A) \[-2\sqrt{3}\,J\]
B) \[5\sqrt{3}\,J\]
C) \[-3\sqrt{2}\,J\]
D) \[3\sqrt{5}\,\,J\]
Correct Answer: A
Solution :
(a) \[-2\sqrt{3}\,J\] Here, length of dipole, \[2a=\,20cm\,=20\times {{10}^{-2}}m\] Charge \[q=\pm 3\times {{10}^{-3}}C,\,\theta =60{}^\circ \] and torque t =6 Nm As \[\tau =6\,Nm\] or \[E=\frac{\tau }{p\,\sin \theta }=\frac{\tau }{q\left( 2a \right)\sin \theta }\] \[\left[ \because \,\,p=q\left( 2a \right) \right]\] \[\therefore \,\,E=\frac{6}{3\times {{10}^{-3}}\times 20\times {{10}^{-2}}\times \sin \,60{}^\circ }\] \[=\frac{{{10}^{5}}}{5\sqrt{3}}N{{C}^{-1}}\] Potential energy of dipole, \[U=-pE\,\cos \theta =-q\left( 2a \right)E\cos \theta \] \[=-3\times {{10}^{-3}}\left( 20\times {{10}^{-2}} \right)\frac{{{10}^{5}}}{5\sqrt{3}}\cos 60{}^\circ \] \[=\frac{-3\times {{10}^{-5}}\times 20\times {{10}^{5}}}{5\sqrt{3}}=-2\sqrt{3}\,J\]
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