play phryzby with physics

The Flux Leaving a Charged Sphere

solved GausssLaw

Q. A uniformly charged conducting sphere of 2.4 mD=2.4m diameter has a surface charge density of 80.0 μC/m²σ=80.0μC/m². What is the total electric flux leaving the surface of the sphere, in N m²/C?

QEΦ = ?
Inputs
D
2.40 m
σ
80 μC/m²
Outputs
fluxΦ?

How it's solved

Solved with GausssLaw

Given

  • D=2.4m
  • σ=80.0μC/m²

Find Φ.

Working
  1. σ=80.0μC/m²=8×10−5C/m²
  2. the surface of a sphere(sphere)A=π⋅D2
  3. A=3.14159265⋅2.42
  4. A=18.0955737m²
  5. charge spread over a surface(sphere)q=σ⋅A
  6. q=8×10−5⋅18.0955737
  7. q=0.00144764589C
  8. gausss law(sphere)Φ=qε0
  9. Φ=0.001447645898.85418781×10−12
  10. Φ=1.635×108N m²/C
  11. Φ≈1.6×108N m²/Cto 2 significant figures
Conclusion

The charge on the sphere is its density times its area, and by Gauss's law all of it sends its flux out through the surface.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :the_flux_leaving_a_charged_sphere do4  kind :numerical5  describes "A uniformly charged conducting sphere of %{diameter} diameter has a surface charge density of " \6            "%{density}. %{flux}, in N m²/C?"7  given diameter: "2.4 m"8  given density: "80.0 μC/m2"9  asks flux: "What is the total electric flux leaving the surface of the sphere"10end
solution.rb
1require_relative "../questions/the_flux_leaving_a_charged_sphere"2require_relative "../science/electric_charges_and_fields/gausss_law"34solve :the_flux_leaving_a_charged_sphere, using: GausssLaw do5  objects :sphere67  take diameter: sphere.diameter8  take density: sphere.surface_charge_density910  flux = sphere.electric_flux11  answer flux:, in: "N m2/C", figures: 21213  conclusion "The charge on the sphere is its density times its area, and by Gauss's law all of it sends its " \14             "flux out through the surface."15end
playground.rb
1require_relative "../solutions/the_flux_leaving_a_charged_sphere"23play_with :the_flux_leaving_a_charged_sphere do4  title "The Flux Leaving a Charged Sphere"5end
The laws it drew on
quantitywrittenunitswithinfrom
piπ—0 … ∞GausssLaw
permittivity of free spaceε0F/m—GausssLaw
field strengthEV/m—GausssLaw
areaAm²—GausssLaw
radiusrm0 … ∞GausssLaw
electric fluxΦN m²/C—GausssLaw
charge insideqC—GausssLaw
diameterDm0 … ∞GausssLaw
surface charge densityσC/m²—GausssLaw
linear charge densityλC/m—GausssLaw
field near a lineElV/m—GausssLaw
field between platesEpV/m—GausssLaw
area of groundAgm²—
radius of groundrgm0 … ∞
area of sphereAm²—
radius of sphererm0 … ∞
electric flux of groundΦgN m²/C—
electric flux of sphereΦN m²/C—
charge inside of groundqgC—
charge inside of sphereqC—
diameter of groundDgm0 … ∞
diameter of sphereDm0 … ∞
surface charge density of groundσgC/m²—
surface charge density of sphereσC/m²—
field near a line of groundEl,gV/m—
linear charge density of groundλgC/m—
field near a line of sphereElV/m—
linear charge density of sphereλC/m—
field between plates of groundEp,gV/m—
field between plates of sphereEpV/m—
Af=π⋅rf2
a round face
Φf=E⋅Af
flux through a face across the field
Φs=qsε0
gausss law
As=π⋅Ds2
the surface of a sphere
qs=σs⋅As
charge spread over a surface
El,l=λl2⋅π⋅ε0⋅rl
field of a long line of charge
Ep,p=σpε0
field between charged plates
Ag=π⋅rg2
a round face(ground)
A=π⋅r2
a round face(sphere)
Φg=E⋅Ag
flux through a face across the field(ground)
Φ=E⋅A
flux through a face across the field(sphere)
Φg=qgε0
gausss law(ground)
Φ=qε0
gausss law(sphere)
Ag=π⋅Dg2
the surface of a sphere(ground)
A=π⋅D2
the surface of a sphere(sphere)
qg=σg⋅Ag
charge spread over a surface(ground)
q=σ⋅A
charge spread over a surface(sphere)
El,g=λg2⋅π⋅ε0⋅rg
field of a long line of charge(ground)
El=λ2⋅π⋅ε0⋅r
field of a long line of charge(sphere)
Ep,g=σgε0
field between charged plates(ground)
Ep=σε0
field between charged plates(sphere)