Electricity-And-Magnetism · Unit 1 · Video 6 · Interactive Practice

Closed Surface Integrals: The Sphere and Its Charge

IKey Formulas

Formula	Name	When to use it
$dA = R^2\sin\theta\,d\theta\,d\phi$	Area element on a sphere	$0 \le \theta \le \pi$ (polar, from the north pole), $0 \le \phi \le 2\pi$
$\oint\!\!\oint_{\text{sphere}} dA = \displaystyle\int_0^{2\pi}\!\!\int_0^{\pi} R^2\sin\theta\,d\theta\,d\phi = 4\pi R^2$	Area of a sphere	The loop marks a closed surface
$Q = \oint\!\!\oint \sigma\,dA = \sigma\,4\pi R^2$	Charge on a uniformly charged sphere	$\sigma$ constant, so it comes out of the integral
$Q = \sigma_1\,2\pi R_1^2 + \sigma_2\,2\pi R_2^2 + \sigma_3\,\pi(R_2^2 - R_1^2)$	Closed surface built from pieces	Constant $\sigma$ on each open piece

Key Insight: When $\sigma$ is constant on a surface, $\iint \sigma\,dA = \sigma \times \text{area}$ , and geometry supplies the area: $4\pi R^2$ for a sphere, $2\pi R^2$ for a hemisphere, $\pi(R_2^2 - R_1^2)$ for a ring. When $\sigma$ varies, it stays inside and you integrate with $dA = R^2\sin\theta\,d\theta\,d\phi$ .

IIThe Area Element on a Sphere

A patch of fixed angular size $d\theta \times d\phi$ has area $R^2\sin\theta\,d\theta\,d\phi$ , so it shrinks toward the poles.

💡 Many math texts swap the names and call the polar angle $\phi$ . Here, as in most physics, $\theta$ is the polar angle measured down from the north pole.

IIIAdding Up Every Patch

The $\theta$ integral turns one pole-to-pole wedge into $2R^2\,d\phi$ ; the $\phi$ integral sweeps wedges around to $4\pi R^2$ .

Progress

IVA Closed Surface Built from Three Pieces

Each open piece carries a constant $\sigma$ , so the total charge is a sum of $\sigma \times$ known area.

σ₁ (µC/m²) 2 σ₂ (µC/m²) 1 σ₃ (µC/m²) -1

💡 Each piece alone is an open surface with an edge, so its integral carries a plain $\iint$ ; only the three together enclose a volume and earn the loop $\oint\!\!\oint$ .

VQuiz Questions

Problem 1 · Charge on a Uniform Sphere

Given: A sphere of radius $R = 0.10\ \text{m}$ carries a uniform surface charge density $\sigma = 2.0\ \mu\text{C/m}^2$ . Find the total charge $Q$ .

✅ Correct!

\sigma

is constant, so

Q = \sigma \times 4\pi R^2 = 2.0 \times 0.126 \approx 0.25\ \mu\text{C}

, with no integral to evaluate.

❌ Not quite.

\pi R^2

is the area of a flat disk of radius

R

, a quarter of the sphere's area

4\pi R^2

❌ Not quite.

2\pi R^2

is a hemisphere. The whole closed sphere has area

4\pi R^2

❌ Check the units. Area goes as

R^2

4\pi R

has units of metres, not square metres.

❌ Not quite. Pull the constant

\sigma

out of

\oint\!\!\oint \sigma\,dA

and multiply by the sphere's area.

Show solution

Because $\sigma$ does not depend on $\theta$ or $\phi$ , it comes out of the integral:

Q = \oint\!\!\oint_{\text{sphere}} \sigma\,dA = \sigma \oint\!\!\oint_{\text{sphere}} dA = \sigma\,4\pi R^2

4\pi R^2 = 4\pi (0.10\ \text{m})^2 = 0.04\pi\ \text{m}^2 \approx 0.126\ \text{m}^2

Q = (2.0\ \mu\text{C/m}^2)(0.04\pi\ \text{m}^2) = 0.08\pi\ \mu\text{C} \approx 0.25\ \mu\text{C}

Problem 2 · One Patch on a Sphere

Given: On a sphere of radius $R = 2\ \text{m}$ , a small patch sits at polar angle $\theta = 30\degree$ (measured from the north pole) and spans $d\theta = d\phi = 0.01\ \text{rad}$ . Find its area $dA$ .

✅ Correct!

dA = R^2\sin\theta\,d\theta\,d\phi = 4 \times \tfrac{1}{2} \times 10^{-4} = 2.0 \times 10^{-4}\ \text{m}^2

, half the size of an equator patch.

❌ Missing the $\sin\theta$ . That is the area of an equator patch. At

\theta = 30\degree

the latitude circle has radius

R\sin\theta = R/2

, so the

\phi

side is half as long.

❌ Angle convention.

\cos 30\degree

would be right if the angle were measured from the equator. Here

\theta

is measured from the north pole, and the latitude radius is

R\sin\theta

❌ Check the powers of $R$ . Both sides of the patch carry a factor of

R

(R\,d\theta)(R\sin\theta\,d\phi)

❌ Not quite. Multiply the two sides

R\,d\theta

and

R\sin\theta\,d\phi

Show solution

The patch's sides are an arc of a line of longitude and an arc of a circle of latitude of radius $R\sin\theta$ :

dA = (R\,d\theta)(R\sin\theta\,d\phi) = R^2\sin\theta\,d\theta\,d\phi

dA = (2\ \text{m})^2 \sin 30\degree\,(0.01)(0.01) = 4 \times 0.5 \times 10^{-4} = 2.0 \times 10^{-4}\ \text{m}^2

The same angular patch at the equator ( $\sin 90\degree = 1$ ) would be $4.0 \times 10^{-4}\ \text{m}^2$ ; near the pole the latitude circles shrink.

Problem 3 · Two Domes and a Ring

Given: The closed surface from the video: a hemisphere of radius $R_1 = 1\ \text{m}$ with $\sigma_1 = 3\ \mu\text{C/m}^2$ , a concentric hemisphere of radius $R_2 = 2\ \text{m}$ with $\sigma_2 = -1\ \mu\text{C/m}^2$ , and the flat annulus joining their rims with $\sigma_3 = 2\ \mu\text{C/m}^2$ . Find the total charge $Q$ .

✅ Correct!

6\pi - 8\pi + 6\pi = 4\pi\ \mu\text{C}

: the negative outer dome is outweighed by the inner dome and the ring.

❌ Check the ring's area. An annulus is a big disk minus a small disk:

\pi(R_2^2 - R_1^2) = 3\pi

, not

\pi(R_2 - R_1)^2 = \pi

❌ Check the dome areas. A hemisphere is half a sphere:

2\pi R^2

, not

4\pi R^2

❌ Check the ring's area. The hole is removed:

\pi R_2^2 - \pi R_1^2

, not

\pi R_2^2 + \pi R_1^2

❌ Not quite. Split into three open pieces, pull each constant

\sigma

out, and use the known areas.

Show solution

Split the closed-surface integral into one integral per piece, and pull out each constant density:

Q = \sigma_1\iint dA_1 + \sigma_2\iint dA_2 + \sigma_3\iint dA_3 = \sigma_1\,2\pi R_1^2 + \sigma_2\,2\pi R_2^2 + \sigma_3\,\pi(R_2^2 - R_1^2)

Inner dome: $3 \times 2\pi(1)^2 = 6\pi\ \mu\text{C}$
Outer dome: $(-1) \times 2\pi(2)^2 = -8\pi\ \mu\text{C}$
Ring: $2 \times \pi(4 - 1) = 6\pi\ \mu\text{C}$

Q = 6\pi - 8\pi + 6\pi = 4\pi\ \mu\text{C} \approx 12.6\ \mu\text{C}

Problem 4 · When σ Varies with θ

Given: A sphere of radius $R$ carries surface charge density $\sigma(\theta) = \sigma_0\cos^2\theta$ , largest at the poles and zero on the equator. Find the total charge $Q$ .

✅ Correct!

\sigma

must stay inside:

Q = \sigma_0 R^2 \cdot 2\pi \int_0^{\pi}\cos^2\theta\sin\theta\,d\theta = \sigma_0 R^2 \cdot 2\pi \cdot \tfrac{2}{3}

❌ The shortcut needs a constant $\sigma$ .

\sigma_0\cos^2\theta

depends on

\theta

, so it cannot come out of the integral;

4\pi\sigma_0R^2

is the charge if

\sigma_0

held everywhere.

❌ Missing the $\sin\theta$ . The area element is

R^2\sin\theta\,d\theta\,d\phi

; without

\sin\theta

the patches near the poles are overcounted.

❌ Not quite.

\cos^2\theta

averages to

\tfrac{1}{2}

over

\theta

, but the sphere has less area near the poles; weighted by

\sin\theta

the average over the sphere is

\tfrac{1}{3}

❌ Not quite. Keep

\sigma(\theta)

inside and integrate with

dA = R^2\sin\theta\,d\theta\,d\phi

Show solution

$\sigma$ varies with $\theta$ , so it stays inside the integral with the full area element:

Q = \oint\!\!\oint \sigma\,dA = \int_0^{2\pi}\!\!\int_0^{\pi} \sigma_0\cos^2\theta\,R^2\sin\theta\,d\theta\,d\phi

Inner integral, with $u = \cos\theta$ , $du = -\sin\theta\,d\theta$ :

\int_0^{\pi}\cos^2\theta\sin\theta\,d\theta = \left[-\tfrac{1}{3}\cos^3\theta\right]_0^{\pi} = \tfrac{1}{3} - \left(-\tfrac{1}{3}\right) = \tfrac{2}{3}

Outer integral: nothing depends on $\phi$ , so it contributes $2\pi$ :

Q = \sigma_0 R^2 \cdot \tfrac{2}{3} \cdot 2\pi = \tfrac{4\pi}{3}\sigma_0 R^2

That is one third of $4\pi\sigma_0R^2$ : the area-weighted average of $\cos^2\theta$ over a sphere is $\tfrac{1}{3}$ .

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