A Parallel Plate Capacitor in Air

solved Capacitance

Q. In a parallel plate capacitor with air between the plates $K = 1$ , each plate has an area of 6 × 10⁻³ m² $A = 6 \times 10^{−3}$ m² and the distance between the plates is 3 mm $d = 3$ mm. Calculate the capacitance of the capacitor, in pF?

Inputs

area

A

0.006 m²

gap

d

3 mm

Outputs

capacitance

C

How it's solved

Solved with Capacitance

Given

$A = 6 \times 10^{−3}$ m²
$d = 3$ mm
$K = 1$ from “with air between the plates”

Find $C$ .

Working

$d = 3$ mm $= 0.003$ m
a parallel plate capacitor(capacitor) $C = \frac{K \cdot ε_{0} \cdot A}{d}$
$C = \frac{1 \cdot 8.85418781 \times 10^{−12} \cdot 0.006}{0.003}$
$C = 17.71$ pF
$C = 1.77083756 \times 10^{−11}$ F $= 17.7083756$ pF
$C \approx 17.7$ pFto 3 significant figures

Conclusion

A parallel plate capacitor holds more the larger its plates and the closer together they are.

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 :a_parallel_plate_capacitor_in_air do4  kind :numerical5  describes "In a parallel plate capacitor with air between the plates, each plate has an area of %{area} and the " \6            "distance between the plates is %{gap}. %{capacitance}, in pF?"7  given area: "6e-3 m2"8  given gap: "3 mm"9  asks capacitance: "Calculate the capacitance of the capacitor"10end

solution.rb

1require_relative "../questions/a_parallel_plate_capacitor_in_air"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :a_parallel_plate_capacitor_in_air, using: Capacitance do5  objects :capacitor67  take area: capacitor.area8  take gap: capacitor.plate_separation910  deduce_from "with air between the plates", capacitor.dielectric_constant => 11112  capacitance = capacitor.capacitance13  answer capacitance:, in: "pF", figures: 31415  conclusion "A parallel plate capacitor holds more the larger its plates and the closer together they are."16end

playground.rb

1require_relative "../solutions/a_parallel_plate_capacitor_in_air"23play_with :a_parallel_plate_capacitor_in_air do4  title "A Parallel Plate Capacitor in Air"5end

The laws it drew on

quantity	written	units	within	from
permittivity of free space	$ε_{0}$	F/m	—	Capacitance
capacitance	$C$	F	0 … ∞	Capacitance
charge	$Q$	C	—	Capacitance
voltage	$V$	V	—	Capacitance
energy stored	$U$	J	—	Capacitance
area	$A$	m²	—	Capacitance
plate separation	$d$	m	—	Capacitance
dielectric constant	$K$	—	—	Capacitance
charge of ground	$Q_{g}$	C	—
capacitance of ground	$C_{g}$	F	0 … ∞
voltage of ground	$V_{g}$	V	—
charge of capacitor	$Q$	C	—
capacitance of capacitor	$C$	F	0 … ∞
voltage of capacitor	$V$	V	—
dielectric constant of ground	$K_{g}$	—	—
area of ground	$A_{g}$	m²	—
plate separation of ground	$d_{g}$	m	—
dielectric constant of capacitor	$K$	—	—
area of capacitor	$A$	m²	—
plate separation of capacitor	$d$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of capacitor	$U$	J	—

Q_{c} = C_{c} \cdot V_{c}

charge on a capacitor

C_{c} = \frac{K_{c} \cdot ε_{0} \cdot A_{c}}{d_{c}}

a parallel plate capacitor

U_{c} = \frac{C_{c} \cdot {V_{c}}^{2}}{2}

energy in a capacitor

Q_{g} = C_{g} \cdot V_{g}

charge on a capacitor(ground)

Q = C \cdot V

charge on a capacitor(capacitor)

C_{g} = \frac{K_{g} \cdot ε_{0} \cdot A_{g}}{d_{g}}

a parallel plate capacitor(ground)

C = \frac{K \cdot ε_{0} \cdot A}{d}

a parallel plate capacitor(capacitor)

U_{g} = \frac{C_{g} \cdot {V_{g}}^{2}}{2}

energy in a capacitor(ground)

U = \frac{C \cdot V^{2}}{2}

energy in a capacitor(capacitor)