Q. An electric dipole consists of two opposite charges of magnitude $q = 1 \times 10^{- 6\ }$C separated by 2.0 cm. the dipole is placed in an external field of $1 \times 10^{5}$ $NC^{- 1}$. What maximum torque does the field exert on the dipole? How much work must an external agent do to turn the dipole end for end, starting from position of alignment ($\theta = 0^{0}$)
A) $4.4 \times 10^{6\ }N - m,\ 3.2 \times 10^{- 4}\ J$
B) $- 2 \times 10^{3\ }N - m,\ - 4 \times 10^{3}\ J$
C) $4 \times 10^{3\ }N - m,2 \times 10^{- 3}\ J$
D) $2 \times 10^{- 3\ }N - m,\ 4 \times 10^{- 3}\ J$
Explanation
(d) $q = \pm 1 \times 10^{- 6}C$
$2a = 2.0cm = 2.0 \times 10^{- 2}m$
$E = 1 \times 10^{5}NC^{- 1},\tau_{\max} = ?$
$W = ?,\ \theta_{1} = 0^{0},\ \theta_{2} = {\ 180_{\ \ \ }^{0}}_{\ \ \ }$
$\tau_{\max} = pE = q(2a)E$
$= 1 \times 10^{- 6} \times 2.0 \times 10^{- 2} \times 1 \times 10^{5} = 2 \times 10^{- 3}\ $Nm
$W = pE\left( \cos\theta_{1} - \cos\theta_{2} \right)$
$= \left( 10^{- 6} \times 2 \times 10^{- 2} \right)\ \left( 10^{5} \right)$ $\left( \cos{0^{0} - \cos 180^{0}} \right)$
$= 4 \times 10^{- 3}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.