An electric dipole of dipole moment 2 × 10-12 C - m is oriented at an angle of 60° with electric field of intensity 3 NC -1 . What work need to be done to turn the dipole so that it becomes anti-parallel to the field?

Question

An electric dipole of dipole moment 2 × 10-12 C - m is oriented at an angle of 60° with electric field of intensity 3 NC -1 . What work need to be done to turn the dipole so that it becomes anti-parallel to the field? (A) 12 × 10-12 J (B) -9 × 10-12 J (C) 9 × 10-12 J (D) -12 × 10-12 J

Tardigrade Admin · Accepted Answer

W =- PE ( cos θ2- cos θ1) = PE ( cos θ1- cos θ2) =2 × 10-12 × 3 ×( cos 60°- cos 180°) =6 × 10-12((1/2)+1) =(3/2) × 6 × 10-12 W =9 × 10-12 J

Q. An electric dipole of dipole moment $2 \times 1 0^{- 12} C - m$ is oriented at an angle of $6 0^{\circ}$ with electric field of intensity $3 N C^{- 1} .$ What work need to be done to turn the dipole so that it becomes anti-parallel to the field?

$12 \times 1 0^{- 12} J$

$- 9 \times 1 0^{- 12} J$

$9 \times 1 0^{- 12} J$

$- 12 \times 1 0^{- 12} J$

$W = - PE (cos θ_{2} - cos θ_{1})$
$= PE (cos θ_{1} - cos θ_{2})$
$= 2 \times 1 0^{- 12} \times 3 \times (cos 6 0^{\circ} - cos 18 0^{\circ})$
$= 6 \times 1 0^{- 12} (\frac{1}{2} + 1)$
$= \frac{3}{2} \times 6 \times 1 0^{- 12}$
$W = 9 \times 1 0^{- 12} J$

Q. An electric dipole of dipole moment 2×10−12C−m is oriented at an angle of 60∘ with electric field of intensity 3NC−1. What work need to be done to turn the dipole so that it becomes anti-parallel to the field?

12×10−12J

−9×10−12J

9×10−12J

−12×10−12J