A Dipole at Thirty Degrees

Q. An electric dipole with dipole moment 4 × 10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N/C. What is the magnitude of the torque acting on the dipole, in N m? expects 1 × 10⁻⁴ N m

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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?

The playground

Inputs

moment

p

4e-09 C m

angle

θ

30.00°

field

E

5e+04 N/C

Outputs

torque

τ

The solution

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 Ruby

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

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

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

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)