Mica Slid In After Disconnecting

solved Capacitance

Q. In a parallel plate capacitor with air between the plates $K_{a} = 1$ , each plate has an area of 6 × 10⁻³ m² $A_{a} = 6 \times 10^{−3}$ m² and the distance between the plates is 3 mm $d_{a} = 3$ mm. It is charged by a 100 V $V_{a} = 100$ V supply. A 3 mm thick mica sheet $d_{m} = d_{a}$ , $A_{m} = A_{a}$ (of dielectric constant = 6 $K_{m} = 6$ ) is inserted between the plates after the supply was disconnected $Q_{m} = Q_{a}$ . What is the potential difference between the plates now, in volts?

Inputs

area

A_{a}

0.006 m²

supply

V_{a}

100 V

constant

K_{m}

Outputs

voltage

V_{m}

How it's solved

Solved with Capacitance

Given

$A_{a} = 6 \times 10^{−3}$ m²
$V_{a} = 100$ V
$K_{m} = 6$
$K_{a} = 1$ from “with air between the plates”
$d_{a} = 3$ mm from “distance between the plates is 3 mm”
$d_{m} = d_{a}$ from “A 3 mm thick mica sheet”
$A_{m} = A_{a}$ from “A 3 mm thick mica sheet”
$Q_{m} = Q_{a}$ from “after the supply was disconnected”

Find $V_{m}$ .

Working

$d_{a} = 3$ mm $= 0.003$ m
a parallel plate capacitor(air) $C_{a} = \frac{K_{a} \cdot ε_{0} \cdot A_{a}}{d_{a}}$
$C_{a} = \frac{1 \cdot 8.85418781 \times 10^{−12} \cdot 0.006}{0.003}$
$C_{a} = 1.77083756 \times 10^{−11}$ F
charge on a capacitor(air) $Q_{a} = C_{a} \cdot V_{a}$
$Q_{a} = 1.77083756 \times 10^{−11} \cdot 100$
$Q_{a} = 1.77083756 \times 10^{−9}$ C
after the supply was disconnected $Q_{m} = Q_{a}$
$Q_{m} = 1.77083756 \times 10^{−9}$
$Q_{m} = 1.77083756 \times 10^{−9}$ C
A 3 mm thick mica sheet (area of mica) $A_{m} = A_{a}$
$A_{m} = 0.006$
$A_{m} = 0.006$ m²
A 3 mm thick mica sheet (plate separation of mica) $d_{m} = d_{a}$
$d_{m} = 0.003$
$d_{m} = 0.003$ m
a parallel plate capacitor(mica) $C_{m} = \frac{K_{m} \cdot ε_{0} \cdot A_{m}}{d_{m}}$
$C_{m} = \frac{6 \cdot 8.85418781 \times 10^{−12} \cdot 0.006}{0.003}$
$C_{m} = 1.06250254 \times 10^{−10}$ F
charge on a capacitor(mica) $Q_{m} = C_{m} \cdot V_{m}$
$1.77083756 \times 10^{−9} = 1.06250254 \times 10^{−10} \cdot V_{m}$
$V_{m} = 16.67$ V
$V_{m} \approx 16.7$ Vto 3 significant figures

Conclusion

Cut off from the supply, the plates keep their charge; the mica multiplies the capacitance, so the voltage falls 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_after_disconnecting 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 charged by a %{supply} supply. A 3 mm thick mica sheet (of " \7            "dielectric constant = %{constant}) is inserted between the plates after the supply was disconnected. " \8            "%{voltage}, in volts?"9  given area: "6e-3 m2"10  given supply: "100 V"11  given constant: "6"12  asks voltage: "What is the potential difference between the plates now"13end

solution.rb

1require_relative "../questions/mica_slid_in_after_disconnecting"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :mica_slid_in_after_disconnecting, using: Capacitance do5  objects :air, :mica67  take area: air.area8  take supply: air.voltage9  take constant: mica.dielectric_constant1011  deduce_from "with air between the plates", air.dielectric_constant => 112  deduce_from "distance between the plates is 3 mm", air.plate_separation => "3 mm"13  deduce_from "A 3 mm thick mica sheet", mica.plate_separation => air.plate_separation, mica.area => air.area14  deduce_from "after the supply was disconnected", mica.charge => air.charge1516  voltage = mica.voltage17  answer voltage:, in: "V", figures: 31819  conclusion "Cut off from the supply, the plates keep their charge; the mica multiplies the capacitance, so the " \20             "voltage falls by the same factor."21end

playground.rb

1require_relative "../solutions/mica_slid_in_after_disconnecting"23play_with :mica_slid_in_after_disconnecting do4  title "Mica Slid In After Disconnecting"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 air	$Q_{a}$	C	—
capacitance of air	$C_{a}$	F	0 … ∞
voltage of air	$V_{a}$	V	—
charge of mica	$Q_{m}$	C	—
capacitance of mica	$C_{m}$	F	0 … ∞
voltage of mica	$V_{m}$	V	—
dielectric constant of ground	$K_{g}$	—	—
area of ground	$A_{g}$	m²	—
plate separation of ground	$d_{g}$	m	—
dielectric constant of air	$K_{a}$	—	—
area of air	$A_{a}$	m²	—
plate separation of air	$d_{a}$	m	—
dielectric constant of mica	$K_{m}$	—	—
area of mica	$A_{m}$	m²	—
plate separation of mica	$d_{m}$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of air	$U_{a}$	J	—
energy stored of mica	$U_{m}$	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_{a} = C_{a} \cdot V_{a}

charge on a capacitor(air)

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

charge on a capacitor(mica)

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

a parallel plate capacitor(ground)

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

a parallel plate capacitor(air)

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

a parallel plate capacitor(mica)

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

energy in a capacitor(ground)

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

energy in a capacitor(air)

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

energy in a capacitor(mica)

d_{m} = d_{a}

A 3 mm thick mica sheet (plate separation of mica)

A_{m} = A_{a}

A 3 mm thick mica sheet (area of mica)

Q_{m} = Q_{a}

after the supply was disconnected