Electricity-And-Magnetism · Unit 1 · Video 7 · Interactive Practice

Electric Flux Through Surfaces and Volume Integrals

IKey Formulas

FormulaNameWhat it says
ΦE≡∬SE⋅dA=∬SEn dA\Phi_E \equiv \iint_S \mathbf{E}\cdot d\mathbf{A} = \iint_S E_n\,dAElectric fluxSum of the normal component of E\mathbf{E} over every patch; dAd\mathbf{A} points outward on a closed surface
En=Ecos⁡αE_n = E\cos\alphaNormal componentα\alpha is the angle between E\mathbf{E} and the normal; En<0E_n \lt 0 when the field enters against dAd\mathbf{A}
∮ ⁣ ⁣∮sphereEn dA=En 4πR2\oint\!\!\oint_{\text{sphere}} E_n\,dA = E_n\,4\pi R^2Constant EnE_n comes outWith En=k/r2E_n = k/r^2 at r=Rr = R: ΦE=4πk\Phi_E = 4\pi k for every radius
Q=∭Vρ dV,dV=dx dy dz=r2sin⁡θ dr dθ dϕQ = \iiint_V \rho\,dV,\quad dV = dx\,dy\,dz = r^2\sin\theta\,dr\,d\theta\,d\phiCharge in a volumeWeight each volume element by the density and add

Key Insight: Wherever symmetry makes the integrand constant on a piece of the surface or volume, it comes outside the integral, and what remains is just an area or a volume.

IIVisualization 1 — The Part of E\mathbf{E} That Crosses the Patch

Only the component of E\mathbf{E} along the normal dAd\mathbf{A} counts: E⋅dA=Ecos⁡α dA\mathbf{E}\cdot d\mathbf{A} = E\cos\alpha\,dA.

IIIVisualization 2 — Flux Through a Sphere Around a Point Charge

On a centred sphere En=k/R2E_n = k/R^2 shrinks exactly as the area 4πR24\pi R^2 grows, so the flux never changes.

IVVisualization 3 — Charge in a Box with ρ=ρ0 x/a\rho = \rho_0\,x/a

Slicing the box into slabs and adding ρ dV\rho\,dV pins the total charge to 12ρ0abc\tfrac{1}{2}\rho_0 abc, half the uniform value.

💡 The box is b=2b = 2 deep with ρ0=1\rho_0 = 1; only the xx integral differs from the uniform case, ∫0aρ0xa dx=12ρ0a\int_0^a \rho_0 \tfrac{x}{a}\,dx = \tfrac{1}{2}\rho_0 a.

VQuiz Questions

Problem 1 · Flux Through a Tilted Patch

Given: A uniform field of magnitude E=200 N/CE = 200\ \text{N/C} crosses a flat patch of area A=0.05 m2A = 0.05\ \text{m}^2, making an angle α=60°\alpha = 60\degree with the patch's normal. Find the flux ΦE\Phi_E through the patch.

✅ Correct! Only En=Ecos⁡60°=100 N/CE_n = E\cos 60\degree = 100\ \text{N/C} crosses the patch, so ΦE=100×0.05=5 N⋅m2/C\Phi_E = 100 \times 0.05 = 5\ \text{N}\cdot\text{m}^2/\text{C}.
❌ Not quite. EA=10EA = 10 would be the flux only if the field crossed head-on. At 60°60\degree to the normal only the normal component Ecos⁡αE\cos\alpha counts.
❌ Close, but check the angle. α\alpha is measured from the normal, so the normal component is Ecos⁡αE\cos\alpha, not Esin⁡αE\sin\alpha (that is the part running along the surface).
❌ Not quite. Flux multiplies the normal component by the area: ΦE=Ecos⁡α⋅A\Phi_E = E\cos\alpha \cdot A.
Show solution

The field is uniform and the patch is flat, so EnE_n is the same everywhere on it and comes out of the integral:

ΦE=∬E⋅dA=En∬dA=Ecos⁡α⋅A\Phi_E = \iint \mathbf{E}\cdot d\mathbf{A} = E_n \iint dA = E\cos\alpha \cdot A ΦE=(200)(cos⁡60°)(0.05)=(200)(0.5)(0.05)=5 N⋅m2/C\Phi_E = (200)(\cos 60\degree)(0.05) = (200)(0.5)(0.05) = 5\ \text{N}\cdot\text{m}^2/\text{C}

Problem 2 · The Closed Cylinder

Given: A field points straight away from a long straight line, with the same strength at the same distance. A closed cylinder of radius R=0.1 mR = 0.1\ \text{m} and length L=0.5 mL = 0.5\ \text{m} encloses the line along its axis, and on the curved side En=1000 N/CE_n = 1000\ \text{N/C}. Find the total flux out of the closed cylinder.

✅ Correct! The caps contribute nothing, and the side gives En⋅2πRL=1000⋅2π(0.1)(0.5)=100πE_n \cdot 2\pi RL = 1000 \cdot 2\pi(0.1)(0.5) = 100\pi.
❌ Close, but… you added the end caps' area 2πR22\pi R^2. On each cap the field runs along the surface, so En=0E_n = 0 there and the caps contribute nothing.
❌ Not quite. πR2L\pi R^2 L is the cylinder's volume, not the area of its curved side. Flux needs an area: 2πRL2\pi R L.
❌ Not quite. En⋅2πRE_n \cdot 2\pi R uses only the circumference; the curved side's area is the circumference times the length, 2πR⋅L2\pi R \cdot L.
❌ Not quite. Break the closed surface into two caps and the curved side, and evaluate each piece.
Show solution

Split the closed surface into its three open pieces:

ΦE=∬cap 1En dA+∬cap 2En dA+∬sideEn dA\Phi_E = \iint_{\text{cap 1}} E_n\,dA + \iint_{\text{cap 2}} E_n\,dA + \iint_{\text{side}} E_n\,dA

On the caps the field is parallel to the surface, so En=0E_n = 0. On the side EnE_n is constant and comes out, leaving the side's area:

ΦE=0+0+En⋅2πRL=1000⋅2π(0.1)(0.5)=100π≈314 N⋅m2/C\Phi_E = 0 + 0 + E_n \cdot 2\pi R L = 1000 \cdot 2\pi (0.1)(0.5) = 100\pi \approx 314\ \text{N}\cdot\text{m}^2/\text{C}

Problem 3 · Charge in a Graded Box

Given: A box with sides a=0.2 ma = 0.2\ \text{m}, b=0.1 mb = 0.1\ \text{m}, c=0.3 mc = 0.3\ \text{m} along xx, yy, zz (one corner at the origin) holds charge of density ρ=ρ0 x/a\rho = \rho_0\,x/a with ρ0=5×10−6 C/m3\rho_0 = 5 \times 10^{-6}\ \text{C/m}^3. Find the total charge QQ.

✅ Correct! Q=12ρ0abc=12(5×10−6)(0.006)=1.5×10−8 CQ = \tfrac{1}{2}\rho_0 abc = \tfrac{1}{2}(5\times10^{-6})(0.006) = 1.5\times10^{-8}\ \text{C}.
❌ Not quite. ρ0abc\rho_0 abc treats the density as uniform. It rises from 00 at x=0x = 0 to ρ0\rho_0 at x=ax = a, so it must stay inside the xx integral.
❌ Not quite. A factor 13\tfrac{1}{3} comes from integrating x2x^2. Here the density is linear in xx: ∫0axa dx=a2\int_0^a \tfrac{x}{a}\,dx = \tfrac{a}{2}.
❌ Close, but… you dropped the 1/a1/a in ρ0x/a\rho_0 x/a: ∫0aρ0xa dx=ρ0a2\int_0^a \rho_0 \tfrac{x}{a}\,dx = \tfrac{\rho_0 a}{2}, not ρ0a22\tfrac{\rho_0 a^2}{2}.
❌ Not quite. Do the xx integral of the density first, then multiply by the yy and zz extents.
Show solution
Q=∫0c∫0b∫0aρ0xa dx dy dzQ = \int_0^c \int_0^b \int_0^a \rho_0 \frac{x}{a}\,dx\,dy\,dz

Innermost (the only one that changes from the uniform case):

∫0aρ0xa dx=ρ0a⋅a22=ρ0a2\int_0^a \rho_0 \frac{x}{a}\,dx = \frac{\rho_0}{a}\cdot\frac{a^2}{2} = \frac{\rho_0 a}{2}

The yy and zz integrals multiply by bb and cc:

Q=12ρ0abc=12(5×10−6)(0.2)(0.1)(0.3)=1.5×10−8 CQ = \tfrac{1}{2}\rho_0 abc = \tfrac{1}{2}(5\times10^{-6})(0.2)(0.1)(0.3) = 1.5\times10^{-8}\ \text{C}

Problem 4 · A Sphere with Rising Density

Given: A solid sphere of radius RR carries charge density ρ=ρ0 r/R\rho = \rho_0\,r/R, rising from 00 at the centre to ρ0\rho_0 at the surface. Find its total charge QQ, using dV=r2sin⁡θ dr dθ dϕdV = r^2\sin\theta\,dr\,d\theta\,d\phi.

✅ Correct! The radial integral gives ρ0R⋅R44\tfrac{\rho_0}{R}\cdot\tfrac{R^4}{4}, and the angles give 2⋅2π=4π2 \cdot 2\pi = 4\pi, so Q=πρ0R3Q = \pi\rho_0 R^3.
❌ Not quite. 43πR3ρ0\tfrac{4}{3}\pi R^3 \rho_0 is the charge if the density were ρ0\rho_0 everywhere. Here it is smaller near the centre, so QQ must come out less.
❌ Close, but… you dropped the sin⁡θ\sin\theta: ∫0πsin⁡θ dθ=2\int_0^\pi \sin\theta\,d\theta = 2, not π\pi.
❌ Not quite. The volume element carries r2r^2. Without it the answer is not even a charge (its units are C/m², not C).
❌ Not quite. Put ρ dV=ρ0rR r2sin⁡θ dr dθ dϕ\rho\,dV = \rho_0 \tfrac{r}{R}\,r^2\sin\theta\,dr\,d\theta\,d\phi into three nested integrals and evaluate each.
Show solution
Q=∫02π∫0π∫0Rρ0rR r2sin⁡θ dr dθ dϕQ = \int_0^{2\pi}\int_0^{\pi}\int_0^{R} \rho_0\frac{r}{R}\,r^2\sin\theta\,dr\,d\theta\,d\phi

The integrand factors, so the three integrals separate:

∫0Rρ0Rr3 dr=ρ0R⋅R44=ρ0R34,∫0πsin⁡θ dθ=2,∫02πdϕ=2π\int_0^R \frac{\rho_0}{R} r^3\,dr = \frac{\rho_0}{R}\cdot\frac{R^4}{4} = \frac{\rho_0 R^3}{4},\qquad \int_0^\pi \sin\theta\,d\theta = 2,\qquad \int_0^{2\pi} d\phi = 2\pi Q=ρ0R34⋅2⋅2π=πρ0R3Q = \frac{\rho_0 R^3}{4}\cdot 2 \cdot 2\pi = \pi\rho_0 R^3

Check: this is 34\tfrac{3}{4} of the uniform value 43πρ0R3\tfrac{4}{3}\pi\rho_0 R^3, sensible because most of a sphere's volume lies near its surface, where ρ\rho is close to ρ0\rho_0.

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