Physics 2 Quiz: Electric Dipoles

What this quiz covers

This quiz focuses on Electric Dipoles, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 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. (correct answer)
2. The student is wrong because the torque on a dipole aligned with $\vec{E}$ is not zero; torque is zero only when $\vec{p}$ is anti-parallel to $\vec{E}$, making the student's entire premise incorrect.
3. The student is wrong because in any non-uniform field the torque cannot be zero for any orientation of $\vec{p}$, so the stated alignment condition is physically impossible.
4. The student correctly identifies both the torque and force conditions; the only flaw is that the student does not specify whether the equilibrium is stable or unstable, not whether equilibrium exists.

Question 2

1. The dipole accelerates in the $-\hat{x}$ direction because systems move to minimize potential energy, and the region of higher field strength has greater energy density, so the dipole is repelled from it.
2. The dipole remains stationary because $\vec{p}$ is along $\hat{z}$ while the field varies along $\hat{x}$; since there is no field gradient in the direction of $\vec{p}$, the forces on $+q$ and $-q$ are equal and opposite, yielding zero net force.
3. The dipole accelerates in the $+\hat{z}$ direction because the positive charge at $z > 0$ experiences a stronger field than the negative charge at $z < 0$, producing a net upward force along the dipole axis.
4. 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 $\vec{F} = \nabla(\vec{p}\cdot\vec{E})$. (correct answer)

Question 3

1. $-pE/2$
2. $+pE$
3. $-pE$
4. $+pE/2$ (correct answer)

Question 4

1. 30° (correct answer)
2. 45°
3. 60°
4. 90°

Question 5

1. antiparallel to field
2. parallel to field (correct answer)
3. perpendicular to field
4. 45 degrees to field

Question 6

1. $-\hat{z}$
2. $+\hat{z}$ (correct answer)
3. $+\hat{y}$
4. $-\hat{x}$

Question 7

1. parallel to $p$; $r^{-2}$
2. opposite to $p$; $r^{-2}$
3. opposite to $p$; $r^{-3}$ (correct answer)
4. parallel to $p$; $r^{-3}$

Question 8

1. The field at P points in the same direction as $\vec{p}$, because the positive charge dominates along the bisector axis at large distances.
2. 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 $\vec{p}$. (correct answer)
3. The field at P points radially outward from the center of the dipole, because the bisector is equidistant from both charges and their tangential contributions vanish.
4. The field at P points parallel to $\vec{p}$ because at large distances the dipole looks like a single positive charge located at the center of mass of the charge distribution.

Question 9

1. The equilibrium at $180°$ is stable because the torque at $\theta$ slightly less than $180°$ acts to restore the dipole back toward $180°$, so it oscillates about that angle indefinitely.
2. The equilibrium at $180°$ is unstable because $U = -pE\cos\theta$ is at a maximum there; after the perturbation the dipole will undergo simple harmonic oscillations about $\theta = 0°$ with angular frequency $\omega = \sqrt{pE/I}$.
3. 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. (correct answer)
4. The equilibrium at $180°$ is neutrally stable because the torque magnitude $|\tau| = pE|\sin\theta|$ is symmetric about both $0°$ and $180°$, so the dipole has no preferred rotational direction after the perturbation.

Question 10

1. $K = 0$, because at $\theta = 0°$ the dipole is aligned with the field and the torque is zero; with no torque acting at the final position, the dipole must be instantaneously at rest.
2. $K = 2pE$, because the potential energy at $90°$ is $+pE$ and at $0°$ is $-pE$, giving a total decrease of $2pE$ that converts entirely to kinetic energy.
3. $K = \tfrac{1}{2}pE$, because only the component of the torque aligned with the angular velocity does work, and averaging $\tau = pE\sin\theta$ over the rotation from $90°$ to $0°$ gives an effective work of $\tfrac{1}{2}pE$.
4. $K = pE$, 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. (correct answer)

Question 11

1. $E_A = E_B$ because both points are at the same distance $r$ from the dipole, and the field magnitude of a dipole depends only on $r$ at large distances.
2. $E_A = 2E_B$ 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. (correct answer)
3. $E_A = \sqrt{2}\, E_B$ because the axial and equatorial fields differ by a factor related to the angular dependence $\sqrt{1+3\cos^2\theta}$ evaluated at $\theta = 0°$ and $\theta = 90°$.
4. $E_A = 4E_B$ because the axial field is proportional to $2p/r^3$ and the equatorial field is proportional to $p/(2r^3)$, so the ratio is $4p/p = 4$.

Question 12

1. $U = -\frac{2p^2}{4\pi\epsilon_0 R^3}$; the configuration is unstable because any perturbation causes the dipoles to rotate toward an anti-parallel arrangement, which would lower the energy even further below this value.
2. $U = +\frac{p^2}{4\pi\epsilon_0 R^3}$; the configuration is unstable because the positive energy means the dipoles repel rotationally, and any perturbation drives them toward anti-parallel alignment to lower the energy.
3. $U = -\frac{2p^2}{4\pi\epsilon_0 R^3}$; 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. (correct answer)
4. $U = -\frac{p^2}{4\pi\epsilon_0 R^3}$; the configuration is stable because the axial field of dipole 1 is parallel to $\vec{p}_2$, the torque on dipole 2 is zero, and small perturbations produce a restoring torque.

Question 13

1. 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. (correct answer)
2. The field falls off as $r^{-3}$ at leading order because the net charge is zero when properly decomposed: the $+2q$ and $-q$ form a net dipole, and the monopole term vanishes.
3. The field falls off as $r^{-3}$ at leading order because the two charges form a pure dipole with moment $p = qd$, and the monopole term is always zero for any two-charge system.
4. The field falls off as $r^{-4}$ at leading order because the system has both a nonzero monopole and a nonzero dipole moment, and the dominant term at large distances arises from their interference, which decays faster than either alone.

Question 14

1. $E_\theta = -\frac{\partial V}{\partial \theta} = \frac{p\sin\theta}{4\pi\epsilon_0 r^2}$, obtained by differentiating $V$ with respect to $\theta$ and negating.
2. $E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$, obtained using the correct spherical-coordinate gradient and negating. (correct answer)
3. $E_\theta = +\frac{1}{r}\frac{\partial V}{\partial \theta} = -\frac{p\sin\theta}{4\pi\epsilon_0 r^3}$, because the field component in the polar direction is co-directional with the potential gradient.
4. $E_\theta = -\frac{\partial V}{\partial r}\sin\theta = \frac{2p\sin\theta\cos\theta}{4\pi\epsilon_0 r^3}$, obtained by projecting the radial derivative of $V$ onto the polar direction.