Just Outside a Charged Sphere

Q. A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field just outside the sphere, in N/C? expects 1.0 × 10⁵ N/C

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The web of concepts

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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?

The playground

Inputs

radius

r_{ps}

12 cm

charge

q_{s}

1.6e-07 C

Outputs

field

E_{q,ps}

The solution

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 Ruby

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

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

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

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)