec{p}$ points upward. Point P lies horizontally to the side. The field from $+q$ points away from $+q$ (diagonally up-and-away from P's perspective), while the field from $-q$ points toward $-q$ (diagonally down-and-away). Because P is equidistant from both charges, the magnitudes are equal. When you decompose these into components, the horizontal components (pointing away from the dipole center) cancel each other exactly. The surviving vertical components both point downward — that is, antiparallel to $\vec{p}$ . This confirms B is correct: the net field at P opposes $\vec{p}$ , with magnitude proportional to $p/r^3$ at large distances. A is wrong because it assumes the positive charge "dominates," ignoring that both charges contribute equally at equal distance — the direction, not magnitude, is what matters here. D makes the same conceptual error: a dipole never reduces to a single charge at large distances; it remains a dipole whose field falls off as $1/r^3$ , not $1/r^2$ . C is wrong because the field does not point radially outward — that would require the tangential components to cancel, but it's actually the radial components that cancel, leaving a tangential (antiparallel to $\vec{p}$ ) net field. As a study tip, always decompose dipole fields into components at the point of interest — direction questions are almost never answerable by intuition alone.
For a dipole in uniform $E$ , torque is half its maximum when the acute angle between $p$ and $E$ is
Answer: 30° Torque equals pE sin(theta), with theta the angle between p and E. Maximum torque occurs at 90 degrees, so half maximum means sin(theta) = 1/2. The acute angle whose sine is 1/2 is 30 degrees. The tempting 60 degrees is wrong because half of 90 degrees is not the same as half the sine value.
The electric potential at a point P located at distance $r$ from the center of an ideal electric dipole $\vec{p}$ is given by $V = \frac{p\cos\theta}{4\pi\epsilon_0 r^2}$ , where $\theta$ is the angle between $\vec{p}$ and the position vector $\vec{r}$ to P. A student wants to find the component of the electric field in the $\hat{\theta}$ direction (the polar direction). Which of the following is correct?
Answer: $E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$ , obtained using the correct spherical-coordinate gradient and negating. Whenever you encounter electric field problems in spherical coordinates, your first instinct should be to recall the full gradient formula — not the Cartesian one. The electric field relates to potential via $\vec{E} = -\nabla V$ , and in spherical coordinates, the gradient has a critical scaling factor: the $\hat{\theta}$ component is $-\frac{1}{r}\frac{\partial V}{\partial \theta}$ , not simply $-\frac{\partial V}{\partial \theta}$ . That factor of $\frac{1}{r}$ exists because $\theta$ is an angular coordinate — a change in angle corresponds to a physical arc length of $r\,d\theta$ , so the spatial rate of change must be divided by $r$ to have the correct units of V/m. Applying this to the dipole potential $V = \frac{p\cos\theta}{4\pi\epsilon_0 r^2}$ , you differentiate with respect to $\theta$ : $\frac{\partial V}{\partial \theta} = -\frac{p\sin\theta}{4\pi\epsilon_0 r^2}$ . Dividing by $r$ and negating gives $E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$ , confirming B is correct. Choice A forgets the $\frac{1}{r}$ factor entirely, leaving the answer with the wrong $r$ -dependence ( $r^{-2}$ instead of $r^{-3}$ ) and wrong units. Choice C drops the negative sign — the field points opposite to the potential gradient, always. Choice D incorrectly projects the radial derivative onto the polar direction using a $\sin\theta$ factor, which has no justification from the gradient formula. A reliable study tip: memorize the spherical gradient in full — $\nabla V = \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial V}{\partial \theta}\hat{\theta} + \frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi}\hat{\phi}$ — and always check that each term has units of V/m before proceeding.
An ideal electric dipole $\vec{p}$ is located at the origin. Point A is on the dipole axis at distance $r$ from the center. Point B is on the perpendicular bisector at the same distance $r$ . Let $E_A$ and $E_B$ be the magnitudes of the electric field at points A and B, respectively. Which of the following is correct?
Answer: $E_A = 2E_B$ because the axial field is $\frac{2p}{4\pi\epsilon_0 r^3}$ and the equatorial field is $\frac{p}{4\pi\epsilon_0 r^3}$ , giving a ratio of exactly 2. When you encounter a dipole field problem, your first instinct should be to recall the two key formulas — one for points on the axis and one for the perpendicular bisector (the equatorial plane). For an ideal dipole of moment $p$ , the axial field (along the dipole axis, $\theta = 0°$ ) is: $E_A = \frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}$ The equatorial field (on the perpendicular bisector, $\theta = 90°$ ) is: $E_B = \frac{1}{4\pi\epsilon_0}\frac{p}{r^3}$ Dividing these directly gives $E_A / E_B = 2$ , confirming that B is correct: $E_A = 2E_B$ . This factor of 2 comes from the geometry of how the two charge contributions add along the axis versus partially cancel on the equator. A is wrong because it assumes dipole field strength depends only on distance — it doesn't. The angular position matters fundamentally; a dipole is inherently anisotropic. Spherical symmetry applies to monopoles, not dipoles. C is wrong but subtly so. The general dipole field magnitude is $E = \frac{p}{4\pi\epsilon_0 r^3}\sqrt{1+3\cos^2\theta}$ . At $\theta = 0°$ , this gives $\sqrt{4} = 2$ ; at $\theta = 90°$ , it gives $\sqrt{1} = 1$ . The ratio is still exactly 2 — not $\sqrt{2}$ . Choice C misapplies the formula by forgetting to evaluate the square root fully. D is wrong because it incorrectly states the equatorial field as $p/(2r^3)$ — that's a fabricated expression with no physical basis. Your takeaway: memorize both dipole field formulas and the factor-of-2 ratio between them. It appears on exams more often than students expect.
What is the potential energy of interaction between the two dipoles, and is their parallel alignment (both pointing in the $+\hat{x}$ direction) stable or unstable?
Answer: $U = -\frac{2p^2}{4\pi\epsilon_0 R^3}$ ; the configuration is stable because this energy is a minimum—a small angular perturbation increases $U$ , and the resulting restoring torque returns the dipoles to alignment. When two dipoles sit end-to-end along the same axis, you need two tools: the formula for the electric field of a dipole along its axis, and the formula for the potential energy of a dipole in an external field. The axial field of dipole 1 at distance $R$ points in the $+\hat{x}$ direction with magnitude $E = \frac{2p}{4\pi\epsilon_0 R^3}$ . The potential energy of dipole 2 in this field is $U = -\vec{p}_2 \cdot \vec{E} = -p \cdot \frac{2p}{4\pi\epsilon_0 R^3} = -\frac{2p^2}{4\pi\epsilon_0 R^3}$ . This is a negative (minimum) energy for parallel alignment, which confirms answer C. Stability follows from the energy landscape: if you perturb either dipole by a small angle $\theta$ , the dot product $\vec{p}_2 \cdot \vec{E}$ decreases, so $U$ rises above the minimum, and the resulting torque $\vec{\tau} = \vec{p} \times \vec{E}$ acts to restore alignment. That is the textbook definition of a stable equilibrium. Answer A gets the energy right but inverts the stability logic. Negative energy does not imply instability — it means you are at a minimum, which is precisely why the configuration is stable. Answer B states a positive energy, which would correspond to anti-parallel alignment on axis, not parallel. It also misidentifies that configuration's stability. Answer D uses the correct sign and correctly notes zero torque at equilibrium, but it drops a factor of 2, writing $-\frac{p^2}{4\pi\epsilon_0 R^3}$ instead of $-\frac{2p^2}{4\pi\epsilon_0 R^3}$ . That factor of 2 is the signature of the axial field versus the equatorial field — memorize both. Study tip: For dipole–dipole problems, always identify whether you're on the axis or the equatorial plane first; the field magnitude differs by exactly a factor of 2, and forgetting this is the most common calculation error on this topic.
A dipole in uniform $E$ is released from rest at 60°. Work done by $E$ before first reaching alignment?
Answer: $+pE/2$ The dipole's potential energy is U = -pE cos(theta). At 60 degrees, U = -pE/2; at alignment, U = -pE. Work done by the field equals the decrease in potential energy: (-pE/2) - (-pE) = +pE/2. The tempting wrong answer is -pE/2, which is the change in potential energy, not the work done by the field.
Which of the following identifies the critical flaw in the student's reasoning?
Answer: The student correctly identifies the torque condition but ignores that a non-uniform field exerts a net force on the dipole even when $\vec{p}$ is aligned with $\vec{E}$ , so the dipole accelerates translationally and is not in static equilibrium. When analyzing dipole behavior, you need to separately evaluate two conditions for true equilibrium: the torque condition (rotational) and the net force condition (translational). A common trap is assuming that zero torque alone means the dipole is stationary. The torque on a dipole is $\vec{\tau} = \vec{p} \times \vec{E}$ , which is indeed zero when $\vec{p}$ is aligned with $\vec{E}$ — so the student's torque reasoning is correct. However, the net force on a dipole in a non-uniform field is $\vec{F} = \nabla(\vec{p} \cdot \vec{E})$ , which is generally nonzero even when the dipole is perfectly aligned. The stronger field on one side pulls the positive charge more than it pushes the negative charge (or vice versa), creating a net translational force. True static equilibrium requires both $\vec{\tau} = 0$ and $\vec{F} = 0$ , and the student only checks one condition. This makes A correct. B is wrong because it reverses the torque physics. Torque is zero when $\vec{p}$ is parallel to $\vec{E}$ (not anti-parallel); the student's premise about torque is actually valid, so this distractor misidentifies the error. C is wrong because torque absolutely can be zero in a non-uniform field — field non-uniformity affects force, not the torque formula, which depends only on local alignment. D is wrong because equilibrium genuinely does not exist here. A net translational force means no static equilibrium at all, stable or unstable — so the flaw goes deeper than merely labeling the type of equilibrium. Study tip: Whenever a question mentions a non-uniform field and a dipole, immediately ask yourself two separate questions: Is torque zero? Is net force zero? Non-uniformity is almost always a signal that the force condition is the hidden piece being tested.
Which of the following correctly characterizes the equilibrium at $\theta = 180°$ and predicts what happens after the perturbation?
Answer: The equilibrium at $180°$ is unstable because $U$ is at a maximum there; after the perturbation the dipole accelerates away from $180°$ and, with no damping, overshoots and oscillates about $\theta = 0°$ with large amplitude—not simple harmonic motion. When analyzing equilibrium stability in rotational systems, ask yourself two things: where is the potential energy extremum, and what happens to the motion after equilibrium is broken? The dipole's potential energy is $U = -pE\cos\theta$ . At $\theta = 180°$ , $\cos(180°) = -1$ , so $U = +pE$ —a maximum. This immediately tells you the equilibrium is unstable: any perturbation causes the system to roll away from that configuration, like a ball balanced on a hilltop. The restoring torque $\tau = -pE\sin\theta$ actually pushes the dipole away from $180°$ rather than back toward it. After the perturbation, the dipole accelerates toward $\theta = 0°$ , gaining kinetic energy. With no damping, energy is conserved—the dipole overshoots $0°$ and swings to large angles on the other side. This is large-amplitude oscillation, not simple harmonic motion. SHM only applies when oscillations are small enough that $\sin\theta \approx \theta$ , which requires proximity to a stable equilibrium (like $\theta = 0°$ ) with small initial displacement. The frequency $\omega = \sqrt{pE/I}$ only describes small oscillations near $0°$ . This makes C correct. A is wrong because $180°$ is unstable, not stable—the torque pushes the dipole away from $180°$ , not back toward it. B correctly identifies the instability but incorrectly applies the SHM formula; after launching from near $180°$ , the dipole arrives at $0°$ with substantial kinetic energy, so oscillations are large-amplitude. D is wrong because torque symmetry doesn't imply neutral stability—the energy landscape clearly has a maximum at $180°$ and a minimum at $0°$ . Study tip: Always check which extremum you're at—stable equilibria sit at potential energy minima, and the SHM formula only applies for small oscillations near those minima.

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