Two Charges Apart

solved PotentialEnergyOfCharges

Q. Two charges of 7 μC $q_{f} = 7$ μC and -2 μC $q_{s} = -2$ μC are placed at (–9 cm, 0, 0) and (9 cm, 0, 0) $r_{fs} = 18$ cm respectively. What is the electrostatic potential energy of the system, in joules?

Inputs

first

q_{f}

7 μC

second

q_{s}

-2 μC

Outputs

energy

U_{fs}

How it's solved

Solved with PotentialEnergyOfCharges

Given

$q_{f} = 7$ μC
$q_{s} = -2$ μC
$r_{fs} = 18$ cm from “(–9 cm, 0, 0) and (9 cm, 0, 0)”

Find $U_{fs}$ .

Working

$q_{f} = 7$ μC $= 7 \times 10^{−6}$ C
$q_{s} = -2$ μC $= −2 \times 10^{−6}$ C
$r_{fs} = 18$ cm $= 0.18$ m
energy of a pair of charges(first, second) $U_{fs} = \frac{k \cdot q_{f} \cdot q_{s}}{r_{fs}}$
$U_{fs} = \frac{8.98755179 \times 10^{9} \cdot 7 \times 10^{−6} \cdot (−2 \times 10^{−6})}{0.18}$
$U_{fs} = −0.699$ J
$U_{fs} \approx −0.70$ Jto 2 significant figures

Conclusion

Opposite charges attract, so the pair has less energy together than apart: the energy is negative, and that much work separates them.

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 :two_charges_apart do4  kind :numerical5  describes "Two charges of %{first} and %{second} are placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. " \6            "%{energy}, in joules?"7  given first: "7 μC"8  given second: "-2 μC"9  asks energy: "What is the electrostatic potential energy of the system"10end

solution.rb

1require_relative "../questions/two_charges_apart"2require_relative "../science/electrostatic_potential_and_capacitance/potential_energy_of_charges"34solve :two_charges_apart, using: PotentialEnergyOfCharges do5  objects :first, :second67  take first: first.charge8  take second: second.charge910  deduce_from "(–9 cm, 0, 0) and (9 cm, 0, 0)", first.distance(from: second) => "18 cm"1112  energy = first.potential_energy(by: second)13  answer energy:, in: "J", figures: 21415  conclusion "Opposite charges attract, so the pair has less energy together than apart: the energy is " \16             "negative, and that much work separates them."17end

playground.rb

1require_relative "../solutions/two_charges_apart"23play_with :two_charges_apart do4  title "Two Charges Apart"5end

The laws it drew on

quantity	written	units	within	from
coulomb constant	$k$	N m² C⁻²	—	PotentialEnergyOfCharges
charge	$q$	C	—	PotentialEnergyOfCharges
distance	$r$	m	—	PotentialEnergyOfCharges
potential energy	$U$	J	—	PotentialEnergyOfCharges
potential energy of ground by first	$U_{gf}$	J	—
charge of ground	$q_{g}$	C	—
charge of first	$q_{f}$	C	—
distance of ground from first	$r_{gf}$	m	—
potential energy of ground by second	$U_{gs}$	J	—
charge of second	$q_{s}$	C	—
distance of ground from second	$r_{gs}$	m	—
potential energy of first by ground	$U_{fg}$	J	—
distance of first from ground	$r_{f}$	m	—
potential energy of first by second	$U_{fs}$	J	—
distance of first from second	$r_{fs}$	m	—
potential energy of second by ground	$U_{sg}$	J	—
distance of second from ground	$r_{s}$	m	—
potential energy of second by first	$U_{sf}$	J	—
distance of second from first	$r_{sf}$	m	—

U_{oo} = \frac{k \cdot q_{o} \cdot q_{o}}{r_{oo}}

energy of a pair of charges

U_{gf} = \frac{k \cdot q_{g} \cdot q_{f}}{r_{gf}}

energy of a pair of charges(ground, first)

U_{gs} = \frac{k \cdot q_{g} \cdot q_{s}}{r_{gs}}

energy of a pair of charges(ground, second)

U_{fg} = \frac{k \cdot q_{f} \cdot q_{g}}{r_{f}}

energy of a pair of charges(first, ground)

U_{fs} = \frac{k \cdot q_{f} \cdot q_{s}}{r_{fs}}

energy of a pair of charges(first, second)

U_{sg} = \frac{k \cdot q_{s} \cdot q_{g}}{r_{s}}

energy of a pair of charges(second, ground)

U_{sf} = \frac{k \cdot q_{s} \cdot q_{f}}{r_{sf}}

energy of a pair of charges(second, first)