Mica Slid In After Disconnecting

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. It is charged by a 100 V supply. A 3 mm thick mica sheet (of dielectric constant = 6) is inserted between the plates after the supply was disconnected. What is the potential difference between the plates now, in volts? expects 16.7 V

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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?

The playground

Inputs

area

A_{a}

0.006 m²

supply

V_{a}

100 V

constant

K_{m}

Outputs

voltage

V_{m}

The solution

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 Ruby

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

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

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

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