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An electric dipole consists of two opposite charges of magnitude $q=1 \times 10^{-6} \mathrm{C}$ separated by $2.0 \mathrm{~cm}$. The dipole is placed in an external field of $1 \times 10^{5} \mathrm{NC}^{-1}$. 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 $\left(\theta=0^{\circ}\right)$ ?
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Verified Answer
The correct answer is:
$2 \times 10^{-3} \mathrm{~N}-\mathrm{m}, 4 \times 10^{-3} \mathrm{~J}$
$$
\begin{array}{l}
\begin{aligned}
2 a &=2.0 \mathrm{~cm}=2.0 \times 10^{-2} \mathrm{~m} \\
E &=1 \times 10^{5} \mathrm{NC}^{-1}, \tau_{\max }=? \\
W=?, \theta_{1}=0^{\circ}, \theta_{2}=180^{\circ} \\
\tau_{\max } &=p E=q(2 a) \mathrm{E} \\
&=1 \times 10^{-6} \times 2.0 \times 10^{-2} \times 1 \times 10^{5} \\
&=2 \times 10^{-3} \mathrm{Nm}
\end{aligned} \\
\qquad \begin{aligned}
W &=p E\left(\cos \theta_{1}-\cos \theta_{2}\right) \\
&=\left(10^{-6} \times 2 \times 10^{-2}\right)\left(10^{5}\right)\left(\cos 0^{\circ}-\cos 180^{\circ}\right) \\
&=4 \times 10^{-3} \mathrm{~J}
\end{aligned}
\end{array}
$$
\begin{array}{l}
\begin{aligned}
2 a &=2.0 \mathrm{~cm}=2.0 \times 10^{-2} \mathrm{~m} \\
E &=1 \times 10^{5} \mathrm{NC}^{-1}, \tau_{\max }=? \\
W=?, \theta_{1}=0^{\circ}, \theta_{2}=180^{\circ} \\
\tau_{\max } &=p E=q(2 a) \mathrm{E} \\
&=1 \times 10^{-6} \times 2.0 \times 10^{-2} \times 1 \times 10^{5} \\
&=2 \times 10^{-3} \mathrm{Nm}
\end{aligned} \\
\qquad \begin{aligned}
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} \mathrm{~J}
\end{aligned}
\end{array}
$$
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