JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Two point charges \(-4 \mu c\) and \(4 \mu c\), constituting an electric dipole, are placed at \((-9,0,0) \mathrm{cm}\) and \((9,0,0) \mathrm{cm}\) in a uniform electric field of strength \(10^4 \mathrm{NC}^{-1}\). The work done on the dipole in rotating it from the equilibrium through \(180^{\circ}\) is :
- A 18.4 mJ
- B 14.4 mJ
- C 12.4 mJ
- D 16.4 mJ
Answer & Solution
Correct Answer
(B) 14.4 mJ
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{U}=-\mathrm{PE} \cos \theta \\ & \mathrm{w}_{\mathrm{ext}}=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\mathrm{PE} \cos 180^{\circ}+\mathrm{PE} \cos 0^{\circ} \\ & \mathrm{w}_{\mathrm{ext}}=2 \mathrm{PE} \\ & =2 \times\left(4…
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