Sharing a Charge

solved Capacitance

Q. A 900 pF $C_{c} = 900$ pF capacitor is charged by a 100 V $V_{c} = 100$ V battery. It is disconnected from the battery $Q_{j} = Q_{c}$ and connected to another $C_{j} = C_{c} + C_{o}$ uncharged 900 pF $C_{o} = 900$ pF capacitor. What is the electrostatic energy stored by the system, in joules?

Inputs

capacitance

C_{c}

900 pF

battery

V_{c}

100 V

other

C_{o}

900 pF

Outputs

energy

U_{j}

How it's solved

Solved with Capacitance

Given

$C_{c} = 900$ pF
$V_{c} = 100$ V
$C_{o} = 900$ pF
$C_{j} = C_{c} + C_{o}$ from “connected to another”
$Q_{j} = Q_{c}$ from “disconnected from the battery”

Find $U_{j}$ .

Working

$C_{c} = 900$ pF $= 9 \times 10^{−10}$ F
$C_{o} = 900$ pF $= 9 \times 10^{−10}$ F
connected to another $C_{j} = C_{c} + C_{o}$
$C_{j} = 9 \times 10^{−10} + 9 \times 10^{−10}$
$C_{j} = 1.8 \times 10^{−9}$ F
charge on a capacitor(charged) $Q_{c} = C_{c} \cdot V_{c}$
$Q_{c} = 9 \times 10^{−10} \cdot 100$
$Q_{c} = 9 \times 10^{−8}$ C
disconnected from the battery $Q_{j} = Q_{c}$
$Q_{j} = 9 \times 10^{−8}$
$Q_{j} = 9 \times 10^{−8}$ C
charge on a capacitor(joined) $Q_{j} = C_{j} \cdot V_{j}$
$9 \times 10^{−8} = 1.8 \times 10^{−9} \cdot V_{j}$
$V_{j} = 50$ V
energy in a capacitor(joined) $U_{j} = \frac{C_{j} \cdot {V_{j}}^{2}}{2}$
$U_{j} = \frac{1.8 \times 10^{−9} \cdot 50^{2}}{2}$
$U_{j} = 2.25 \times 10^{−6}$ J
$U_{j} \approx 2.2 \times 10^{−6}$ Jto 2 significant figures

Conclusion

The charge spreads over twice the capacitance at half the voltage, so only half the energy is left; the rest is lost as the charge rushes across.

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 :sharing_a_charge do4  kind :numerical5  describes "A %{capacitance} capacitor is charged by a %{battery} battery. It is disconnected from the battery and " \6            "connected to another uncharged %{other} capacitor. %{energy}, in joules?"7  given capacitance: "900 pF"8  given battery: "100 V"9  given other: "900 pF"10  asks energy: "What is the electrostatic energy stored by the system"11end

solution.rb

1require_relative "../questions/sharing_a_charge"2require_relative "../science/electrostatic_potential_and_capacitance/capacitance"34solve :sharing_a_charge, using: Capacitance do5  objects :charged, :other, :joined67  take capacitance: charged.capacitance8  take battery: charged.voltage9  take other: other.capacitance1011  deduce_from "connected to another", joined.capacitance => charged.capacitance + other.capacitance12  deduce_from "disconnected from the battery", joined.charge => charged.charge1314  energy = joined.energy_stored15  answer energy:, in: "J", figures: 21617  conclusion "The charge spreads over twice the capacitance at half the voltage, so only half the energy " \18             "is left; the rest is lost as the charge rushes across."19end

playground.rb

1require_relative "../solutions/sharing_a_charge"23play_with :sharing_a_charge do4  title "Sharing a Charge"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 charged	$Q_{c}$	C	—
capacitance of charged	$C_{c}$	F	0 … ∞
voltage of charged	$V_{c}$	V	—
charge of other	$Q_{o}$	C	—
capacitance of other	$C_{o}$	F	0 … ∞
voltage of other	$V_{o}$	V	—
charge of joined	$Q_{j}$	C	—
capacitance of joined	$C_{j}$	F	0 … ∞
voltage of joined	$V_{j}$	V	—
dielectric constant of ground	$K_{g}$	—	—
area of ground	$A_{g}$	m²	—
plate separation of ground	$d_{g}$	m	—
dielectric constant of charged	$K_{c}$	—	—
area of charged	$A_{c}$	m²	—
plate separation of charged	$d_{c}$	m	—
dielectric constant of other	$K_{o}$	—	—
area of other	$A_{o}$	m²	—
plate separation of other	$d_{o}$	m	—
dielectric constant of joined	$K_{j}$	—	—
area of joined	$A_{j}$	m²	—
plate separation of joined	$d_{j}$	m	—
energy stored of ground	$U_{g}$	J	—
energy stored of charged	$U_{c}$	J	—
energy stored of other	$U_{o}$	J	—
energy stored of joined	$U_{j}$	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} = C_{c} \cdot V_{c}

charge on a capacitor(charged)

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

charge on a capacitor(other)

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

charge on a capacitor(joined)

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

a parallel plate capacitor(ground)

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

a parallel plate capacitor(charged)

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

a parallel plate capacitor(other)

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

a parallel plate capacitor(joined)

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

energy in a capacitor(ground)

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

energy in a capacitor(charged)

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

energy in a capacitor(other)

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

energy in a capacitor(joined)

C_{j} = C_{c} + C_{o}

connected to another

Q_{j} = Q_{c}

disconnected from the battery