Mica Slid In with the Supply Connected

solved Capacitance

Q. In a parallel plate capacitor with air between the plates, 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. It is connected to a 100 V $V = 100$ V supply. A 3 mm thick mica sheet $d = 3$ mm (of dielectric constant = 6 $K = 6$ ) is inserted between the plates while the voltage supply remained connected. What is the charge on each plate now, in coulombs?

Inputs

area

A

0.006 m²

supply

V

100 V

constant

K

Outputs

charge

Q

How it's solved

Solved with Capacitance

Given

$A = 6 \times 10^{−3}$ m²
$V = 100$ V
$K = 6$
$d = 3$ mm from “A 3 mm thick mica sheet”

Find $Q$ .

Working

$d = 3$ mm $= 0.003$ m
a parallel plate capacitor(capacitor) $C = \frac{K \cdot ε_{0} \cdot A}{d}$
$C = \frac{6 \cdot 8.85418781 \times 10^{−12} \cdot 0.006}{0.003}$
$C = 1.06250254 \times 10^{−10}$ F
charge on a capacitor(capacitor) $Q = C \cdot V$
$Q = 1.06250254 \times 10^{−10} \cdot 100$
$Q = 1.063 \times 10^{−8}$ C
$Q \approx 1.06 \times 10^{−8}$ Cto 3 significant figures

Conclusion

The mica fills the gap and multiplies the capacitance by its dielectric constant; with the supply still holding the voltage, the charge grows by the same factor.

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 :mica_slid_in_with_the_supply_connected 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 3 mm. It is connected to a %{supply} supply. A 3 mm thick mica sheet " \7            "(of dielectric constant = %{constant}) is inserted between the plates while the voltage supply " \8            "remained connected. %{charge}, in coulombs?"9  given area: "6e-3 m2"10  given supply: "100 V"11  given constant: "6"12  asks charge: "What is the charge on each plate now"13end

solution.rb

1require_relative "../questions/mica_slid_in_with_the_supply_connected"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :mica_slid_in_with_the_supply_connected, using: Capacitance do5  objects :capacitor67  take area: capacitor.area8  take supply: capacitor.voltage9  take constant: capacitor.dielectric_constant1011  deduce_from "A 3 mm thick mica sheet", capacitor.plate_separation => "3 mm"1213  charge = capacitor.charge14  answer charge:, in: "C", figures: 31516  conclusion "The mica fills the gap and multiplies the capacitance by its dielectric constant; with the supply " \17             "still holding the voltage, the charge grows by the same factor."18end

playground.rb

1require_relative "../solutions/mica_slid_in_with_the_supply_connected"23play_with :mica_slid_in_with_the_supply_connected do4  title "Mica Slid In with the Supply Connected"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)