Dipoles Turning in a Field

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, is polarised (at low temperature) by applying a strong electrostatic field of magnitude 1 × 10⁶ V/m. The direction of the field is suddenly changed by an angle of 60°. What is the heat released in aligning the dipoles along the new direction of the field, in joules? expects 3.0 J

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

moment

p

6e-06 C m

field

E

1e+06 V/m

turn

θ_{0}

60.00°

Outputs

heat

Q

The solution

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 Ruby

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

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

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

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)