Sharing a Charge

Q. A 900 pF capacitor is charged by a 100 V battery. It is disconnected from the battery and connected to another uncharged 900 pF capacitor. What is the electrostatic energy stored by the system, in joules? expects 2.2 × 10⁻⁶ J

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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?

The playground

Inputs

capacitance

C_{c}

900 pF

battery

V_{c}

100 V

other

C_{o}

900 pF

Outputs

energy

U_{j}

The solution

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 Ruby

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

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

1require_relative "../solutions/sharing_a_charge"23play_with :sharing_a_charge do4  title "Sharing a Charge"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 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