An electric dipole has a fixed dipole moment | vedclass

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Question

(C) The torque experienced by an electric dipole in an electric field is given by $\vec{T} = \vec{P} 	imes \vec{E}$.Let the dipole moment be $\vec{P}

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The torque experienced by an electric dipole in an electric field is given by $\vec{T} = \vec{P} 	imes \vec{E}$.Let the dipole moment be $\vec{P} = P \cos 	heta \hat{i} + P \sin 	heta \hat{j}$.For the first electric field $\vec{E_1} = E\hat{i}$:$\vec{T_1} = (P \cos 	heta \hat{i} + P \sin 	heta \hat{j}) 	imes (E\hat{i}) = PE \cos 	heta (\hat{i} 	imes \hat{i}) + PE \sin 	heta (\hat{j} 	imes \hat{i}) = 0 - PE \sin 	heta \hat{k} = -PE \sin 	heta \hat{k}$.Given $\vec{T_1} = 	au \hat{k}$,so $	au = -PE \sin 	heta$.For the second electric field $\vec{E_2} = \sqrt{3}E\hat{j}$:$\vec{T_2} = (P \cos 	heta \hat{i} + P \sin 	heta \hat{j}) 	imes (\sqrt{3}E\hat{j}) = \sqrt{3}PE \cos 	heta (\hat{i} 	imes \hat{j}) + \sqrt{3}PE \sin 	heta (\hat{j} 	imes \hat{j}) = \sqrt{3}PE \cos 	heta \hat{k} + 0 = \sqrt{3}PE \cos 	heta \hat{k}$.Given $\vec{T_2} = -\vec{T_1} = -(	au \hat{k}) = -(-PE \sin 	heta \hat{k}) = PE \sin 	heta \hat{k}$.Equating the two expressions for $\vec{T_2}$:$\sqrt{3}PE \cos 	heta \hat{k} = PE \sin 	heta \hat{k}$.$	an 	heta = \sqrt{3}$.Therefore,$	heta = 60^{\circ}$.

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