Mica Slid In with the Supply Connected

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 connected to a 100 V supply. A 3 mm thick mica sheet (of dielectric constant = 6) is inserted between the plates while the voltage supply remained connected. What is the charge on each plate now, in coulombs? expects 1.06 × 10⁻⁸ C

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

area

A

0.006 m²

supply

V

100 V

constant

K

Outputs

charge

Q

The solution

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 Ruby

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

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

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

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)