Two Copper Spheres Fifty Centimetres Apart

solved CoulombsLaw

Q. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm $r_{ab} = 50$ cm, and the charge on each $q_{b} = q_{a}$ is 6.5 × 10⁻⁷ C $q_{a} = 6.5 \times 10^{−7}$ C. The radii of A and B are negligible compared to the distance of separation. What is the mutual force of electrostatic repulsion, in newtons?

Inputs

apart

r_{ab}

50 cm

charge

q_{a}

6.5e-07 C

Outputs

force

F_{e,ab}

How it's solved

Solved with CoulombsLaw

Given

$r_{ab} = 50$ cm
$q_{a} = 6.5 \times 10^{−7}$ C
$q_{b} = q_{a}$ from “the charge on each”

Find $F_{e,ab}$ .

Working

$r_{ab} = 50$ cm $= 0.5$ m
the charge on each $q_{b} = q_{a}$
$q_{b} = 6.5 \times 10^{−7}$
$q_{b} = 6.5 \times 10^{−7}$ C
coulombs law(a, b) $F_{e,ab} = \frac{k \cdot q_{a} \cdot q_{b}}{{r_{ab}}^{2}}$
$F_{e,ab} = \frac{8.98755179 \times 10^{9} \cdot 6.5 \times 10^{−7} \cdot 6.5 \times 10^{−7}}{{0.5}^{2}}$
$F_{e,ab} = 0.01519$ N
$F_{e,ab} \approx 0.015$ Nto 2 significant figures

Conclusion

Spheres small beside their distance act as point charges, so Coulomb's law gives the push between 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_copper_spheres_fifty_centimetres_apart do4  kind :numerical5  describes "Two insulated charged copper spheres A and B have their centres separated by a distance of %{apart}, " \6            "and the charge on each is %{charge}. The radii of A and B are negligible compared to the distance of " \7            "separation. %{force}, in newtons?"8  given apart: "50 cm"9  given charge: "6.5e-7 C"10  asks force: "What is the mutual force of electrostatic repulsion"11end

solution.rb

1require_relative "../questions/two_copper_spheres_fifty_centimetres_apart"2require_relative "../science/electric_charges_and_fields/coulombs_law"34solve :two_copper_spheres_fifty_centimetres_apart, using: CoulombsLaw do5  objects :a, :b67  take apart: a.separation(from: b)8  take charge: a.charge910  deduce_from "the charge on each", b.charge => a.charge1112  force = a.electric_force(by: b)13  answer force:, in: "N", figures: 21415  conclusion "Spheres small beside their distance act as point charges, so Coulomb's law gives the push " \16             "between them."17end

playground.rb

1require_relative "../solutions/two_copper_spheres_fifty_centimetres_apart"23play_with :two_copper_spheres_fifty_centimetres_apart do4  title "Two Copper Spheres Fifty Centimetres Apart"5end

The laws it drew on

quantity	written	units	within	from
charge	$q$	C	—	CoulombsLaw
coulomb constant	$k$	N m² C⁻²	—	CoulombsLaw
separation	$r$	m	0 … ∞	CoulombsLaw
electric force	$F_{e}$	N	—	CoulombsLaw
electric force on ground by a	$F_{e,ga}$	N	—
charge of ground	$q_{g}$	C	—
charge of a	$q_{a}$	C	—
separation of ground from a	$r_{ga}$	m	0 … ∞
electric force on ground by b	$F_{e,gb}$	N	—
charge of b	$q_{b}$	C	—
separation of ground from b	$r_{gb}$	m	0 … ∞
electric force on a by ground	$F_{e,ag}$	N	—
separation of a from ground	$r_{a}$	m	0 … ∞
electric force on a by b	$F_{e,ab}$	N	—
separation of a from b	$r_{ab}$	m	0 … ∞
electric force on b by ground	$F_{e,bg}$	N	—
separation of b from ground	$r_{b}$	m	0 … ∞
electric force on b by a	$F_{e,ba}$	N	—
separation of b from a	$r_{ba}$	m	0 … ∞

F_{e,oo} = \frac{k \cdot q_{o} \cdot q_{o}}{{r_{oo}}^{2}}

coulombs law

F_{e,ga} = \frac{k \cdot q_{g} \cdot q_{a}}{{r_{ga}}^{2}}

coulombs law(ground, a)

F_{e,gb} = \frac{k \cdot q_{g} \cdot q_{b}}{{r_{gb}}^{2}}

coulombs law(ground, b)

F_{e,ag} = \frac{k \cdot q_{a} \cdot q_{g}}{{r_{a}}^{2}}

coulombs law(a, ground)

F_{e,ab} = \frac{k \cdot q_{a} \cdot q_{b}}{{r_{ab}}^{2}}

coulombs law(a, b)

F_{e,bg} = \frac{k \cdot q_{b} \cdot q_{g}}{{r_{b}}^{2}}

coulombs law(b, ground)

F_{e,ba} = \frac{k \cdot q_{b} \cdot q_{a}}{{r_{ba}}^{2}}

coulombs law(b, a)

q_{b} = q_{a}

the charge on each