本卷参照期中模拟卷（Mock Test）的难度和题型编写，覆盖第 2–5 章，共 4 题，建议用时 90 分钟。简答题参考答案使用简短英语。

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Question 1 (25 marks) — Vector Analysis and Boundary Conditions

(a) (5 marks) Describe the concepts of flux and divergence of a vector field. State the mathematical definition of each and explain their physical relationship.

(b) (5 marks) Describe the concepts of circulation and curl of a vector field. State the mathematical definition of each and explain their physical relationship.

(c) (5 marks) Given the scalar field $\varphi = 2x^2y + y^3 - z^2$, find the gradient $\nabla\varphi$ at the point $P(1, 2, 3)$ in Cartesian coordinates.

(d) (10 marks) Two dielectric media meet at the XOZ plane ($y = 0$). Medium 1 ($y > 0$): $\varepsilon_1 = 3\varepsilon_0$; Medium 2 ($y < 0$): $\varepsilon_2 = 9\varepsilon_0$. There is no free charge on the interface. The electric field in Medium 1 is $\vec{E}_1 = \vec{e}_x \cdot 3 + \vec{e}_y \cdot 12 + \vec{e}_z \cdot 6$ (V/m). Find $\vec{E}_2$ in Medium 2.

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Question 2 (25 marks) — Electrostatic Field Fundamentals

(a) (12 marks) The electric potential in a region is $\varphi = 3x^2y - y^3 + 2z$ (SI units). Find:

1. The electric field $\vec{E}$ at the point $(1, 2, 3)$;
2. The volume charge density $\rho_v$ at the point $(1, 2, 3)$.

Use $\varepsilon_0 = 8.854 \times 10^{-12}$ F/m.

(b) (8 marks) Prove that for a linear, homogeneous, isotropic dielectric with permittivity $\varepsilon$, the polarization volume charge density is:

$\rho_p = -\left(1 - \frac{\varepsilon_0}{\varepsilon}\right)\rho_f$

where $\rho_f$ is the free volume charge density. State the key assumptions used in the proof.

(c) (5 marks) Given four electrostatic field line diagrams (labeled 1–4), identify the type of source near the centre for each:

1. Field lines radiate radially outward in all directions from a point, with equal spacing at equal distances.
2. Field lines form closed curves emerging from one point and returning to a nearby point, resembling a “figure-eight” pattern in the plane.
3. Field lines radiate outward uniformly in all radial directions from a central axis (cylindrical symmetry).
4. Field lines converge radially inward toward a point from all directions.

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Question 3 (25 marks) — Method of Images

(a) (8 marks) A point charge $q$ is placed at distance $d$ from the centre of a grounded conducting sphere of radius $a$ ($d > a$). Determine the image charge $q'$ and its position $b$. Using these results, derive the induced surface charge density $\sigma(\theta)$ on the sphere, where $\theta$ is the polar angle measured from the line connecting the sphere centre to $q$.

(b) (8 marks) Two semi-infinite grounded conducting planes meet at an angle of $60°$ ($\pi/3$) along the $z$-axis. A point charge $+q$ is placed at position $(x_0, y_0) = (\sqrt{3}, 1)$ in the $xy$-plane (inside the wedge). Determine the number of image charges required, and give the positions and magnitudes of all image charges.

(c) (5 marks) A grounded conducting plane ($z = 0$) has a conducting hemispherical boss of radius $a$ on it (hemisphere in $z > 0$, centred at the origin). A point charge $q$ is placed at $(0, 0, d)$ on the $z$-axis ($d > a$). Find all image charges needed to solve for the potential in the region $z > 0$, $r > a$.

(d) (4 marks) Describe one practical engineering application of the method of images. Explain how the technique is applied and what it simplifies.

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Question 4 (25 marks) — Steady Current in Lossy Dielectrics

A parallel-plate capacitor has two flat electrodes of area $S$ separated by distance $d$. The space between the electrodes is filled with two lossy dielectric layers, each of thickness $d/2$, stacked vertically (series configuration):

• Layer 1 (top, $0 < y < d/2$): permittivity $\varepsilon_1$, conductivity $\sigma_1$
• Layer 2 (bottom, $d/2 < y < d$): permittivity $\varepsilon_2$, conductivity $\sigma_2$

The top electrode ($y = 0$) is at potential $V_0$; the bottom electrode ($y = d$) is grounded ($\varphi = 0$). Assume steady-state DC conditions.

(a) (6 marks) State the boundary conditions at the dielectric–dielectric interface ($y = d/2$) and at the electrode–dielectric surfaces.

(b) (8 marks) Find the electric field $\vec{E}$ in each layer.

(c) (5 marks) Calculate the leakage resistance $R$ of this structure.

(d) (6 marks) Determine if there is free charge on the dielectric interface at $y = d/2$. If yes, calculate the surface charge density $\rho_s$.

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Reference Solutions

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Question 1 — Solutions

(a) Flux and Divergence

Flux is the net flow of a vector field through a surface, defined as $\Phi = \oint_S \vec{F} \cdot d\vec{S}$. For a closed surface, positive net flux indicates a net source enclosed (more field lines exit than enter); negative net flux indicates a net sink.

Divergence is a scalar quantity at each point measuring the local outward flow per unit volume: $\text{div}\,\vec{F} = \nabla \cdot \vec{F}$. Positive divergence means the point acts as a source; negative means a sink; zero means the field is solenoidal (divergence-free).

The divergence theorem connects them:

$\oint_S \vec{F} \cdot d\vec{S} = \int_V (\nabla \cdot \vec{F})\, dV$

The total flux through any closed surface equals the volume integral of divergence inside. In electrostatics, $\nabla \cdot \vec{D} = \rho_f$ (Gauss’s law) — electric flux originates from free charges.

(b) Circulation and Curl

Circulation is the line integral of a vector field around a closed contour: $C = \oint_C \vec{F} \cdot d\vec{l}$. Nonzero circulation indicates the field has a rotational component along that path.

Curl is a vector quantity at each point measuring the local tendency to rotate: $\text{curl}\,\vec{F} = \nabla \times \vec{F}$. Its direction gives the axis of maximum rotation (right-hand rule), and its magnitude gives the maximum circulation per unit area.

Stokes’ theorem connects them:

$\oint_C \vec{F} \cdot d\vec{l} = \int_S (\nabla \times \vec{F}) \cdot d\vec{S}$

The circulation around any closed contour equals the flux of the curl through any surface bounded by that contour. In electrostatics, $\nabla \times \vec{E} = 0$ (conservative field) — the circulation around any closed loop is zero, which is why we can define a scalar potential $\varphi$ via $\vec{E} = -\nabla\varphi$.

(c) Gradient

The gradient in Cartesian coordinates:

$\nabla\varphi = \vec{e}_x \frac{\partial\varphi}{\partial x} + \vec{e}_y \frac{\partial\varphi}{\partial y} + \vec{e}_z \frac{\partial\varphi}{\partial z}$

Computing each partial derivative of $\varphi = 2x^2y + y^3 - z^2$:

$\frac{\partial\varphi}{\partial x} = 4xy, \quad \frac{\partial\varphi}{\partial y} = 2x^2 + 3y^2, \quad \frac{\partial\varphi}{\partial z} = -2z$

At $P(1, 2, 3)$:

$\nabla\varphi\big|_P = \vec{e}_x(4 \cdot 1 \cdot 2) + \vec{e}_y(2 \cdot 1^2 + 3 \cdot 2^2) + \vec{e}_z(-2 \cdot 3)$

$\boxed{\nabla\varphi\big|_P = 8\vec{e}_x + 14\vec{e}_y - 6\vec{e}_z}$

(d) Boundary Conditions at Dielectric Interface

The interface is the plane $y = 0$, so the unit normal is $\vec{n} = \vec{e}_y$.

Tangential components ($x$ and $z$ directions): From $E_{1t} = E_{2t}$ (tangential $\vec{E}$ is continuous):

$E_{2x} = E_{1x} = 3 \text{ V/m}, \quad E_{2z} = E_{1z} = 6 \text{ V/m}$

Normal component ($y$ direction): With $\rho_s = 0$, $D_{1n} = D_{2n}$, i.e. $\varepsilon_1 E_{1y} = \varepsilon_2 E_{2y}$:

$E_{2y} = \frac{\varepsilon_1}{\varepsilon_2} E_{1y} = \frac{3\varepsilon_0}{9\varepsilon_0} \cdot 12 = 4 \text{ V/m}$

$\boxed{\vec{E}_2 = 3\vec{e}_x + 4\vec{e}_y + 6\vec{e}_z \text{ (V/m)}}$

Discussion: The normal component of $\vec{E}$ decreased (from 12 to 4 V/m) because $\vec{D}$ is continuous but $\varepsilon_2 > \varepsilon_1$. Going from a low-$\varepsilon$ to a high-$\varepsilon$ medium, the normal $\vec{E}$ component is reduced (the field “refracts” away from the normal).

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Question 2 — Solutions

(a) Electric Field and Volume Charge Density

Electric field from $\vec{E} = -\nabla\varphi$:

$E_x = -\frac{\partial\varphi}{\partial x} = -6xy$

$E_y = -\frac{\partial\varphi}{\partial y} = -(3x^2 - 3y^2) = -3x^2 + 3y^2$

$E_z = -\frac{\partial\varphi}{\partial z} = -2$

At $(1, 2, 3)$:

$\vec{E} = (-6 \cdot 1 \cdot 2)\vec{e}_x + (-3 \cdot 1 + 3 \cdot 4)\vec{e}_y + (-2)\vec{e}_z$

$\boxed{\vec{E} = -12\vec{e}_x + 9\vec{e}_y - 2\vec{e}_z \text{ (V/m)}}$

Volume charge density from Poisson’s equation $\nabla^2\varphi = -\rho_v/\varepsilon_0$:

$\frac{\partial^2\varphi}{\partial x^2} = 6y, \quad \frac{\partial^2\varphi}{\partial y^2} = -6y, \quad \frac{\partial^2\varphi}{\partial z^2} = 0$

$\nabla^2\varphi = 6y + (-6y) + 0 = 0$

The $y$-dependent terms cancel exactly: $\nabla^2\varphi = 0$ everywhere. The potential satisfies Laplace’s equation, so the region is source-free. However, note that this only holds inside the charge-free region — charges must exist on the boundaries to maintain this potential distribution.

$\boxed{\rho_v = 0 \text{ everywhere (charge-free region)}}$

Remark: If the potential had been $\varphi = 3x^2y + y^3 + 2z$ (note the sign change on $y^3$), then $\nabla^2\varphi = 6y + 6y + 0 = 12y$, giving $\rho_v = -12y\varepsilon_0$, and at $(1,2,3)$: $\rho_v = -24\varepsilon_0 \approx -2.13 \times 10^{-10}$ C/m$^3$. Always check whether the Laplacian actually vanishes.

(b) Proof: Polarization Volume Charge Density

In a linear, homogeneous, isotropic dielectric:

• Constitutive relation: $\vec{D} = \varepsilon\vec{E}$
• Polarization: $\vec{P} = \vec{D} - \varepsilon_0\vec{E} = (\varepsilon - \varepsilon_0)\vec{E}$

From Gauss’s law: $\nabla \cdot \vec{D} = \rho_f$

The polarization charge density is: $\rho_p = -\nabla \cdot \vec{P}$

$\rho_p = -\nabla \cdot [(\varepsilon - \varepsilon_0)\vec{E}] = -(\varepsilon - \varepsilon_0)\nabla \cdot \vec{E}$

Here $\varepsilon$ is pulled out because the dielectric is homogeneous (spatially uniform), so it is not affected by the divergence operator.

Since $\nabla \cdot \vec{E} = \nabla \cdot (\vec{D}/\varepsilon) = \rho_f/\varepsilon$:

$\rho_p = -(\varepsilon - \varepsilon_0)\frac{\rho_f}{\varepsilon}$

$\boxed{\rho_p = -\left(1 - \frac{\varepsilon_0}{\varepsilon}\right)\rho_f}$

Key assumptions: (1) linear ($\vec{P}$ proportional to $\vec{E}$); (2) homogeneous ($\varepsilon$ independent of position); (3) isotropic ($\varepsilon$ is a scalar, not a tensor).

(c) Field Line Identification

1. Diverging radial lines from a point → Positive point charge (source). Field lines originate from the charge and extend radially outward; field decays as $1/r^2$.

2. Closed curved lines emerging from one point and returning to a nearby point → Electric dipole. Field lines start at the positive charge and end at the negative charge; far-field decays as $1/r^3$.

3. Radially outward lines from a central axis, uniform along the axis → Infinite line of charge. Cylindrical symmetry; field decays as $1/\rho$ (cylindrical coordinate).

4. Converging radial lines toward a point from all directions → Negative point charge (sink). Field lines come from infinity and terminate on the charge; field decays as $1/r^2$.

Identification criteria: Symmetry (spherical/cylindrical/axial), direction (outward = positive source, inward = negative sink), and density decay rate ($1/r^2$, $1/r$, or $1/r^3$) reveal the source type.

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Question 3 — Solutions

(a) Point Charge + Grounded Conducting Sphere

Image charge: To make the sphere surface an equipotential ($\varphi = 0$), place a single image charge inside the sphere:

$\boxed{q' = -\frac{a}{d}q, \quad b = \frac{a^2}{d}}$

The image $q'$ is on the line from the centre to $q$, at distance $b$ from the centre ($b < a$ since $d > a$).

Derivation of surface charge density: At a point on the sphere with polar angle $\theta$, the distances to $q$ (at distance $d$) and $q'$ (at distance $b$) are:

$R_1 = \sqrt{a^2 + d^2 - 2ad\cos\theta}, \quad R_2 = \sqrt{a^2 + b^2 - 2ab\cos\theta}$

Substituting $b = a^2/d$:

$R_2 = \sqrt{a^2 + \frac{a^4}{d^2} - \frac{2a^3}{d}\cos\theta} = \frac{a}{d}\sqrt{d^2 + a^2 - 2ad\cos\theta} = \frac{a}{d}R_1$

The surface charge density is $\sigma = -\varepsilon_0 \dfrac{\partial\varphi}{\partial r}\bigg|_{r=a}$. For two point charges, the radial derivative at $r = a$ gives:

$\sigma(\theta) = \frac{q}{4\pi}\left[\frac{a - d\cos\theta}{R_1^3}\right] + \frac{q'}{4\pi}\left[\frac{a - b\cos\theta}{R_2^3}\right]$

Substituting $q' = -aq/d$ and $R_2 = (a/d)R_1$:

$\frac{q'}{R_2^3} = \frac{-aq/d}{(a/d)^3 R_1^3} = \frac{-qd^2}{a^2 R_1^3}$

$\sigma(\theta) = \frac{q}{4\pi R_1^3}\left[(a - d\cos\theta) - \frac{d^2}{a^2}\left(a - \frac{a^2}{d}\cos\theta)\right)\right]$

$= \frac{q}{4\pi R_1^3}\left[(a - d\cos\theta) - \frac{d^2}{a} + d\cos\theta\right]$

$= \frac{q}{4\pi R_1^3}\left[a - \frac{d^2}{a}\right] = \frac{q(a^2 - d^2)}{4\pi a R_1^3}$

$\boxed{\sigma(\theta) = -\frac{q(d^2 - a^2)}{4\pi a(a^2 + d^2 - 2ad\cos\theta)^{3/2}}}$

The induced charge is negative (opposite sign to $q$), as expected. The total induced charge equals $q' = -aq/d$.

(b) Two Semi-Infinite Grounded Conducting Planes at $\pi/3$

Number of image charges: The general rule for two planes meeting at angle $\alpha = \pi/n$ ($n$ integer) requires $2n - 1$ image charges.

For $\alpha = 60° = \pi/3$: $n = 3$, so we need $\boxed{5}$ image charges.

Setup: The charge $q_0$ at $(\sqrt{3}, 1)$ has polar coordinates $(r, \phi) = (2, 30°)$.

• Plane 1: $\phi = 0°$ (positive $x$-axis, $y = 0$, $x > 0$)
• Plane 2: $\phi = 60°$ ($y = x\tan 60° = x\sqrt{3}$)

All 6 charges (original + 5 images) lie on a circle of radius 2, arranged as a regular hexagon with alternating signs:

Charge	Polar coords	Cartesian coords	Magnitude	How obtained
$q_0$ (original)	$(2, 30°)$	$(\sqrt{3},\, 1)$	$+q$	Original charge
$q_1$	$(2, -30°)$	$(\sqrt{3},\, -1)$	$-q$	Reflection of $q_0$ across Plane 1
$q_2$	$(2, 90°)$	$(0,\, 2)$	$-q$	Reflection of $q_0$ across Plane 2
$q_3$	$(2, 150°)$	$(-\sqrt{3},\, 1)$	$+q$	Reflection of $q_1$ across Plane 2
$q_4$	$(2, -90°)$	$(0,\, -2)$	$+q$	Reflection of $q_2$ across Plane 1
$q_5$	$(2, 210°)$	$(-\sqrt{3},\, -1)$	$-q$	Cross-reflection (same point reached from either plane)

Sign pattern: Alternating $+q, -q$ around the hexagon.

Verification: On Plane 1 ($y = 0$): every charge at $(r, \phi)$ with magnitude $Q$ has a mirror at $(r, -\phi)$ with magnitude $-Q$; these cancel on the plane. On Plane 2 ($\phi = 60°$): every charge at $(r, 60° + \delta)$ has a mirror at $(r, 60° - \delta)$ with magnitude $-Q$; these also cancel. Both boundary conditions ($\varphi = 0$) are satisfied simultaneously.

(c) Hemisphere on Grounded Plane

Two boundary conditions must be satisfied simultaneously:

1. Plane ($z = 0$): $\varphi = 0$
2. Hemisphere ($r = a$, $z > 0$): $\varphi = 0$

Three image charges are needed:

Image	Charge	Position	Purpose
$q_1$	$-q$	$(0,\, 0,\, -d)$	Image of $q$ w.r.t. plane $z = 0$
$q_2$	$-\dfrac{a}{d}q$	$(0,\, 0,\, a^2/d)$	Image of $q$ w.r.t. sphere $r = a$
$q_3$	$+\dfrac{a}{d}q$	$(0,\, 0,\, -a^2/d)$	Image of $q_2$ w.r.t. plane $z = 0$

Verification:

• Plane condition ($z = 0$): $(q, q_1)$ are a mirror pair with equal and opposite charges; $(q_2, q_3)$ likewise. Each pair creates zero potential on the plane. Check.
• Hemisphere condition ($r = a$): $(q, q_2)$ form a standard grounded sphere image pair — their combined potential is zero on $r = a$. $(q_1, q_3)$: since $q_1 = -q$ at distance $d$ and $q_3 = +(a/d)q$ at distance $a^2/d$, we have $q_3 = -(a/d)q_1$ at distance $a^2/d$ — also a valid sphere image pair for $q_1$. Check.

Valid region: $z > 0$ and $r > a$ (outside the hemisphere, above the plane).

(d) Practical Application

Overhead power transmission line capacitance calculation.

An overhead transmission line (conductor at height $h$ above ground) forms a capacitor with the earth. The earth, being a reasonably good conductor, can be approximated as a grounded infinite conducting plane. By the method of images, the effect of the earth is replaced by an imaginary conductor carrying line charge density $-\rho_l$ at depth $h$ below the surface.

This transforms the problem into calculating the capacitance between two parallel line charges separated by distance $2h$, which has the analytical solution:

$C_{\text{unit length}} = \frac{\pi\varepsilon_0}{\ln(2h/a)}$

where $a$ is the conductor radius. Without the method of images, one would need to solve Laplace’s equation with the irregular boundary of the ground surface — far more complex.

Other applications: (1) EMC/shielding analysis — modeling ground plane effects on electromagnetic interference; (2) IC interconnect parasitic capacitance — computing wire-to-substrate capacitance; (3) Lightning protection — estimating field enhancement near a lightning rod above ground.

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Question 4 — Solutions

(a) Boundary Conditions

At the dielectric–dielectric interface ($y = d/2$):

The interface normal is $\vec{n} = \vec{e}_y$ (pointing from Layer 2 toward Layer 1).

1. Normal $\vec{J}$ continuous: From $\nabla \cdot \vec{J} = 0$ in steady state (pillbox across the interface):

$J_{1y} = J_{2y} \quad \Longrightarrow \quad \sigma_1 E_1 = \sigma_2 E_2$

2. Tangential $\vec{E}$ continuous: From $\nabla \times \vec{E} = 0$ (rectangular contour across the interface):

$E_{1t} = E_{2t}$

For this parallel-plate geometry, $\vec{E}$ is purely in the $y$-direction, so there are no tangential components — this condition is automatically satisfied.

3. Normal $\vec{D}$ discontinuity: $D_{1y} - D_{2y} = \rho_s$ (free surface charge exists at the interface):

$\varepsilon_1 E_1 - \varepsilon_2 E_2 = \rho_s$

At the electrode surfaces:

• Top electrode ($y = 0$, $\varphi = V_0$): surface charge $\rho_s(0) = \varepsilon_1 E_1$
• Bottom electrode ($y = d$, $\varphi = 0$): surface charge $\rho_s(d) = \varepsilon_2 E_2$

(b) Electric Field in Each Layer

Let $E_1$ and $E_2$ denote the magnitudes of $\vec{E}$ in Layers 1 and 2 (both pointing in the $-\vec{e}_y$ direction, from high to low potential).

From $\vec{J}$ continuity:

\sigma_1 E_1 = \sigma_2 E_2 \quad \Rightarrow \quad E_2 = \frac{\sigma_1}{\sigma_2} E_1 \tag{1}

From the voltage constraint:

E_1 \cdot \frac{d}{2} + E_2 \cdot \frac{d}{2} = V_0 \tag{2}

Substituting (1) into (2):

$E_1 \cdot \frac{d}{2}\left(1 + \frac{\sigma_1}{\sigma_2}\right) = V_0$

$\boxed{E_1 = \frac{2V_0}{d} \cdot \frac{\sigma_2}{\sigma_1 + \sigma_2}, \quad E_2 = \frac{2V_0}{d} \cdot \frac{\sigma_1}{\sigma_1 + \sigma_2}}$

In vector form: $\vec{E}_1 = -E_1\vec{e}_y$ and $\vec{E}_2 = -E_2\vec{e}_y$.

Discussion: The field is stronger in the layer with lower conductivity (higher resistivity). This is analogous to resistive voltage division: the higher-resistance layer drops more voltage per unit thickness.

(c) Leakage Resistance

The current density in steady state:

$J = \sigma_1 E_1 = \frac{2V_0}{d} \cdot \frac{\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$

Total leakage current: $I = J \cdot S$

$R = \frac{V_0}{I} = \frac{d}{2S} \cdot \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2} = \frac{d}{2S}\left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right)$

$\boxed{R = \frac{d}{2S\sigma_1} + \frac{d}{2S\sigma_2}}$

This is the series combination of two resistors: $R_1 = (d/2)/(\sigma_1 S)$ and $R_2 = (d/2)/(\sigma_2 S)$, as expected for a series geometry.

(d) Free Surface Charge Density at the Interface

From the normal $\vec{D}$ boundary condition:

$\rho_s = D_{1y} - D_{2y} = \varepsilon_1 E_1 - \varepsilon_2 E_2$

Substituting the electric fields from part (b):

$\rho_s = \varepsilon_1 \cdot \frac{2V_0}{d} \cdot \frac{\sigma_2}{\sigma_1 + \sigma_2} - \varepsilon_2 \cdot \frac{2V_0}{d} \cdot \frac{\sigma_1}{\sigma_1 + \sigma_2}$

$\boxed{\rho_s = \frac{2V_0}{d(\sigma_1 + \sigma_2)}\left(\varepsilon_1\sigma_2 - \varepsilon_2\sigma_1\right)}$

Discussion:

• $\rho_s = 0$ when $\varepsilon_1/\sigma_1 = \varepsilon_2/\sigma_2$ (both materials have the same relaxation time $\tau = \varepsilon/\sigma$).
• $\rho_s > 0$ when $\varepsilon_1/\sigma_1 > \varepsilon_2/\sigma_2$ (Layer 1 has a longer relaxation time).
• Physically, even in DC steady state, free charge accumulates at the dielectric interface because the different $\varepsilon/\sigma$ ratios cause unequal charge flux into the interface region. This is a key distinction from the purely electrostatic (no conduction) case, where interface charge is zero when there is no externally applied surface charge.

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Appendix: Reference Formulas

Coordinate Systems

	Cartesian $(x, y, z)$	Cylindrical $(\rho, \phi, z)$	Spherical $(r, \theta, \phi)$
$\nabla f$	$\vec{e}_x\dfrac{\partial f}{\partial x} + \vec{e}_y\dfrac{\partial f}{\partial y} + \vec{e}_z\dfrac{\partial f}{\partial z}$	$\vec{e}_\rho\dfrac{\partial f}{\partial \rho} + \vec{e}_\phi\dfrac{1}{\rho}\dfrac{\partial f}{\partial \phi} + \vec{e}_z\dfrac{\partial f}{\partial z}$	$\vec{e}_r\dfrac{\partial f}{\partial r} + \vec{e}_\theta\dfrac{1}{r}\dfrac{\partial f}{\partial \theta} + \vec{e}_\phi\dfrac{1}{r\sin\theta}\dfrac{\partial f}{\partial \phi}$

Divergence

$\nabla \cdot \vec{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$

$\nabla \cdot \vec{A} = \frac{1}{\rho}\frac{\partial(\rho A_\rho)}{\partial \rho} + \frac{1}{\rho}\frac{\partial A_\phi}{\partial \phi} + \frac{\partial A_z}{\partial z}$

$\nabla \cdot \vec{A} = \frac{1}{r^2}\frac{\partial(r^2 A_r)}{\partial r} + \frac{1}{r\sin\theta}\frac{\partial(\sin\theta\, A_\theta)}{\partial \theta} + \frac{1}{r\sin\theta}\frac{\partial A_\phi}{\partial \phi}$

Curl

$\nabla \times \vec{A} = \begin{vmatrix} \vec{e}_x & \vec{e}_y & \vec{e}_z \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}$

$\nabla \times \vec{A} = \frac{1}{\rho}\begin{vmatrix} \vec{e}_\rho & \rho\vec{e}_\phi & \vec{e}_z \\ \dfrac{\partial}{\partial \rho} & \dfrac{\partial}{\partial \phi} & \dfrac{\partial}{\partial z} \\ A_\rho & \rho A_\phi & A_z \end{vmatrix}$

$\nabla \times \vec{A} = \frac{1}{r^2\sin\theta}\begin{vmatrix} \vec{e}_r & r\vec{e}_\theta & r\sin\theta\,\vec{e}_\phi \\ \dfrac{\partial}{\partial r} & \dfrac{\partial}{\partial \theta} & \dfrac{\partial}{\partial \phi} \\ A_r & rA_\theta & r\sin\theta\, A_\phi \end{vmatrix}$

Laplacian

$\nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}$

$\nabla^2 f = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial f}{\partial \rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2}$

$\nabla^2 f = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial f}{\partial \theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial \phi^2}$