Dipoles Turning in a Field

solved DipoleInAField

Q. A molecule of a substance has a permanent electric dipole moment of magnitude 10⁻²⁹ C m. A mole of this substance, with a total dipole moment of 6 × 10⁻⁶ C m $p = 6 \times 10^{−6}$ C m, is polarised (at low temperature) by applying a strong electrostatic field of magnitude 1 × 10⁶ V/m $E = 1 \times 10^{6}$ V/m. The direction of the field is suddenly changed by an angle of 60° $θ_{0} = 60 °$ . What is the heat released in aligning the dipoles along the new direction $θ_{1} = 0 °$ of the field, in joules?

Inputs

moment

p

6e-06 C m

field

E

1e+06 V/m

turn

θ_{0}

60.00°

Outputs

heat

Q

How it's solved

Solved with DipoleInAField

Given

$p = 6 \times 10^{−6}$ C m
$E = 1 \times 10^{6}$ V/m
$θ_{0} = 60 °$
$θ_{1} = 0 °$ from “aligning the dipoles along the new direction”

Find $Q$ .

Working

energy of a dipole in a field(dipoles, aligning, 1) $U_{0} = - p \cdot E \cdot \cos θ_{0}$
$U_{0} = - 6 \times 10^{−6} \cdot 1 \times 10^{6} \cdot \cos 60 °$
$U_{0} = −3$ J
energy of a dipole in a field(dipoles, aligning, 2) $U_{1} = - p \cdot E \cdot \cos θ_{1}$
$U_{1} = - 6 \times 10^{−6} \cdot 1 \times 10^{6} \cdot \cos 0 °$
$U_{1} = −6$ J
energy lost as heat(dipoles, aligning) $Q = U_{0} - U_{1}$
$Q = (−3) - (−6)$
$Q = 3$ J
$Q \approx 3.0$ Jto 2 significant figures

Conclusion

The dipoles start at sixty degrees to the new field and turn into line with it, dropping to lower energy; the drop comes out as heat.

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 :dipoles_turning_in_a_field do4  kind :numerical5  describes "A molecule of a substance has a permanent electric dipole moment of magnitude 10⁻²⁹ C m. A mole of this " \6            "substance, with a total dipole moment of %{moment}, is polarised (at low temperature) by applying a " \7            "strong electrostatic field of magnitude %{field}. The direction of the field is suddenly changed by an " \8            "angle of %{turn}. %{heat} in aligning the dipoles along the new direction of the field, in joules?"9  given moment: "6e-6 C m"10  given field: "1e6 V/m"11  given turn: "60 deg"12  asks heat: "What is the heat released"13end

solution.rb

1require_relative "../questions/dipoles_turning_in_a_field"2require_relative "../science/electrostatic_potential_and_capacitance/dipole_in_a_field"34solve :dipoles_turning_in_a_field, using: DipoleInAField do5  objects :dipoles6  dipoles.goes(:aligning)78  take moment: dipoles.dipole_moment9  take field: E10  take turn: dipoles.angle_to_the_field(at: aligning.start)1112  deduce_from "aligning the dipoles along the new direction", dipoles.angle_to_the_field(at: aligning.end) => "0 deg"1314  heat = dipoles.heat_given_out(during: aligning)15  answer heat:, in: "J", figures: 21617  conclusion "The dipoles start at sixty degrees to the new field and turn into line with it, dropping to " \18             "lower energy; the drop comes out as heat."19end

playground.rb

1require_relative "../solutions/dipoles_turning_in_a_field"23play_with :dipoles_turning_in_a_field do4  title "Dipoles Turning in a Field"5end

The laws it drew on

quantity	written	units	within	from
field strength	$E$	V/m	—	DipoleInAField
dipole moment	$p$	C m	—	DipoleInAField
angle to the field	$θ$	—	0 … 3.142	DipoleInAField
potential energy	$U$	J	—	DipoleInAField
heat given out	$Q$	J	—	DipoleInAField
charge	$q$	C	—	DipoleInAField
length of the dipole	$l$	m	—	DipoleInAField
torque	$τ$	J	—	DipoleInAField
potential energy of dipoles at start of aligning	$U_{0}$	J	—
dipole moment of dipoles	$p$	C m	—
angle to the field of dipoles at start of aligning	$θ_{0}$	—	0 … 3.142
potential energy of dipoles at end of aligning	$U_{1}$	J	—
angle to the field of dipoles at end of aligning	$θ_{1}$	—	0 … 3.142
dipole moment of ground	$p_{g}$	C m	—
charge of ground	$q_{g}$	C	—
length of the dipole of ground	$l_{g}$	m	—
charge of dipoles	$q$	C	—
length of the dipole of dipoles	$l$	m	—
torque on ground	$τ_{g}$	J	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
torque on ground at start of aligning	$τ_{0,g}$	J	—
angle to the field of ground at start of aligning	$θ_{0,g}$	—	0 … 3.142
torque on ground at end of aligning	$τ_{1,g}$	J	—
angle to the field of ground at end of aligning	$θ_{1,g}$	—	0 … 3.142
torque on dipoles	$τ$	J	—
angle to the field of dipoles	$θ$	—	0 … 3.142
torque on dipoles at start of aligning	$τ_{0}$	J	—
torque on dipoles at end of aligning	$τ_{1}$	J	—
heat given out of dipoles during aligning	$Q$	J	—

p_{d} = q_{d} \cdot l_{d}

moment of a pair of charges

τ_{d} (moment) = p_{d} \cdot E \cdot \sin θ_{d} (moment)

torque on a dipole

U_{0} = - p \cdot E \cdot \cos θ_{0}

energy of a dipole in a field(dipoles, aligning, 1)

U_{1} = - p \cdot E \cdot \cos θ_{1}

energy of a dipole in a field(dipoles, aligning, 2)

p_{g} = q_{g} \cdot l_{g}

moment of a pair of charges(ground)

p = q \cdot l

moment of a pair of charges(dipoles)

τ_{g} = p_{g} \cdot E \cdot \sin θ_{g}

torque on a dipole(ground)

τ_{0,g} = p_{g} \cdot E \cdot \sin θ_{0,g}

torque on a dipole(ground, start of aligning)

τ_{1,g} = p_{g} \cdot E \cdot \sin θ_{1,g}

torque on a dipole(ground, end of aligning)

τ = p \cdot E \cdot \sin θ

torque on a dipole(dipoles)

τ_{0} = p \cdot E \cdot \sin θ_{0}

torque on a dipole(dipoles, start of aligning)

τ_{1} = p \cdot E \cdot \sin θ_{1}

torque on a dipole(dipoles, end of aligning)

Q = U_{0} - U_{1}

energy lost as heat(dipoles, aligning)