Question

An electric dipole consists of two opposite charges of magnitude $q=1\times 10^{-6}C$ separated by $2.0cm$. The dipole is placed in an external field of $1\times 10^{5}N{C}^{-1}$. What maximum torque does the field exert on the dipole? How much work must an external agent do to turn the dipole end to end, starting from position of alignment $\left(\theta =0^{\circ }\right)?$

• 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. $2\times 10^{-3}N-m,4\times 10^{-3}J$

Answer 1

Answer: (D)

$q=\pm 1\times 10^{-6}C$
$2a=2.0cm=2.0\times 10^{-2}m$
$E=1\times 10^{5}N{C}^{-1},{\tau }_{\max }=?$
$W=?,{\theta }_1=0^{\circ },{\theta }_2=180^{\circ }$
${\tau }_{\max }=pE=q(2a)E$
$=1\times 10^{-6}\times 2.0\times 10^{-2}\times 1\times 10^5$
$=2\times 10^{-3}Nm$
$W=pE\left(\cos {\theta }_1-\cos {\theta }_2\right)$
$=\left(10^{-6}\times 2\times 10^{-2}\right)\left(10^5\right)\left(\cos 0^{\circ }-\cos 180^{\circ }\right)$
$=4\times 10^{-3}J$