The Charge on Each Plate

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. If this capacitor is connected to a 100 V $V = 100$ V supply, what is the charge on each plate of the capacitor, in coulombs?

Inputs

area

A

0.006 m²

gap

d

3 mm

supply

V

100 V

Outputs

charge

Q

How it's solved

Solved with Capacitance

Given

$A = 6 \times 10^{−3}$ m²
$d = 3$ mm
$V = 100$ V
$K = 1$ from “with air between the plates”

Find $Q$ .

Working

$d = 3$ mm $= 0.003$ m
a parallel plate capacitor(capacitor) $C = \frac{K \cdot ε_{0} \cdot A}{d}$
$C = \frac{1 \cdot 8.85418781 \times 10^{−12} \cdot 0.006}{0.003}$
$C = 1.77083756 \times 10^{−11}$ F
charge on a capacitor(capacitor) $Q = C \cdot V$
$Q = 1.77083756 \times 10^{−11} \cdot 100$
$Q = 1.771 \times 10^{−9}$ C
$Q \approx 1.77 \times 10^{−9}$ Cto 3 significant figures

Conclusion

The plates' area and gap fix the capacitance, and the supply's voltage times it is the charge drawn onto each plate.

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_charge_on_each_plate 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}. If this capacitor is connected to a %{supply} supply, " \7            "%{charge}, in coulombs?"8  given area: "6e-3 m2"9  given gap: "3 mm"10  given supply: "100 V"11  asks charge: "what is the charge on each plate of the capacitor"12end

solution.rb

1require_relative "../questions/the_charge_on_each_plate"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :the_charge_on_each_plate, using: Capacitance do5  objects :capacitor67  take area: capacitor.area8  take gap: capacitor.plate_separation9  take supply: capacitor.voltage1011  deduce_from "with air between the plates", capacitor.dielectric_constant => 11213  charge = capacitor.charge14  answer charge:, in: "C", figures: 31516  conclusion "The plates' area and gap fix the capacitance, and the supply's voltage times it is the charge " \17             "drawn onto each plate."18end

playground.rb

1require_relative "../solutions/the_charge_on_each_plate"23play_with :the_charge_on_each_plate do4  title "The Charge on Each Plate"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)