Physics 2 Quiz: Selecting Eandm Principles

What this quiz covers

This quiz focuses on Selecting Eandm Principles, 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 argument is correct in its conclusion but wrong in its premise: the voltage, not the charge, is held constant by the battery. Since $V = \mathcal{E}$ is fixed and $C$ increases, applying $U = \frac{1}{2}C\mathcal{E}^2$ shows that $U$ increases by a factor of $\kappa$. The formula $U = Q^2/(2C)$ is valid but inapplicable here without first recognizing that $Q$ is not constant.
2. The argument is entirely correct: the battery fixes the charge $Q$ on the plates, and since $C$ increases, $U = Q^2/(2C)$ shows that $U$ decreases. The student correctly identifies the battery as a charge reservoir that maintains constant $Q$ regardless of changes in capacitance.
3. The argument reaches the wrong conclusion. The battery holds $V = \mathcal{E}$ fixed, so the charge increases to $Q' = \kappa C_0 \mathcal{E}$ when the dielectric is inserted. Using $U = \frac{1}{2}C'V^2$ with $C' = \kappa C_0$ shows $U$ increases by factor $\kappa$. The form $Q^2/(2C)$ is algebraically valid but misapplied here because the student incorrectly treated $Q$ as the conserved quantity rather than $V$. (correct answer)
4. The argument is wrong in both premise and conclusion. The battery holds neither $Q$ nor $V$ constant during the insertion process; instead, it maintains a constant power output $P = \mathcal{E}I$. Finding the correct energy change requires integrating the power delivered over the insertion time, and static energy formulas like $U = Q^2/(2C)$ cannot be applied to a dynamic process.

Question 2

1. The challenge identifies a real dependency but mischaracterizes it as circular reasoning. Both methods require knowing $B$ as an intermediate step — that is sequential computation, not circularity. The inductance method computes $L = \mu_0 N^2 A/\ell$ from $B$ and then $U = \frac{1}{2}LI^2$; the volume integration computes $u = B^2/(2\mu_0)$ and integrates over the torus volume. Both are valid routes to the same answer; the choice is one of efficiency, not logical validity. (correct answer)
2. The challenge is entirely correct. $U = \frac{1}{2}LI^2$ is definitionally circular because $L$ is defined through $\Phi = NBA$, which requires $B$ first. The energy density integration $U = \frac{B^2}{2\mu_0} \cdot V_{\text{torus}}$ is the only logically independent method, because it derives the stored energy directly from the field without invoking any circuit-level definition that presupposes $B$.
3. The challenge is incorrect. The formula $U = \frac{1}{2}LI^2$ can be derived from circuit theory alone by integrating instantaneous power: $U = \int_0^I \mathcal{E}\,dq = \int_0^I L I'\,dI' = \frac{1}{2}LI^2$. Since this derivation does not reference $\vec{B}$ at all, the inductance method is completely independent of field theory. The volume integration and the inductance method are thus two genuinely independent principles that happen to yield the same stored energy.
4. The challenge is partially correct: $U = \frac{1}{2}LI^2$ is circular only when $L$ is computed analytically from field theory via $\Phi = LI$. If $L$ is instead determined experimentally — for example, from the resonant frequency of an LC circuit — then $U = \frac{1}{2}LI^2$ is a valid, non-circular approach. For a purely analytical problem where $L$ must be derived from $B$, the volume integration is the only non-circular method available.

Question 3

1. Approaches I and II are equivalent and both correct; Approach III is incorrect because the series-capacitor model applies only when the plates are physically separate conductors, not when they share a continuous gap with a dielectric interface inside it.
2. All three approaches are equivalent and yield the same result. Approach III is the most efficient because it bypasses computing $\vec{E}$ and integrating, directly applying the series-capacitor formula. The key insight is that $\vec{D}$ is continuous across the dielectric interface (no free charge there), making all three methods mutually consistent. (correct answer)
3. Approaches I and III are both correct, but Approach II is invalid: integrating $\vec{E}$ across a dielectric interface is only permissible when the field is uniform throughout the entire gap, a condition that fails here since $\kappa \neq 1$ causes the field magnitudes to differ in each region.
4. Approach I alone is fundamentally necessary; Approaches II and III are shortcuts valid only in the special case $\kappa = 1$. When a dielectric is present, bound surface charges at the interface alter the field in a way that the voltage integral and series formula cannot properly capture without first solving for $\vec{E}$ from boundary conditions.

Question 4

1. Methods I and III are equivalent and both correct; Method II is invalid because the force between current-carrying wires is a magnetic force, not derivable from a scalar potential energy — magnetic forces do no work on moving charges and therefore cannot be expressed as the gradient of any energy function.
2. All three methods are valid and yield the same result. Method I is most efficient for this geometry. Method II is valid because the macroscopic force can be obtained from the gradient of magnetic energy at constant current. Method III is a longer but rigorous route, since although $\vec{A}$ diverges logarithmically, its curl converges to the correct finite field $\vec{B}$. (correct answer)
3. Only Method I is correct. Method II fails because mutual inductance is defined only for closed finite circuits, making $m$ per unit length undefined for infinite wires. Method III is invalid because the vector potential $\vec{A}$ for an infinite wire diverges logarithmically, and a divergent $\vec{A}$ cannot be used to compute $\vec{B}$ via the curl operation.
4. Methods I and II are correct, but Method III is invalid: the curl of a logarithmically divergent vector potential is mathematically undefined for an infinite wire. Although $\vec{A}$ appears in the Lagrangian formulation of electrodynamics, its spatial divergence prevents the relation $\vec{B} = \nabla \times \vec{A}$ from being evaluated in this geometry.

Question 5

1. The claim is entirely correct. Using $\vec{E} = -\nabla V$ is always more fundamental than direct vector integration because the gradient operation encodes all spatial information about the field without requiring component-by-component decomposition, making it the preferred method in any geometry with a known potential.
2. The first part is correct: scalar integration for $V$ is simpler and exact, since every element $dq$ is equidistant from the axial point. The second part is partially wrong: direct vector integration of $d\vec{E}$ is equally valid and, on the axis, no more complex than taking the gradient, because azimuthal symmetry eliminates radial components. Neither method is 'more fundamental'; the choice is one of efficiency. (correct answer)
3. The first part is correct, but the second part is entirely wrong: $\vec{E} = -\nabla V$ is only valid in regions where no free charges are present, so it cannot be applied on the axis of the ring because the ring itself introduces a singularity in $V$ at $r = 0$, invalidating differentiation there.
4. Both parts of the claim are wrong. Direct vector integration of $d\vec{E}$ is simpler than scalar integration of $dV$ because the field directly gives observable forces, while potential requires an additional differentiation step that introduces extra complexity and can amplify errors from any approximations made during integration.

Question 6

1. Use the work-energy theorem: $W = \int_{r_1}^{r_2} F\, dr = kQq\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = \frac{1}{2}mv^2$. Energy methods are optimal here because the Coulomb force is conservative, eliminating the need for vector decomposition, and the result is exact — not an approximation. (correct answer)
2. Apply Newton's second law $F = kQq/r^2 = ma$ to get acceleration as a function of $r$, then integrate $a\,dt$ twice to obtain displacement and velocity. This is necessary because the force varies with position, making energy methods an approximation rather than an exact result.
3. Use conservation of energy: $\frac{1}{2}mv^2 = kQq\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$, since the potential energy decreases as the like charges separate and the kinetic energy must compensate. The energy principle is preferred over force methods because it is a scalar equation and avoids vector calculus.
4. Compute the electric potential $V$ at $r_1$ and $r_2$ using $V = kQ/r$, find $\Delta V = V(r_2) - V(r_1)$, then apply $W = q\Delta V$ and set equal to $\frac{1}{2}mv^2$. Potential is the preferred principle because it separates the source field from the test charge, whereas energy methods conflate the two and introduce errors when $q$ is small.

Question 7

1. The result is correct and Route 2 is the superior approach because it directly computes the potential via a single scalar integral, completely avoiding the two-region field integration required in Route 1. The integrand $k\rho(4\pi r^2 dr)/r$ correctly gives $dV$ for a shell of radius $r$ and thickness $dr$, and both routes yield the same final answer of $2\pi k \rho R^2$.
2. The result is incorrect. Route 2 contains a conceptual error: the formula $dV = k\,dq/r$ gives the potential at a field point due to a source at distance $r$, but here $r$ serves simultaneously as the shell radius and as the distance from the center — this double role introduces an error. Route 1 gives a different numerical answer, confirming that Route 2 is flawed.
3. The result is incorrect. Route 2 gives $V(0) = 2\pi k\rho R^2$, but Route 1 gives $V(0) = \frac{3kQ}{2R} = 2\pi k\rho R^2$. Since the two answers agree numerically, Route 2 must contain a subtle error in the integrand that coincidentally produces the correct answer for this particular geometry, even though it is not generally valid.
4. The result is correct and equals $\frac{3kQ}{2R}$ where $Q = \frac{4}{3}\pi R^3 \rho$. Route 1 requires integrating $\vec{E}$ over two regions with different expressions, while Route 2 is a single scalar integral. At the center, every shell of radius $r$ is at an unambiguous distance $r$ from the field point, making the integrand exact. Both routes are valid, but Route 2 is more direct. (correct answer)

Question 8

1. The student's approach is valid only if $q$ moves along a path of constant $\theta$, because along such a path the dipole potential is purely radial. This ensures that the torque on the dipole from $q$ has no moment arm to cause rotation, so angular momentum is conserved and the dipole orientation is unchanged throughout the motion.
2. The student's approach is always valid regardless of dipole constraints. The electric force on $q$ is conservative because it derives from a scalar potential, and conservative forces always conserve the mechanical energy of a test charge moving through the field. The dipole's possible motion belongs to a separate energy budget that does not affect the energy equation for $q$.
3. The student's approach is valid when the dipole is rigidly constrained in position and orientation. In that case, constraint forces maintain the dipole's state, and the total mechanical energy of $q$ — kinetic energy plus electrostatic potential energy $qV$ — is conserved. The critic's concern is physically valid for a free dipole but irrelevant when the dipole cannot move. (correct answer)
4. The student's approach is valid only in the limit $q \rightarrow 0$, because then the back-reaction force on the dipole vanishes identically and the dipole potential is undisturbed. For any finite $q$, the dipole gains kinetic energy (translational and rotational) at the expense of $q$, so the formula $\Delta KE_q = -q\Delta V$ systematically underestimates the true final speed of $q$.

Question 9

1. Use superposition of fields: the external field contributes $V_{\text{ext}} = -E_0 R$ at the top and $+E_0 R$ at the bottom of the sphere. Since the conductor must be an equipotential, the surface potential equals the average of these two values, giving $V = 0$. This averaging is valid because the external field is antisymmetric about the equatorial plane and the induced charges are assumed to distribute symmetrically.
2. Use the energy principle: the induced charges rearrange until the total electrostatic energy is minimized. At the energy minimum, the net force on every surface charge element vanishes, which requires the tangential field at the surface to be zero. Setting the tangential component of $\vec{E}_0$ equal to the tangential component of the induced field at $r = R$ yields a surface potential of $V = E_0 R$.
3. Use the conductor boundary condition: since the sphere is isolated and uncharged, the net induced charge is zero. Because the potential of an isolated conductor is determined solely by its total free charge, Gauss's Law gives $V = kQ_{\text{net}}/R = 0$. The presence of the external field does not alter this result because the induced dipole layer carries no net charge.
4. Apply the mean-value theorem for harmonic functions: since the potential satisfies Laplace's equation in the charge-free exterior, the potential at the center of the sphere equals the average of $V$ over the sphere's surface. In the unperturbed external field $V_{\text{ext}} = -E_0 z$, the average over the sphere is zero by antisymmetry about $z = 0$. Placing an uncharged conductor does not change this average (the induced dipole contributes zero net average by symmetry), so the surface potential — which is uniform on the conductor — must be zero. (correct answer)

Question 10

1. Use Gauss's Law to find $\vec{E}$ in each region, then integrate $\int_{b}^{\infty} \vec{E} \cdot d\vec{\ell}$ to get the potential difference, and multiply by $q_0$. This approach is necessary because the charge distribution is not uniform, so energy methods alone cannot give the correct result.
2. Use the electric potential $V$ at the outer surface, found by superposition of the point charge and the induced surface charges, then apply $W = q_0(V_b - V_\infty)$. This is most efficient because spherical symmetry allows $V$ to be written by inspection without integrating the field, and potential is a scalar requiring no vector decomposition. (correct answer)
3. Apply energy conservation by computing the total electrostatic potential energy $U$ stored in the system, then set the work equal to the change in $U$. This is optimal because work-energy methods always bypass field integrals in problems involving conductors, making them universally superior to potential methods.
4. Use Gauss's Law to find $\vec{E}$ only outside the shell ($r > b$), then integrate from $b$ to $\infty$. This is the most direct approach because the conductor screens all interior charge, so the exterior field is zero and the integral is trivially evaluated without any computation.