Just Outside a Charged Sphere

solved ElectricField

Q. A spherical conductor of radius 12 cm $r_{ps} = 12$ cm has a charge of 1.6 × 10⁻⁷ C $q_{s} = 1.6 \times 10^{−7}$ C distributed uniformly on its surface. What is the electric field just outside the sphere, in N/C?

Inputs

radius

r_{ps}

12 cm

charge

q_{s}

1.6e-07 C

Outputs

field

E_{q,ps}

How it's solved

Solved with ElectricField

Given

$r_{ps} = 12$ cm
$q_{s} = 1.6 \times 10^{−7}$ C

Find $E_{q,ps}$ .

Working

$r_{ps} = 12$ cm $= 0.12$ m
field of a point charge(point, sphere) $E_{q,ps} = \frac{k \cdot q_{s}}{{r_{ps}}^{2}}$
$E_{q,ps} = \frac{8.98755179 \times 10^{9} \cdot 1.6 \times 10^{−7}}{{0.12}^{2}}$
$E_{q,ps} = 99860$ N/C
$E_{q,ps} = 99861.6866$ V/m $= 99861.6866$ N/C
$E_{q,ps} \approx 1.0 \times 10^{5}$ N/Cto 2 significant figures

Conclusion

Outside a uniformly charged sphere the field is as if all the charge sat at its centre, so just outside it is that of a point charge a radius away.

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 :just_outside_a_charged_sphere do4  kind :numerical5  describes "A spherical conductor of radius %{radius} has a charge of %{charge} distributed uniformly on its " \6            "surface. %{field} just outside the sphere, in N/C?"7  given radius: "12 cm"8  given charge: "1.6e-7 C"9  asks field: "What is the electric field"10end

solution.rb

1require_relative "../questions/just_outside_a_charged_sphere"2require_relative "../science/electric_charges_and_fields/electric_field"34solve :just_outside_a_charged_sphere, using: ElectricField do5  objects :sphere, :point67  take radius: point.separation(from: sphere)8  take charge: sphere.charge910  field = point.field_of_a_charge(by: sphere)11  answer field:, in: "N/C", figures: 21213  conclusion "Outside a uniformly charged sphere the field is as if all the charge sat at its centre, so just " \14             "outside it is that of a point charge a radius away."15end

playground.rb

1require_relative "../solutions/just_outside_a_charged_sphere"23play_with :just_outside_a_charged_sphere do4  title "Just Outside a Charged Sphere"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 sphere	$F_{s}$	N	—
charge of sphere	$q_{s}$	C	—
force of the field on point	$F_{p}$	N	—
charge of point	$q_{p}$	C	—
field of a charge of ground by sphere	$E_{q,gs}$	V/m	—
separation of ground from sphere	$r_{gs}$	m	0 … ∞
field of a charge of ground by point	$E_{q,gp}$	V/m	—
separation of ground from point	$r_{gp}$	m	0 … ∞
field of a charge of sphere by ground	$E_{q,sg}$	V/m	—
separation of sphere from ground	$r_{s}$	m	0 … ∞
field of a charge of sphere by point	$E_{q,sp}$	V/m	—
separation of sphere from point	$r_{sp}$	m	0 … ∞
field of a charge of point by ground	$E_{q,pg}$	V/m	—
separation of point from ground	$r_{p}$	m	0 … ∞
field of a charge of point by sphere	$E_{q,ps}$	V/m	—
separation of point from sphere	$r_{ps}$	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_{s} = q_{s} \cdot E

force on a charge in a field(sphere)

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

force on a charge in a field(point)

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

field of a point charge(ground, sphere)

E_{q,gp} = \frac{k \cdot q_{p}}{{r_{gp}}^{2}}

field of a point charge(ground, point)

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

field of a point charge(sphere, ground)

E_{q,sp} = \frac{k \cdot q_{p}}{{r_{sp}}^{2}}

field of a point charge(sphere, point)

E_{q,pg} = \frac{k \cdot q_{g}}{{r_{p}}^{2}}

field of a point charge(point, ground)

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

field of a point charge(point, sphere)