An electric dipole of dipole moment 6.0 times | vedclass

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(B) The work done $W$ in rotating an electric dipole in a uniform electric field is given by the change in potential energy: $W = \Delta U = U_f - U_i

Vedclass · Accepted Answer

$18$

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(B) The work done $W$ in rotating an electric dipole in a uniform electric field is given by the change in potential energy: $W = \Delta U = U_f - U_i$.Potential energy of a dipole is $U = -pE \cos 	heta$.Initially,the dipole is along the electric field,so $	heta_i = 0^{\circ}$. Thus,$U_i = -pE \cos(0^{\circ}) = -pE$.Finally,the dipole is rotated by $180^{\circ}$,so $	heta_f = 180^{\circ}$. Thus,$U_f = -pE \cos(180^{\circ}) = pE$.The work done is $W = pE - (-pE) = 2pE$.Given $p = 6.0 	imes 10^{-6} \, Cm$ and $E = 1.5 	imes 10^3 \, NC^{-1}$.$W = 2 	imes (6.0 	imes 10^{-6}) 	imes (1.5 	imes 10^3) = 18 	imes 10^{-3} \, J = 18 \, mJ$.

An electric dipole of dipole moment $6.0 \times 10^{-6} \, Cm$ is placed in a uniform electric field of $1.5 \times 10^3 \, NC^{-1}$ in such a way that the dipole moment is along the electric field. The work done in rotating the dipole by $180^{\circ}$ in this field will be $......... \, mJ$.

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