A Dipole at Thirty Degrees

solved DipoleInAField

Q. An electric dipole with dipole moment 4 × 10⁻⁹ C m $p = 4 \times 10^{−9}$ C m is aligned at 30° $θ = 30 °$ with the direction of a uniform electric field of magnitude 5 × 10⁴ N/C $E = 5 \times 10^{4}$ N/C. What is the magnitude of the torque acting on the dipole, in N m?

Inputs

moment

p

4e-09 C m

angle

θ

30.00°

field

E

5e+04 N/C

Outputs

torque

τ

How it's solved

Solved with DipoleInAField

Given

$p = 4 \times 10^{−9}$ C m
$θ = 30 °$
$E = 5 \times 10^{4}$ N/C

Find $τ$ .

Working

$E = 5 \times 10^{4}$ N/C $= 50000$ V/m
torque on a dipole(dipole) $τ = p \cdot E \cdot \sin θ$
$τ = 4 \times 10^{−9} \cdot 50000 \cdot \sin 30 °$
$τ = 1 \times 10^{−4}$ N m
$τ = 1 \times 10^{−4}$ J $= 1 \times 10^{−4}$ N m
$τ \approx 1 \times 10^{−4}$ N mto 1 significant figure

Conclusion

The field pulls the two charges of the dipole opposite ways, turning it toward the field; the turning grows with the sine of the angle between them.

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 :a_dipole_at_thirty_degrees do4  kind :numerical5  describes "An electric dipole with dipole moment %{moment} is aligned at %{angle} with the direction of a uniform " \6            "electric field of magnitude %{field}. %{torque}, in N m?"7  given moment: "4e-9 C m"8  given angle: "30 deg"9  given field: "5e4 N/C"10  asks torque: "What is the magnitude of the torque acting on the dipole"11end

solution.rb

1require_relative "../questions/a_dipole_at_thirty_degrees"2require_relative "../science/electrostatic_potential_and_capacitance/dipole_in_a_field"34solve :a_dipole_at_thirty_degrees, using: DipoleInAField do5  objects :dipole67  take moment: dipole.dipole_moment8  take angle: dipole.angle_to_the_field9  take field: E1011  torque = dipole.torque12  answer torque:, in: "N m", figures: 11314  conclusion "The field pulls the two charges of the dipole opposite ways, turning it toward the field; the " \15             "turning grows with the sine of the angle between them."16end

playground.rb

1require_relative "../solutions/a_dipole_at_thirty_degrees"23play_with :a_dipole_at_thirty_degrees do4  title "A Dipole at Thirty Degrees"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
dipole moment of ground	$p_{g}$	C m	—
charge of ground	$q_{g}$	C	—
length of the dipole of ground	$l_{g}$	m	—
dipole moment of dipole	$p$	C m	—
charge of dipole	$q$	C	—
length of the dipole of dipole	$l$	m	—
torque on ground	$τ_{g}$	J	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
torque on dipole	$τ$	J	—
angle to the field of dipole	$θ$	—	0 … 3.142

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

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

moment of a pair of charges(ground)

p = q \cdot l

moment of a pair of charges(dipole)

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

torque on a dipole(ground)

τ = p \cdot E \cdot \sin θ

torque on a dipole(dipole)