Electricity-And-Magnetism · Unit 2 · Video 3 · Interactive Practice

Vector Fields, Sources, Sinks and Divergence

IKey Formulas

Formula	Name	Meaning
$\vec v(x,y,z) = v_x\,\hat\imath + v_y\,\hat\jmath + v_z\,\hat k$	Velocity field	One vector per point; each component is a scalar field
$\nabla = \hat\imath\,\dfrac{\partial}{\partial x} + \hat\jmath\,\dfrac{\partial}{\partial y} + \hat k\,\dfrac{\partial}{\partial z}$	Del operator	A vector of partial derivatives
$\nabla\cdot\vec v = \dfrac{\partial v_x}{\partial x} + \dfrac{\partial v_y}{\partial y} + \dfrac{\partial v_z}{\partial z}$	Divergence	Net outflow per unit volume (a scalar)
$\nabla\cdot\vec v > 0$ , $\ \nabla\cdot\vec v < 0$ , $\ \nabla\cdot\vec v = 0$	Source, sink, neither	Incompressible flow: fluid created, removed, or neither

Key Insight: Divergence is the flux out of a tiny closed surface divided by the volume it encloses. Outward arrows alone do not make a source: $\vec v = (x\,\hat\imath + y\,\hat\jmath)/(x^2 + y^2)$ has $\nabla\cdot\vec v = 0$ at every point except the origin.

IIVisualization 1 — Divergence as Net Outflow from a Tiny Box

In a plane flow ( $v_z = 0$ ), the net outflow through a small box's faces per unit area equals $\nabla\cdot\vec v$ .

Field Box side h 1

IIIVisualization 2 — Outward Arrows Are Not Enough

Fluid created between two circles equals the difference of the flows across them; for speed $1/r$ that difference is zero.

Radial flow

IVVisualization 3 — A Source in a Steady Downward Flow

A source of strength $Q$ in a downward stream of speed $U$ fills a rounded cap with its own fluid.

Source strength Q 4

💡 $Q$ is the area of fluid the source creates per second, so far downstream that fluid, moving at speed $U$ , fills a band of width $Q/U$ .

VQuiz Questions

Problem 1 · Compute a Divergence

Given: $\vec v = 3x\,\hat\imath - y\,\hat\jmath + 2z\,\hat k$ — find $\nabla\cdot\vec v$ .

✅ Correct!

3 + (-1) + 2 = 4

, the same at every point, so every point is a source.

❌ Close, but… keep the sign:

\partial v_y/\partial y = \partial(-y)/\partial y = -1

, not

+1

❌ Not quite. The divergence is a scalar: the three partial derivatives are added, not listed as a vector.

❌ Not quite. Differentiate each component with respect to its own variable:

\partial v_x/\partial x

\partial v_y/\partial y

\partial v_z/\partial z

Show solution

With $v_x = 3x$ , $v_y = -y$ , $v_z = 2z$ :

\nabla\cdot\vec v = \frac{\partial (3x)}{\partial x} + \frac{\partial (-y)}{\partial y} + \frac{\partial (2z)}{\partial z} = 3 - 1 + 2 = 4

The divergence is the scalar $4$ everywhere: fluid is created at every point at the same rate per unit volume.

Problem 2 · Outward in the Plane, but a Source?

Given: $\vec v = 2x\,\hat\imath + 2y\,\hat\jmath - 4z\,\hat k$ . In the plane $z = 0$ every arrow points straight away from the $z$ -axis. Classify the point $(1, 1, 0)$ .

✅ Correct! The outflow in

x

and

y

is exactly fed by the inflow along

z

2 + 2 - 4 = 0

❌ Close, but…

2 + 2 = 4

counts only the faces perpendicular to

x

and

y

. Fluid also flows in along

z

\partial v_z/\partial z = -4

❌ Not quite.

-4

is only the

z

-term. The

x

- and

y

-terms add

2 + 2

❌ Not quite. Adding magnitudes ignores the sign of

\partial v_z/\partial z = -4

: inflow along

z

cancels outflow.

Show solution

\nabla\cdot\vec v = \frac{\partial (2x)}{\partial x} + \frac{\partial (2y)}{\partial y} + \frac{\partial (-4z)}{\partial z} = 2 + 2 - 4 = 0

The divergence is zero everywhere, including at $(1, 1, 0)$ , so the point is neither a source nor a sink. Fluid pours in along the $z$ -direction at exactly the rate it spreads out in $x$ and $y$ ; the direction of the arrows in one plane cannot tell you this.

Problem 3 · Divergence at a Point

Given: $\vec v = x^2 y\,\hat\imath + y z^2\,\hat\jmath + x z\,\hat k$ — find $\nabla\cdot\vec v$ at $(1, -2, 3)$ and classify the point.

What is $\nabla\cdot\vec v$ at $(1, -2, 3)$ ?

What is the point $(1, -2, 3)$ ?

✅ Correct!

\nabla\cdot\vec v = 2xy + z^2 + x = -4 + 9 + 1 = 6 > 0

: a source.

❌ Close, but…

\partial(x^2 y)/\partial x = 2xy

, not

xy

: the factor

2

was dropped.

❌ Not quite. That sums

\partial v_x/\partial y + \partial v_y/\partial z + \partial v_z/\partial x

. The divergence pairs each component with its own variable.

❌ Not quite.

2xy = 2(1)(-2) = -4

is negative; keep its sign.

❌ Check the divergence. Differentiate

x^2y

x

yz^2

y

xz

z

, then substitute.

❌ Check the classification. The sign of

\nabla\cdot\vec v

decides it: positive means fluid is created there.

Show solution

Step 1: the three partial derivatives

\frac{\partial (x^2 y)}{\partial x} = 2xy,\qquad \frac{\partial (y z^2)}{\partial y} = z^2,\qquad \frac{\partial (x z)}{\partial z} = x

Step 2: add and substitute $(x, y, z) = (1, -2, 3)$ :

\nabla\cdot\vec v = 2xy + z^2 + x = 2(1)(-2) + 3^2 + 1 = -4 + 9 + 1 = 6

Step 3: classify. $6 > 0$ , so $(1, -2, 3)$ is a source.

Problem 4 · A Point Source in Three Dimensions

Given: a flow spreads from the origin in all three dimensions, $\vec v = c\,r^n\,\hat r$ with $c > 0$ , where $r$ is the distance from the origin. Find the power $n$ for which $\nabla\cdot\vec v = 0$ at every point except the origin.

✅ Correct! Speed

\propto 1/r^2

against sphere area

4\pi r^2

: the same flux crosses every sphere.

❌ Close, but…

1/r

is the answer in a plane, where a circle's length grows as

r

. A sphere's area grows as

r^2

❌ Not quite. At constant speed the flux

4\pi r^2 c

grows with

r

, so fluid is created everywhere:

\nabla\cdot\vec v = 2c/r

❌ Not quite. Then the flux

4\pi c/r

shrinks with

r

: fluid is removed everywhere,

\nabla\cdot\vec v = -c/r^4

Show solution

The flux out through a sphere of radius $r$ is speed times area:

\Phi(r) = c\,r^n \cdot 4\pi r^2 = 4\pi c\, r^{n+2}

No fluid is created between two spheres exactly when $\Phi$ is the same for every $r$ , so $n + 2 = 0$ , giving $n = -2$ . In components, $\vec v = c\,(x, y, z)/r^3$ , and

\nabla\cdot\vec v = c\left(\frac{3}{r^3} - \frac{3(x^2 + y^2 + z^2)}{r^5}\right) = 0 \quad (r \ne 0).

All the fluid is created at the origin, which is the only source.

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