Eighteen Centimetres from 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 at a point 18 cm from the centre of the sphere, in N/C? expects 4.4 × 10⁴ N/C

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Q. A spherical conductor of radius 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 at a point 18 cm $r_{ps} = 18$ cm from the centre of the sphere, in N/C?

The playground

Inputs

charge

q_{s}

1.6e-07 C

distance

r_{ps}

18 cm

Outputs

field

E_{q,ps}

The solution

Solved with ElectricField

Given

$q_{s} = 1.6 \times 10^{−7}$ C
$r_{ps} = 18$ cm

Find $E_{q,ps}$ .

Working

$r_{ps} = 18$ cm $= 0.18$ m
field of a point charge(point, sphere) $E_{q,ps} = \frac{k \cdot q_{s}}{{r_{ps}}^{2}}$
$E_{q,ps} = \frac{8.98755179 \times 10^{9} \cdot 1.6 \times 10^{−7}}{{0.18}^{2}}$
$E_{q,ps} = 44380$ N/C
$E_{q,ps} = 44382.9718$ V/m $= 44382.9718$ N/C
$E_{q,ps} \approx 4.4 \times 10^{4}$ N/Cto 2 significant figures

Conclusion

Outside the sphere its charge acts as if gathered at the centre, so the field falls off as the square of the distance from there.

The Ruby

1require_relative "../physics"23question :eighteen_centimetres_from_a_charged_sphere do4  kind :numerical5  describes "A spherical conductor of radius 12 cm has a charge of %{charge} distributed uniformly on its surface. " \6            "%{field} at a point %{distance} from the centre of the sphere, in N/C?"7  given charge: "1.6e-7 C"8  given distance: "18 cm"9  asks field: "What is the electric field"10end

1require_relative "../questions/eighteen_centimetres_from_a_charged_sphere"2require_relative "../science/electric_charges_and_fields/electric_field"34solve :eighteen_centimetres_from_a_charged_sphere, using: ElectricField do5  objects :sphere, :point67  take charge: sphere.charge8  take distance: point.separation(from: sphere)910  field = point.field_of_a_charge(by: sphere)11  answer field:, in: "N/C", figures: 21213  conclusion "Outside the sphere its charge acts as if gathered at the centre, so the field falls off as the " \14             "square of the distance from there."15end

1require_relative "../solutions/eighteen_centimetres_from_a_charged_sphere"23play_with :eighteen_centimetres_from_a_charged_sphere do4  title "Eighteen Centimetres from 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)