A Twelve Picofarad Capacitor

solved Capacitance

Q. A 12 pF $C = 12$ pF capacitor is connected to a 50 V $V = 50$ V battery. How much electrostatic energy is stored in the capacitor, in joules?

Inputs

capacitance

C

12 pF

battery

V

50 V

Outputs

energy

U

How it's solved

Solved with Capacitance

Given

$C = 12$ pF
$V = 50$ V

Find $U$ .

Working

$C = 12$ pF $= 1.2 \times 10^{−11}$ F
energy in a capacitor(capacitor) $U = \frac{C \cdot V^{2}}{2}$
$U = \frac{1.2 \times 10^{−11} \cdot 50^{2}}{2}$
$U = 1.5 \times 10^{−8}$ J
$U \approx 1.5 \times 10^{−8}$ Jto 2 significant figures

Conclusion

Charging a capacitor takes work against the charge already on it, and that work is stored as half the capacitance times the square of the voltage.

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_twelve_picofarad_capacitor do4  kind :numerical5  describes "A %{capacitance} capacitor is connected to a %{battery} battery. %{energy}, in joules?"6  given capacitance: "12 pF"7  given battery: "50 V"8  asks energy: "How much electrostatic energy is stored in the capacitor"9end

solution.rb

1require_relative "../questions/a_twelve_picofarad_capacitor"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :a_twelve_picofarad_capacitor, using: Capacitance do5  objects :capacitor67  take capacitance: capacitor.capacitance8  take battery: capacitor.voltage910  energy = capacitor.energy_stored11  answer energy:, in: "J", figures: 21213  conclusion "Charging a capacitor takes work against the charge already on it, and that work is stored as " \14             "half the capacitance times the square of the voltage."15end

playground.rb

1require_relative "../solutions/a_twelve_picofarad_capacitor"23play_with :a_twelve_picofarad_capacitor do4  title "A Twelve Picofarad Capacitor"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 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)