Answer: $\langle 20, 30, 10 \rangle$

Compute the gradient of T: $\nabla T = \langle -4x, -6y, -2z \rangle$. At $(1,1,1)$: $\nabla T = \langle -4, -6, -2 \rangle$. Apply Fourier's Law: $\mathbf{F} = -k\nabla T = -5\langle -4, -6, -2 \rangle = \langle 20, 30, 10 \rangle$. Why distractors fail: Option A applies $\mathbf{F} = k\nabla T$ (missing the negative sign). Option C gives $\nabla T$ itself without multiplying by $-k$. Option D gives $-\nabla T$ without the factor of $k = 5$.

Answer: Compute $\nabla T$, then integrate $-k\nabla T \cdot \hat{\mathbf{n}}$ over the sphere using $dS = R^2 \sin\phi\,d\phi\,d\theta$

Identify what must be computed: Total outward heat flow is $\iint_S \mathbf{F} \cdot d\mathbf{S}$ where $\mathbf{F} = -k\nabla T$. This requires the gradient of $T$ and the oriented surface element of the sphere. Correct approach (Option A): Compute $\nabla T$ (most naturally in spherical coordinates, since $T$ depends on $\phi$), then integrate $-k\nabla T \cdot \hat{\mathbf{n}}$ over the sphere using the spherical surface element $dS = R^2\sin\phi\,d\phi\,d\theta$. This correctly accounts for both the heat flux direction and the geometry of the sphere. Why distractors fail: Option B: The Divergence Theorem converts to a volume integral inside the ball, but $T$ is only defined on the surface, not throughout the interior, so this approach is not directly applicable without additional assumptions. Option C: It drops the surface area element and treats the problem as one-dimensional, losing the $\sin\phi$ Jacobian and the $\theta$ integration. Option D: Stokes' Theorem converts a surface integral of curl into a line integral — heat flux is not a curl, so this theorem does not apply.

Answer: The Divergence Theorem combined with Laplace's equation $\nabla^2 T = 0$

Express total heat flux: The total outward heat flux is $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S (-k\nabla T) \cdot d\mathbf{S}$. Apply the Divergence Theorem: By the Divergence Theorem, $\iint_S (-k\nabla T) \cdot d\mathbf{S} = \iiint_V \nabla \cdot (-k\nabla T)\,dV = -k\iiint_V \nabla^2 T\,dV$. Use the steady-state condition: In steady state with no internal sources, $T$ satisfies Laplace's equation $\nabla^2 T = 0$, so the volume integral is zero. Hence the total outward flux is zero. Why distractors fail: Option A: Stokes' Theorem relates a line integral to a surface integral of curl, not divergence — it is not the right tool here. Option C: Green's Theorem is a 2D result, inappropriate for a closed surface in 3D. Option D: The gradient theorem relates a line integral to endpoint values of a potential; it does not address flux through a closed surface.

Answer: Magnetic monopoles do not exist; magnetic field lines always form closed loops

Meaning of zero divergence: The equation $\nabla \cdot \mathbf{B} = 0$ means the magnetic field has no sources or sinks. Geometrically, the net magnetic flux through any closed surface is zero. Physical interpretation: This implies there are no isolated magnetic charges (monopoles). Every magnetic field line that enters a closed surface must also leave it, forming closed loops. Why distractors fail: Option A describes charge conservation (the continuity equation), not Gauss's law for magnetism. Option C describes Faraday's Law ($\nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t$). Option D describes Gauss's Law for electricity ($\nabla \cdot \mathbf{E} = \rho/\varepsilon_0$).

Answer: $\frac{Q}{\varepsilon_0}$

Apply Gauss's Law directly: Gauss's Law states $\iint_S \mathbf{E} \cdot d\mathbf{S} = Q_{\text{enc}}/\varepsilon_0$. Since the charge $Q$ is at the center and fully enclosed, $Q_{\text{enc}} = Q$. Compute the flux: The total electric flux is simply $\Phi_E = Q/\varepsilon_0$, regardless of the radius $R$. The result depends only on the enclosed charge. Why distractors fail: Option A multiplies by $\varepsilon_0$ instead of dividing. Option B gives the electric field magnitude at the surface, not the total flux. Option D gives the electric potential at the surface, not the flux.

Answer: Heat flows from regions of high temperature to regions of low temperature

Direction of the gradient: The gradient $\nabla T$ points in the direction of greatest temperature increase. Role of the negative sign: The negative sign reverses this direction, so $\mathbf{F} = -k\nabla T$ points from hot to cold. This matches the physical observation that heat naturally flows down the temperature gradient. Why distractors fail: Option A reverses the correct direction of heat flow. Option B conflates gradient magnitude with direction. Option D is true (heat flux is perpendicular to isotherms) but does not specifically explain the role of the negative sign.

Answer: Since $\nabla \cdot \mathbf{E} = 0$ and $\nabla \times \mathbf{E} = \mathbf{0}$, it is a valid electrostatic field in a charge-free region

Check divergence (Gauss's Law): $\nabla \cdot \mathbf{E} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(xz) + \frac{\partial}{\partial z}(xy) = 0 + 0 + 0 = 0$. This is consistent with a charge-free region ($\rho = 0$). Check curl (Faraday's Law for statics): $\nabla \times \mathbf{E} = \left\langle \frac{\partial}{\partial y}(xy) - \frac{\partial}{\partial z}(xz),\ \frac{\partial}{\partial z}(yz) - \frac{\partial}{\partial x}(xy),\ \frac{\partial}{\partial x}(xz) - \frac{\partial}{\partial y}(yz) \right\rangle = \langle x - x,\ y - y,\ z - z \rangle = \mathbf{0}$. A valid electrostatic field must be irrotational. Conclusion and distractors: Both conditions are satisfied, so Option B is correct. Option A falsely claims the curl is nonzero. Option C falsely claims the divergence is nonzero. Option D applies an incorrect criterion — the Laplacian of $\mathbf{E}$ is not the standard test for electrostatic validity.

Answer: They relate the divergence and curl of the electric and magnetic fields to charge densities and current densities

Content of Maxwell's equations: Maxwell's four equations use the divergence and curl operators to describe how electric and magnetic fields are generated by charges and currents and how they interact with each other. Why the correct answer works: Option B accurately captures the essence: Gauss's laws give $\nabla \cdot \mathbf{E}$ and $\nabla \cdot \mathbf{B}$, while Faraday's law and Ampère–Maxwell law give $\nabla \times \mathbf{E}$ and $\nabla \times \mathbf{B}$, all in terms of sources (charges and currents). Why distractors fail: Option A incorrectly substitutes gravitational for electromagnetic fields. Option C conflates Maxwell's equations with conservation laws for mass/energy. Option D is wrong because the magnetic field $\mathbf{B}$ is not the gradient of a scalar potential (it is divergence-free, and its behavior is described by a vector potential).

Answer: The force is directed toward the origin (attractive)

Interpret the unit vector: The unit vector $\hat{\mathbf{r}}$ points radially outward from the origin toward the mass $m$. Effect of the negative sign: Multiplying by $-1$ reverses the direction of $\hat{\mathbf{r}}$, so the force points inward toward the origin. This makes gravity an attractive force, pulling $m$ toward $M$. Why distractors fail: Option A is wrong because the sign has nothing to do with how force varies with distance (that is governed by the $1/r^2$ factor). Option C is wrong because $G$ is a positive constant. Option D is wrong because the force is purely radial, not perpendicular.

Answer: $GM\left(\frac{1}{b} - \frac{1}{a}\right)$

Set up the work integral: Work done by the gravitational field along a radial path is $W = \int_a^b \mathbf{F} \cdot d\mathbf{r}$. Since $\mathbf{F} = -\frac{GM}{r^2}\hat{\mathbf{r}}$ and $d\mathbf{r} = dr\,\hat{\mathbf{r}}$, we get $W = \int_a^b -\frac{GM}{r^2}\,dr$. Evaluate the integral: $W = -GM\left[-\frac{1}{r}\right]_a^b = -GM\left(-\frac{1}{b} + \frac{1}{a}\right) = GM\left(\frac{1}{b} - \frac{1}{a}\right)$. Interpret and check distractors: Since $a < b$, $1/b < 1/a$, so $W < 0$. This makes sense: moving a mass away from the source against gravity requires work input, so the field does negative work. Option A has the sign reversed. Option C uses an incorrect $\ln$ antiderivative. Option D uses $1/r^2$ instead of the correct $1/r$ antiderivative.

Answer: The student's result is correct; by symmetry, $\mathbf{E}$ is perpendicular to the plane, so the flux through the curved side is zero

Analyze the symmetry: For a uniformly charged infinite plane, the electric field $\mathbf{E}$ must be perpendicular to the plane everywhere by symmetry. Evaluate flux contributions: On the curved surface of the cylinder, $\mathbf{E}$ is parallel to the surface (perpendicular to the outward normal), so $\mathbf{E} \cdot d\mathbf{S} = 0$ there. The only nonzero contributions come from the two flat end caps, each contributing $EA$, giving total flux $2EA = \sigma A / \varepsilon_0$, hence $E = \sigma / (2\varepsilon_0)$. Why distractors fail: Option A incorrectly assumes the curved side contributes flux. Option B is wrong — cylindrical symmetry of the Gaussian surface is the standard and correct choice for planar charge distributions. Option D is wrong because Gauss's Law applies to any charge distribution.

Answer: The total charge enclosed within the surface $S$

Recall the meaning of the subscript: The subscript 'enc' stands for 'enclosed.' Gauss's Law relates the electric flux through a closed surface to the total charge contained inside that surface. Why the correct answer works: Option B correctly identifies $Q_{\text{enc}}$ as the total charge enclosed by the closed surface $S$. Charges outside $S$ do not contribute to the net flux. Why distractors fail: Option A confuses charge on the surface with charge inside. Option C describes a surface charge density $\sigma$, not the enclosed charge. Option D is incorrect because external charges produce zero net flux through a closed surface.