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

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$8 	imes 10^{-3} \ C$

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(B) The torque $	au$ experienced by an electric dipole in a uniform electric field is given by the formula: $	au = pE \sin 	heta$,where $p = q 	imes (2a)$ is the dipole moment,$E$ is the electric field,and $	heta$ is the angle between the dipole axis and the electric field.Given:Length of dipole $(2a) = 2 \ cm = 0.02 \ m = 2 	imes 10^{-2} \ m$Electric field $(E) = 10^{5} \ N/C$Angle $(	heta) = 60^{\circ}$Torque $(	au) = 8 \sqrt{3} \ Nm$Substituting the values into the formula:$8 \sqrt{3} = (q 	imes 2 	imes 10^{-2}) 	imes 10^{5} 	imes \sin 60^{\circ}$$8 \sqrt{3} = q 	imes 2 	imes 10^{3} 	imes \frac{\sqrt{3}}{2}$$8 \sqrt{3} = q 	imes \sqrt{3} 	imes 10^{3}$$q = \frac{8 \sqrt{3}}{\sqrt{3} 	imes 10^{3}} = 8 	imes 10^{-3} \ C$.Thus,the magnitude of the charge is $8 	imes 10^{-3} \ C$.

An electric dipole of length $2 \ cm$ is placed with its axis making an angle of $60^{\circ}$ to a uniform electric field of $10^{5} \ N/C$. If it experiences a torque of $8 \sqrt{3} \ Nm$,calculate the magnitude of the charge on the dipole. (Given: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$)

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