A dipole consisting of an electron and a proton separated by a distance of 4 × 10 − 10 m is situated in an electric field of intensity 3 × 10 5 NC − 1 at an angle of 30 o with the field. Calculate the dipole moment and the torque acting on it. Charge on an electron = 1 .602 × 10 − 19 C .

Given, Magnitude of each charge of the dipole q = e = 1 .602 × 10 − 19 C Separation between charges d = 4 × 10 − 10 m Intensity of the electric field E = 3 × 10 5 NC − 1 Angle between the electric field and the dipole moment θ = 30 ∘ Magnetic dipole moment p = q × d = 1 .602 × 10 − 19 × 4 × 10 − 10 ≃ 6 .41 × 10 − 29 Cm Torque on the dipole τ = pEsin θ = 6 .41 × 10 − 29 × 3 × 10 5 × sin 30 ∘ Torque on the dipole = 6 .41 × 10 − 29 × 3 × 10 5 × 1 2 So the torque on the dipole = 9 .615 × 10 − 24 Nm

A dipole consisting of an electron and a proton separated by a distance of 4 × 10 − 10 m is situated in an electric field of intensity 3 × 10 5 NC − 1 at an angle of 30 o with the field. Calculate the dipole moment and the torque acting on it. Charge on an electron = 1 .602 × 10 − 19 C .

Given, Magnitude of each charge of the dipole q = e = 1 .602 × 10 − 19 C Separation between charges d = 4 × 10 − 10 m Intensity of the electric field E = 3 × 10 5 NC − 1 Angle between the electric field and the dipole moment θ = 30 ∘ Magnetic dipole moment p = q × d = 1 .602 × 10 − 19 × 4 × 10 − 10 ≃ 6 .41 × 10 − 29 Cm Torque on the dipole τ = pEsin θ = 6 .41 × 10 − 29 × 3 × 10 5 × sin 30 ∘ Torque on the dipole = 6 .41 × 10 − 29 × 3 × 10 5 × 1 2 So the torque on the dipole = 9 .615 × 10 − 24 Nm

An electric dipole is placed at an angle of 60 o with an electric field of magnitude 4 × 10 5 NC − 1 . It experiences a torque of 8 3 Nm . If the length of the dipole is 4 cm , determine the magnitude of either charge of the dipole.

Given, Torque on the dipole τ = 8 3 Nm Magnitude of the field strength E = 4 × 10 5 NC − 1 Length of the dipole d = 4 cm = 4 × 10 − 2 m Angle between the electric field and the dipole moment θ = 60 ∘ Torque on the dipole τ = pE sin θ = qdE sin θ ⇒ q = τ dE sin θ = 8 3 0 .04 × 4 × 10 5 × sin 60 ∘ = 8 3 0 .04 × 4 × 10 5 × 3 2 = 10 − 3 C Magnitude of the either charge of the dipole q = 10 − 3 C

The electric field in a certain region of space 5 ı ^ + 4 j ^ − 4 k ^ × 10 5 NC − 1 . Calculate electric flux due to this field over an area of 2 i ^ - j ^ × 10 - 2 m 2 .

Given, Electric field vector E → = 5 ı ^ + 4 j ^ − 4 k ^ × 10 5 NC − 1 Surface area vector S → = 2 i ^ - j ^ × 10 - 2 m 2 We know, electric flux is defined as number of electric lines of force passing through in a given area. Mathematically, electric flux is the dot product of electric field and surface area. Electric flux ϕ E = E → . S → = 5 ı ^ + 4 j ^ − 4 k ^ × 10 5 . 2 i ^ - j ^ × 10 - 2 = [ 5 × 2 + 4 × ( − 1 ) − 0 ] × 10 3 N m 2 C − 1 = 6 × 10 3 N m 2 C − 1