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Topic 1 of 9

Vector Calculus for EM

Electromagnetism is written in the language of vector calculus. Gauss's theorem relates the flux of a vector field through a closed surface to its divergence inside; Stokes's theorem relates the circulation around a closed loop to the curl inside. Mastering these theorems makes Maxwell's equations transparent.

Key Concepts

Gradient

\nabla f=(\partial f/\partial x)\hat x+(\partial f/\partial y)\hat y+(\partial f/\partial z)\hat z

points in the direction of steepest ascent of

f

and has magnitude equal to the rate of increase. The electric field

\vec E=-\nabla V

Divergence

\nabla\cdot\vec A=\partial A_x/\partial x+\partial A_y/\partial y+\partial A_z/\partial z

measures the source strength per unit volume.

\nabla\cdot\vec E=\rho/\varepsilon_0

(Gauss's law in differential form).

Curl

\nabla\times\vec A

measures the rotation per unit area.

\nabla\times\vec E=-\partial\vec B/\partial t

(Faraday's law in differential form). A conservative field has zero curl.

Stokes's & Gauss's Theorems

Gauss:

\oint_S \vec A\cdot d\vec a=\int_V(\nabla\cdot\vec A)d\tau

. Stokes:

\oint_C\vec A\cdot d\vec l=\int_S(\nabla\times\vec A)\cdot d\vec a

. These connect differential to integral forms of Maxwell's equations.

Key Equations

Gauss's theorem

\oint_S \vec{A}\cdot d\vec{a} = \int_V (\nabla\cdot\vec{A})\,d\tau

Converts a surface integral to a volume integral.

Stokes's theorem

\oint_C \vec{A}\cdot d\vec{l} = \int_S (\nabla\times\vec{A})\cdot d\vec{a}

Converts a line integral to a surface integral.

Helmholtz theorem

\vec{A} = -\nabla V + \nabla\times\vec{W}

Any vector field is the sum of an irrotational and a solenoidal part.

Worked Example

Applying the Divergence Theorem

Problem

For the vector field $\vec A=r\hat r$ in spherical coordinates (so $A_r=r$ ), verify the divergence theorem over a sphere of radius $R$ .

Solution

Divergence: $\nabla\cdot\vec A=(1/r^2)\partial(r^2\cdot r)/\partial r=(1/r^2)\cdot3r^2=3$ . Volume integral: $\int_V3\,d\tau=3\cdot(4\pi R^3/3)=4\pi R^3$ .

Surface integral: on sphere $r=R$ , $\vec A\cdot d\vec a=R\cdot4\pi R^2\cdot dr/|dr|$ ... $\vec A\cdot\hat r=R$ , area element $da=R^2\sin\theta\,d\theta\,d\phi$ : $\oint R\,da=R\cdot4\pi R^2=4\pi R^3$ . ✓

Answer Both sides give

4\pi R^3

— the theorem is verified.

Practice

Exercises

7 problems

1 of 7

Explore the gradient field of $f = x^2+y^2+z^2$. Arrows show how $\nabla f$ points uphill — radially outward, growing with $r$. Find |∇f| at the point (1, 2, 2).

gradient field

Explore test radius r |∇f| = 3.00

The gradient always points uphill — radially outward for f=r². Arrows get longer farther from origin since |∇f| = 2r.

|∇f| at (1,2,2)

|∇f| =

Formula: |∇f| = 2√(x²+y²+z²) = 2r

2 of 7

Watch the pulsing Gaussian box: more field exits each face than enters (field grows with position). Find ∇·A for $\vec A = (x, y, z)$.

divergence

Every grid point is a source — arrows get longer away from origin (A grows). The net outward flux per unit volume = divergence.

∇·A for A=(x,y,z)

∇·A =

Formula: ∇·A = ∂Ax/∂x + ∂Ay/∂y + ∂Az/∂z

3 of 7

For $\vec A=y\hat x-x\hat y$ , find $|\nabla\times\vec A|$ .

Unlock Exercise 3