Question

An electric dipole of dipole moment $2 \times 10^{-12} C - m$ is oriented at an angle of $60^{\circ}$ 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. $12 \times 10^{-12} J$
• B. $-9 \times 10^{-12} J$
• C. $9 \times 10^{-12} J$
• D. $-12 \times 10^{-12} J$

Answer 1

Answer: (C)

$W =- PE \left(\cos \theta_{2}-\cos \theta_{1}\right) $
$= PE \left(\cos \theta_{1}-\cos \theta_{2}\right) $
$=2 \times 10^{-12} \times 3 \times\left(\cos 60^{\circ}-\cos 180^{\circ}\right) $
$=6 \times 10^{-12}\left(\frac{1}{2}+1\right) $
$=\frac{3}{2} \times 6 \times 10^{-12} $
$W =9 \times 10^{-12} J $