Flux Through a Square

solved GausssLaw

Q. Consider a uniform electric field E = 3 × 10³ N/C $E = 3 \times 10^{3}$ N/C along x. What is the flux of this field through a square of 10 cm on a side $A = 100$ cm² whose plane is parallel to the yz plane, in N m²/C?

Inputs

field

E

3000 N/C

Outputs

flux

Φ

How it's solved

Solved with GausssLaw

Given

$E = 3 \times 10^{3}$ N/C
$A = 100$ cm² from “a square of 10 cm on a side”

Find $Φ$ .

Working

$E = 3 \times 10^{3}$ N/C $= 3000$ V/m
$A = 100$ cm² $= 0.01$ m²
flux through a face across the field(square) $Φ = E \cdot A$
$Φ = 3000 \cdot 0.01$
$Φ = 30$ N m²/C
$Φ \approx 30$ N m²/Cto 2 significant figures

Conclusion

A plane parallel to the yz plane stands square across a field along x, so the whole of its area catches the field.

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 :flux_through_a_square do4  kind :numerical5  describes "Consider a uniform electric field E = %{field} along x. %{flux} of this field through a square of " \6            "10 cm on a side whose plane is parallel to the yz plane, in N m²/C?"7  given field: "3e3 N/C"8  asks flux: "What is the flux"9end

solution.rb

1require_relative "../questions/flux_through_a_square"2require_relative "../science/electric_charges_and_fields/gausss_law"34solve :flux_through_a_square, using: GausssLaw do5  objects :square67  take field: E89  deduce_from "a square of 10 cm on a side", square.area => "100 cm2"1011  flux = square.electric_flux12  answer flux:, in: "N m2/C", figures: 21314  conclusion "A plane parallel to the yz plane stands square across a field along x, so the whole of its area " \15             "catches the field."16end

playground.rb

1require_relative "../solutions/flux_through_a_square"23play_with :flux_through_a_square do4  title "Flux Through a Square"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	GausssLaw
permittivity of free space	$ε_{0}$	F/m	—	GausssLaw
field strength	$E$	V/m	—	GausssLaw
area	$A$	m²	—	GausssLaw
radius	$r$	m	0 … ∞	GausssLaw
electric flux	$Φ$	N m²/C	—	GausssLaw
charge inside	$q$	C	—	GausssLaw
diameter	$D$	m	0 … ∞	GausssLaw
surface charge density	$σ$	C/m²	—	GausssLaw
linear charge density	$λ$	C/m	—	GausssLaw
field near a line	$E_{l}$	V/m	—	GausssLaw
field between plates	$E_{p}$	V/m	—	GausssLaw
area of ground	$A_{g}$	m²	—
radius of ground	$r_{g}$	m	0 … ∞
area of square	$A$	m²	—
radius of square	$r$	m	0 … ∞
electric flux of ground	$Φ_{g}$	N m²/C	—
electric flux of square	$Φ$	N m²/C	—
charge inside of ground	$q_{g}$	C	—
charge inside of square	$q$	C	—
diameter of ground	$D_{g}$	m	0 … ∞
diameter of square	$D$	m	0 … ∞
surface charge density of ground	$σ_{g}$	C/m²	—
surface charge density of square	$σ$	C/m²	—
field near a line of ground	$E_{l,g}$	V/m	—
linear charge density of ground	$λ_{g}$	C/m	—
field near a line of square	$E_{l}$	V/m	—
linear charge density of square	$λ$	C/m	—
field between plates of ground	$E_{p,g}$	V/m	—
field between plates of square	$E_{p}$	V/m	—

A_{f} = π \cdot {r_{f}}^{2}

a round face

Φ_{f} = E \cdot A_{f}

flux through a face across the field

Φ_{s} = \frac{q_{s}}{ε_{0}}

gausss law

A_{s} = π \cdot {D_{s}}^{2}

the surface of a sphere

q_{s} = σ_{s} \cdot A_{s}

charge spread over a surface

E_{l,l} = \frac{λ_{l}}{2 \cdot π \cdot ε_{0} \cdot r_{l}}

field of a long line of charge

E_{p,p} = \frac{σ_{p}}{ε_{0}}

field between charged plates

A_{g} = π \cdot {r_{g}}^{2}

a round face(ground)

A = π \cdot r^{2}

a round face(square)

Φ_{g} = E \cdot A_{g}

flux through a face across the field(ground)

Φ = E \cdot A

flux through a face across the field(square)

Φ_{g} = \frac{q_{g}}{ε_{0}}

gausss law(ground)

Φ = \frac{q}{ε_{0}}

gausss law(square)

A_{g} = π \cdot {D_{g}}^{2}

the surface of a sphere(ground)

A = π \cdot D^{2}

the surface of a sphere(square)

q_{g} = σ_{g} \cdot A_{g}

charge spread over a surface(ground)

q = σ \cdot A

charge spread over a surface(square)

E_{l,g} = \frac{λ_{g}}{2 \cdot π \cdot ε_{0} \cdot r_{g}}

field of a long line of charge(ground)

E_{l} = \frac{λ}{2 \cdot π \cdot ε_{0} \cdot r}

field of a long line of charge(square)

E_{p,g} = \frac{σ_{g}}{ε_{0}}

field between charged plates(ground)

E_{p} = \frac{σ}{ε_{0}}

field between charged plates(square)