The Field Midway Between Two Charges

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. What is the electric field at the midpoint O $r_{oa} = \frac{r_{ab}}{2}$ , $r_{ob} = \frac{r_{ab}}{2}$ of the line AB joining the two charges $E = E_{q,oa} - E_{q,ob}$ , in N/C?

Inputs

charge a

q_{a}

3 μC

charge b

q_{b}

-3 μC

apart

r_{ab}

20 cm

Outputs

field

E

How it's solved

Solved with ElectricField

Given

$q_{a} = 3$ μC
$q_{b} = -3$ μC
$r_{ab} = 20$ cm
$r_{oa} = \frac{r_{ab}}{2}$ 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 $E$ .

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 ${field}_{strength} = E_{q,oa} - E_{q,ob}$
${field}_{strength} = 2.69626554 \times 10^{6} - (−2.69626554 \times 10^{6})$
$E = 5.393 \times 10^{6}$ N/C
$E = 5.39253108 \times 10^{6}$ V/m $= 5.39253108 \times 10^{6}$ N/C
$E \approx 5.4 \times 10^{6}$ N/Cto 2 significant figures

Conclusion

The positive charge pushes away from A and the negative charge pulls toward B, so at the midpoint both fields point the same way, from A to B, and add.

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 :the_field_midway_between_two_charges do4  kind :numerical5  describes "Two point charges qA = %{charge_a} and qB = %{charge_b} are located %{apart} apart in vacuum. " \6            "%{field} at the midpoint O of the line AB joining the two charges, in N/C?"7  given charge_a: "3 μC"8  given charge_b: "-3 μC"9  given apart: "20 cm"10  asks field: "What is the electric field"11end

solution.rb

1require_relative "../questions/the_field_midway_between_two_charges"2require_relative "../science/electric_charges_and_fields/electric_field"34solve :the_field_midway_between_two_charges, using: ElectricField do5  objects :a, :b, :o67  take charge_a: a.charge8  take charge_b: b.charge9  take apart: a.separation(from: b)1011  deduce_from "the midpoint O", o.separation(from: a) => a.separation(from: b) / 2,12                                o.separation(from: b) => a.separation(from: b) / 213  deduce_from "the line AB joining the two charges", E => o.field_of_a_charge(by: a) - o.field_of_a_charge(by: b)1415  field = E16  answer field:, in: "N/C", figures: 21718  conclusion "The positive charge pushes away from A and the negative charge pulls toward B, so at the midpoint " \19             "both fields point the same way, from A to B, and add."20end

playground.rb

1require_relative "../solutions/the_field_midway_between_two_charges"23play_with :the_field_midway_between_two_charges do4  title "The Field Midway Between Two Charges"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