A Capacitor Filled and Squeezed

Q. A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6, in pF? expects 96 pF

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. 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?

The playground

Inputs

air

C_{b}

8 pF

constant

K_{a}

Outputs

filled

C_{a}

The solution

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 Ruby

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

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

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

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