A Parallel Plate Capacitor in Air

Q. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor, in pF? expects 17.7 pF

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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?

The playground

Inputs

area

A

0.006 m²

gap

d

3 mm

Outputs

capacitance

C

The solution

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 Ruby

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

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

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

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)