Two point-charges of + 1 μC and - 1 μC are kept at a distance of 2 cm . They form an electric dipole. This dipole is situated in a uniform electric; field of 1 × 10 5 NC - 1 , with dipole moment vector p making an angle of 30 ∘ with the field. Calculate (i) the magnitude of electric dipole moment.

Given, q 1 = + 1 × 10 - 6 C q 2 = - 1 × 10 - 6 C Dipole length = 2 l = 2 cm = 2 × 10 - 2 m Electric field intensity = E = 1 × 10 5 NC - 1 Dipole moment makes angle 30 ° i . e . , θ = 30 ° Electric dipole moment = p = 2 l · q = 2 × 10 - 2 × 1 × 10 - 6 = 2 × 10 - 8 Cm

Two point-charges of + 1 μC and - 1 μC are kept at a distance of 2 cm . They form an electric dipole. This dipole is situated in a uniform electric; field of 1 × 10 5 NC - 1 , with dipole moment vector p making an angle of 30 ∘ with the field. Calculate (i) the magnitude of electric dipole moment.

Given, q 1 = + 1 × 10 - 6 C q 2 = - 1 × 10 - 6 C Dipole length = 2 l = 2 cm = 2 × 10 - 2 m Electric field intensity = E = 1 × 10 5 NC - 1 Dipole moment makes angle 30 ° i . e . , θ = 30 ° Electric dipole moment = p = 2 l · q = 2 × 10 - 2 × 1 × 10 - 6 = 2 × 10 - 8 Cm

Two point-charges of + 1 μC and - 1 μC are kept at a distance of 2 cm . They form an electric dipole. This dipole is situated in a uniform electric field of 1 × 10 5 NC - 1 , with dipole moment vector p making an angle of 30 ∘ with the field. Calculate, (ii) the torque acting on the dipole.

Given, q 1 = + 1 × 10 - 6 C q 2 = - 1 × 10 - 6 C Dipole length = 2 l = 2 cm = 2 × 10 - 2 m Electric field intensity = E = 1 × 10 5 NC - 1 Dipole moment makes angle 30 ° i . e . , θ = 30 ° Torque on dipole = t = PE sin θ = 2 × 10 - 8 × 1 × 10 5 × sin 30 ° = 2 × 10 - 3 × 1 2 = 1 × 10 - 3 Nm

Two point-charges of + 1 μC and - 1 μC are kept at a distance of 2 cm . They form an electric dipole. This dipole is situated in a uniform electric; field of 1 × 10 5 NC - 1 , with dipole moment vector p making an angle of 30 ∘ with the field. Calculate (iii) When would be the torque maximum? How much?

Given, q 1 = + 1 × 10 - 6 C q 2 = - 1 × 10 - 6 C Dipole length = 2 l = 2 cm = 2 × 10 - 2 m Electric field intensity = E = 1 × 10 5 NC - 1 Dipole moment makes angle 30 ° i . e . , θ = 30 ° For t → t m a x m , sinθ = m a x m = 1 = sin 90 ° ⇒ θ = 90 ° ∴ t m a x m = PEsin 90 ° = 2 × 10 - 8 × 1 × 10 5 = 2 × 10 - 3 Nm It is maximum when it is perpendicular to electric field.

An electric dipole, when held at 30 ∘ with respect to a uniform electric field of 10 4 N C , experiences a torque of 9 × 10 - 26 Nm . Calculate the moment of the dipole.

Given, Angle between dipole moment and the electric field intensity = θ = 30 ° Electric field intensity = E = 10 4 NC - 1 Torque experienced = t = 9 × 10 - 26 Nm We know that t = PE sin θ ⇒ P = t Esin θ = 9 × 10 - 26 10 4 × sin 30 ° = 9 × 10 - 26 10 4 × 1 2 = 18 × 10 - 30 Cm = 1 . 8 × 10 - 29 Cm .