An electric dipole is placed at an angle 60^{ | vedclass

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Question

(C) The torque $	au$ experienced by an electric dipole in an external electric field is given by the formula: $	au = pE \sin 	heta$,where $p = q

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$10^{-3}$

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(C) The torque $	au$ experienced by an electric dipole in an external electric field is given by the formula: $	au = pE \sin 	heta$,where $p = q 	imes (2a)$ is the dipole moment.Given:$	au = 8 \sqrt{3} \,Nm$$E = 4 	imes 10^5 \,N/C$$	heta = 60^{\circ}$Length of dipole $(2a) = 4 \,cm = 4 	imes 10^{-2} \,m$Substituting the values into the formula:$8 \sqrt{3} = (q 	imes 4 	imes 10^{-2}) 	imes (4 	imes 10^5) 	imes \sin 60^{\circ}$Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have:$8 \sqrt{3} = q 	imes 16 	imes 10^3 	imes \frac{\sqrt{3}}{2}$$8 \sqrt{3} = q 	imes 8 \sqrt{3} 	imes 10^3$$1 = q 	imes 10^3$$q = 10^{-3} \,C$.

An electric dipole is placed at an angle $60^{\circ}$ with an electric field of strength $4 \times 10^5 \,N/C$. It experiences a torque equal to $8 \sqrt{3} \,Nm$. Calculate the charge on the dipole,if the dipole length is $4 \,cm$.

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