A Test Charge at the Midpoint

solved ElectricField

Q. Two point charges qA = 3 μC $q_{a} = 3$ μC and qB = -3 μC $q_{b} = -3$ μC are located 20 cm $r_{ab} = 20$ cm apart in vacuum. A negative test charge of magnitude 1.5 × 10⁻⁹ C $q_{o} = 1.5 \times 10^{−9}$ C is placed at the midpoint O $r_{oa} = \frac{r_{ab}}{2}$ , $r_{ob} = \frac{r_{ab}}{2}$ of the line AB joining the two charges $E = E_{q,oa} - E_{q,ob}$ . What is the force experienced by the test charge, in newtons?

Inputs

charge a

q_{a}

3 μC

charge b

q_{b}

-3 μC

apart

r_{ab}

20 cm

test

q_{o}

1.5e-09 C

Outputs

force

F_{o}

How it's solved

Solved with ElectricField

Given

$q_{a} = 3$ μC
$q_{b} = -3$ μC
$r_{ab} = 20$ cm
$q_{o} = 1.5 \times 10^{−9}$ C
$r_{oa} = \frac{r_{ab}}{2}$ from “the midpoint O”
$r_{ob} = \frac{r_{ab}}{2}$ from “the midpoint O”
$E = E_{q,oa} - E_{q,ob}$ from “the line AB joining the two charges”

Find $F_{o}$ .

Working

$q_{a} = 3$ μC $= 3 \times 10^{−6}$ C
$q_{b} = -3$ μC $= −3 \times 10^{−6}$ C
$r_{ab} = 20$ cm $= 0.2$ m
the midpoint O (separation of o from a) $r_{oa} = \frac{r_{ab}}{2}$
$r_{oa} = \frac{0.2}{2}$
$r_{oa} = 0.1$ m
field of a point charge(o, a) $E_{q,oa} = \frac{k \cdot q_{a}}{{r_{oa}}^{2}}$
$E_{q,oa} = \frac{8.98755179 \times 10^{9} \cdot 3 \times 10^{−6}}{{0.1}^{2}}$
$E_{q,oa} = 2.69626554 \times 10^{6}$ V/m
the midpoint O (separation of o from b) $r_{ob} = \frac{r_{ab}}{2}$
$r_{ob} = \frac{0.2}{2}$
$r_{ob} = 0.1$ m
field of a point charge(o, b) $E_{q,ob} = \frac{k \cdot q_{b}}{{r_{ob}}^{2}}$
$E_{q,ob} = \frac{8.98755179 \times 10^{9} \cdot (−3 \times 10^{−6})}{{0.1}^{2}}$
$E_{q,ob} = −2.69626554 \times 10^{6}$ V/m
the line AB joining the two charges ${field}_{strength} = E_{q,oa} - E_{q,ob}$
${field}_{strength} = 2.69626554 \times 10^{6} - (−2.69626554 \times 10^{6})$
$E = 5.39253108 \times 10^{6}$ V/m
force on a charge in a field(o) $F_{o} = q_{o} \cdot E$
$F_{o} = 1.5 \times 10^{−9} \cdot 5.39253108 \times 10^{6}$
$F_{o} = 0.008089$ N
$F_{o} \approx 0.0081$ Nto 2 significant figures

Conclusion

The two fields add at the midpoint, and a charge there feels the field times its charge; being negative, it is pushed against the field, toward A.

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 :a_test_charge_at_the_midpoint do4  kind :numerical5  describes "Two point charges qA = %{charge_a} and qB = %{charge_b} are located %{apart} apart in vacuum. A " \6            "negative test charge of magnitude %{test} is placed at the midpoint O of the line AB joining the two " \7            "charges. %{force}, in newtons?"8  given charge_a: "3 μC"9  given charge_b: "-3 μC"10  given apart: "20 cm"11  given test: "1.5e-9 C"12  asks force: "What is the force experienced by the test charge"13end

solution.rb

1require_relative "../questions/a_test_charge_at_the_midpoint"2require_relative "../science/electric_charges_and_fields/electric_field"34solve :a_test_charge_at_the_midpoint, using: ElectricField do5  objects :a, :b, :o67  take charge_a: a.charge8  take charge_b: b.charge9  take apart: a.separation(from: b)10  take test: o.charge1112  deduce_from "the midpoint O", o.separation(from: a) => a.separation(from: b) / 2,13                                o.separation(from: b) => a.separation(from: b) / 214  deduce_from "the line AB joining the two charges", E => o.field_of_a_charge(by: a) - o.field_of_a_charge(by: b)1516  force = o.force_of_the_field17  answer force:, in: "N", figures: 21819  conclusion "The two fields add at the midpoint, and a charge there feels the field times its charge; being " \20             "negative, it is pushed against the field, toward A."21end

playground.rb

1require_relative "../solutions/a_test_charge_at_the_midpoint"23play_with :a_test_charge_at_the_midpoint do4  title "A Test Charge at the Midpoint"5end

The laws it drew on

quantity	written	units	within	from
charge	$q$	C	—	ElectricField
coulomb constant	$k$	N m² C⁻²	—	ElectricField
field strength	$E$	V/m	—	ElectricField
separation	$r$	m	0 … ∞	ElectricField
field of a charge	$E_{q}$	V/m	—	ElectricField
force of the field	$F$	N	—	ElectricField
force of the field on ground	$F_{g}$	N	—
charge of ground	$q_{g}$	C	—
force of the field on a	$F_{a}$	N	—
charge of a	$q_{a}$	C	—
force of the field on b	$F_{b}$	N	—
charge of b	$q_{b}$	C	—
force of the field on o	$F_{o}$	N	—
charge of o	$q_{o}$	C	—
field of a charge of ground by a	$E_{q,ga}$	V/m	—
separation of ground from a	$r_{ga}$	m	0 … ∞
field of a charge of ground by b	$E_{q,gb}$	V/m	—
separation of ground from b	$r_{gb}$	m	0 … ∞
field of a charge of ground by o	$E_{q,go}$	V/m	—
separation of ground from o	$r_{go}$	m	0 … ∞
field of a charge of a by ground	$E_{q,ag}$	V/m	—
separation of a from ground	$r_{a}$	m	0 … ∞
field of a charge of a by b	$E_{q,ab}$	V/m	—
separation of a from b	$r_{ab}$	m	0 … ∞
field of a charge of a by o	$E_{q,ao}$	V/m	—
separation of a from o	$r_{ao}$	m	0 … ∞
field of a charge of b by ground	$E_{q,bg}$	V/m	—
separation of b from ground	$r_{b}$	m	0 … ∞
field of a charge of b by a	$E_{q,ba}$	V/m	—
separation of b from a	$r_{ba}$	m	0 … ∞
field of a charge of b by o	$E_{q,bo}$	V/m	—
separation of b from o	$r_{bo}$	m	0 … ∞
field of a charge of o by ground	$E_{q,og}$	V/m	—
separation of o from ground	$r_{o}$	m	0 … ∞
field of a charge of o by a	$E_{q,oa}$	V/m	—
separation of o from a	$r_{oa}$	m	0 … ∞
field of a charge of o by b	$E_{q,ob}$	V/m	—
separation of o from b	$r_{ob}$	m	0 … ∞

F_{b} = q_{b} \cdot E

force on a charge in a field

E_{q,ps} = \frac{k \cdot q_{s}}{{r_{ps}}^{2}}

field of a point charge

F_{g} = q_{g} \cdot E

force on a charge in a field(ground)

F_{a} = q_{a} \cdot E

force on a charge in a field(a)

F_{b} = q_{b} \cdot E

force on a charge in a field(b)

F_{o} = q_{o} \cdot E

force on a charge in a field(o)

E_{q,ga} = \frac{k \cdot q_{a}}{{r_{ga}}^{2}}

field of a point charge(ground, a)

E_{q,gb} = \frac{k \cdot q_{b}}{{r_{gb}}^{2}}

field of a point charge(ground, b)

E_{q,go} = \frac{k \cdot q_{o}}{{r_{go}}^{2}}

field of a point charge(ground, o)

E_{q,ag} = \frac{k \cdot q_{g}}{{r_{a}}^{2}}

field of a point charge(a, ground)

E_{q,ab} = \frac{k \cdot q_{b}}{{r_{ab}}^{2}}

field of a point charge(a, b)

E_{q,ao} = \frac{k \cdot q_{o}}{{r_{ao}}^{2}}

field of a point charge(a, o)

E_{q,bg} = \frac{k \cdot q_{g}}{{r_{b}}^{2}}

field of a point charge(b, ground)

E_{q,ba} = \frac{k \cdot q_{a}}{{r_{ba}}^{2}}

field of a point charge(b, a)

E_{q,bo} = \frac{k \cdot q_{o}}{{r_{bo}}^{2}}

field of a point charge(b, o)

E_{q,og} = \frac{k \cdot q_{g}}{{r_{o}}^{2}}

field of a point charge(o, ground)

E_{q,oa} = \frac{k \cdot q_{a}}{{r_{oa}}^{2}}

field of a point charge(o, a)

E_{q,ob} = \frac{k \cdot q_{b}}{{r_{ob}}^{2}}

field of a point charge(o, b)

r_{oa} = \frac{r_{ab}}{2}

the midpoint O (separation of o from a)

r_{ob} = \frac{r_{ab}}{2}

the midpoint O (separation of o from b)

{field}_{strength} = E_{q,oa} - E_{q,ob}

the line AB joining the two charges