Question

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 to end, starting from position of alignment $\left(\theta=0^{\circ}\right) ?$

• A. $4.4 \times10^{6}N-m, 3.2 \times10^{-4}J$
• B. $- 2\times10^{-3}N-m,-4\times10^{3} J$
• C. $4\times10^{3}N-m, 2\times10^{-3} J$
• D. $2\times10^{-3} N-m, 4\times10^{-3}J$

Answer 1

Answer: (D)

$q=\pm 1 \times 10^{-6} C$
$2 a=2.0\, cm =2.0 \times 10^{-2} m$
$E=1 \times 10^{5} NC ^{-1}, \tau_{\max }=?$
$W=?, \theta_{1}=0^{\circ}, \theta_{2}=180^{\circ}$
$\tau_{\max } =p E=q(2 a) E$
$=1 \times 10^{-6} \times 2.0 \times 10^{-2} \times 1 \times 10^{5}$
$=2 \times 10^{-3} Nm$
$W= p E\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$=\left(10^{-6} \times 2 \times 10^{-2}\right)\left(10^{5}\right)\left(\cos 0^{\circ}-\cos 180^{\circ}\right)$
$=4 \times 10^{-3} J$