Question

14. An electric dipole of length 2 cm is placed with its axis making angle of 30° to a uniform electric
field of 10⁵ N/C. It experiences a torque of 17.32 Nm. The magnitude of charge on the dipole is​

Answer 1

Answer:

we know, torque of electric dipole is the cross product of dipole moment and electric field.

i.e., \bf{\tau=P\times E}τ=P×E

magnitude of torque, |\tau|=|P||E|sin\theta∣τ∣=∣P∣∣E∣sinθ ,

given, |\tau|=17.3Nm,|E|=10^5N/C∣τ∣=17.3Nm,∣E∣=10

5

N/C and \theta=30^{\circ}θ=30

∘

so, 17.3N/m = |P| × 10^5 × sin30°

or, 3.46 × 10^-4 = |P|

so, magnitude of dipole moment, |P| = 3.46 × 10^-4 Cm

now, potential energy = -P.E

= -|P|.|E|cos30°

= -3.46 × 10^-4 × 10^5 × √3/2

= - 3 × 10¹

= -30 J

Answer 2

Answer:

we know, torque of electric dipole is the cross product of dipole moment and electric field.

i.e., \bf{\tau=P\times E}τ=P×E

magnitude of torque, |\tau|=|P||E|sin\theta∣τ∣=∣P∣∣E∣sinθ ,

given, |\tau|=17.3Nm,|E|=10^5N/C∣τ∣=17.3Nm,∣E∣=10

5

N/C and \theta=30^{\circ}θ=30

∘

so, 17.3N/m = |P| × 10^5 × sin30°

or, 3.46 × 10^-4 = |P|

so, magnitude of dipole moment, |P| = 3.46 × 10^-4 Cm

now, potential energy = -P.E

= -|P|.|E|cos30°

= -3.46 × 10^-4 × 10^5 × √3/2

= - 3 × 10¹

= -30 J