A Capacitor Filled and Squeezed

solved Capacitance

Q. A parallel plate capacitor with air between the plates $K_{b} = 1$ has a capacitance of 8 pF $C_{b} = 8$ pF. What will be the capacitance if the distance between the plates is reduced by half $d_{a} = \frac{d_{b}}{2}$ , and the space between them $A_{a} = A_{b}$ is filled with a substance of dielectric constant 6 $K_{a} = 6$ , in pF?

Inputs

air

C_{b}

8 pF

constant

K_{a}

Outputs

filled

C_{a}

How it's solved

Solved with Capacitance

Given

$C_{b} = 8$ pF
$K_{a} = 6$
$K_{b} = 1$ from “with air between the plates”
$d_{a} = \frac{d_{b}}{2}$ from “reduced by half”
$A_{a} = A_{b}$ from “the space between them”

Find $C_{a}$ .

Working

$C_{b} = 8$ pF $= 8 \times 10^{−12}$ F
a parallel plate capacitor(before) $C_{b} = \frac{K_{b} \cdot ε_{0} \cdot A_{b}}{d_{b}}$
reduced by half $d_{a} = \frac{d_{b}}{2}$
the space between them $A_{a} = A_{b}$
a parallel plate capacitor(after) $C_{a} = \frac{K_{a} \cdot ε_{0} \cdot A_{a}}{d_{a}}$
$C_{a} = \frac{5.31251269 \times 10^{−11} \cdot A_{a}}{d_{a}}$
$C_{a} = 96$ pF
$C_{a} = 9.6 \times 10^{−11}$ F $= 96$ pF
$C_{a} \approx 96$ pFto 2 significant figures

Conclusion

Halving the gap doubles the capacitance and the dielectric multiplies it again by its constant; the plates' size never needs to be known.

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 :a_capacitor_filled_and_squeezed do4  kind :numerical5  describes "A parallel plate capacitor with air between the plates has a capacitance of %{air}. %{filled} if the " \6            "distance between the plates is reduced by half, and the space between them is filled with a substance " \7            "of dielectric constant %{constant}, in pF?"8  given air: "8 pF"9  given constant: "6"10  asks filled: "What will be the capacitance"11end

solution.rb

1require_relative "../questions/a_capacitor_filled_and_squeezed"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :a_capacitor_filled_and_squeezed, using: Capacitance do5  objects :before, :after67  take air: before.capacitance8  take constant: after.dielectric_constant910  deduce_from "with air between the plates", before.dielectric_constant => 111  deduce_from "reduced by half", after.plate_separation => before.plate_separation / 212  deduce_from "the space between them", after.area => before.area1314  filled = after.capacitance15  answer filled:, in: "pF", figures: 21617  conclusion "Halving the gap doubles the capacitance and the dielectric multiplies it again by its constant; the " \18             "plates' size never needs to be known."19end

playground.rb

1require_relative "../solutions/a_capacitor_filled_and_squeezed"23play_with :a_capacitor_filled_and_squeezed do4  title "A Capacitor Filled and Squeezed"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 before	$Q_{b}$	C	—
capacitance of before	$C_{b}$	F	0 … ∞
voltage of before	$V_{b}$	V	—
charge of after	$Q_{a}$	C	—
capacitance of after	$C_{a}$	F	0 … ∞
voltage of after	$V_{a}$	V	—
dielectric constant of ground	$K_{g}$	—	—
area of ground	$A_{g}$	m²	—
plate separation of ground	$d_{g}$	m	—
dielectric constant of before	$K_{b}$	—	—
area of before	$A_{b}$	m²	—
plate separation of before	$d_{b}$	m	—
dielectric constant of after	$K_{a}$	—	—
area of after	$A_{a}$	m²	—
plate separation of after	$d_{a}$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of before	$U_{b}$	J	—
energy stored of after	$U_{a}$	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_{b} = C_{b} \cdot V_{b}

charge on a capacitor(before)

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

charge on a capacitor(after)

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

a parallel plate capacitor(ground)

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

a parallel plate capacitor(before)

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

a parallel plate capacitor(after)

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

energy in a capacitor(ground)

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

energy in a capacitor(before)

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

energy in a capacitor(after)

d_{a} = \frac{d_{b}}{2}

reduced by half

A_{a} = A_{b}

the space between them