JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 23)
An electric dipole of dipole moment $6 \times 10^{-6} \mathrm{Cm}$ is placed in uniform electric field of magnitude $10^{6} \mathrm{V} / \mathrm{m}$. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be _______ J.
Answer
12
Explanation
$$\begin{aligned}
& \mathrm{p}=6 \times 10^{-6} \mathrm{Cm} \\
& \mathrm{E}=10^6 \mathrm{v} / \mathrm{m} \\
& \mathrm{~W}=\Delta \mathrm{U}=-\mathrm{pE}\left(\cos \theta_{\mathrm{f}}-\cos \theta_{\mathrm{i}}\right) \\
& \mathrm{W}=2 \mathrm{pE}=12 \mathrm{~J}
\end{aligned}$$
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