Two Copper Spheres Fifty Centimetres Apart

Q. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm, and the charge on each is 6.5 × 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? expects 0.015 N

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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?

The playground

Inputs

apart

r_{ab}

50 cm

charge

q_{a}

6.5e-07 C

Outputs

force

F_{e,ab}

The solution

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 Ruby

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

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

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

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