The Field Midway Between Two Charges

Q. Two point charges qA = 3 μC and qB = -3 μC are located 20 cm apart in vacuum. What is the electric field at the midpoint O of the line AB joining the two charges, in N/C? expects 5.4 × 10⁶ N/C

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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?

The playground

Inputs

charge a

q_{a}

3 μC

charge b

q_{b}

-3 μC

apart

r_{ab}

20 cm

Outputs

field

E

The solution

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 Ruby

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

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

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

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