An electric dipole of length 3 text{ cm} is p | vedclass

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

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$1.2 	imes 10^{-2} 	ext{ C}$

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(B) The torque $	au$ experienced by an electric dipole in a uniform electric field $E$ is given by the formula $	au = pE \sin 	heta$, where $p$ is the dipole moment and $	heta$ is the angle between the dipole axis and the electric field.Given:Length of dipole $2a = 3 	ext{ cm} = 3 	imes 10^{-2} 	ext{ m}$Electric field $E = 5 	imes 10^4 	ext{ N/C}$Angle $	heta = 60^\circ$Torque $	au = 9\sqrt{3} 	ext{ Nm}$We know $p = q 	imes (2a)$, where $q$ is the magnitude of the charge.Substituting the values into the torque formula:$9\sqrt{3} = q 	imes (3 	imes 10^{-2}) 	imes (5 	imes 10^4) 	imes \sin 60^\circ$$9\sqrt{3} = q 	imes 1500 	imes \frac{\sqrt{3}}{2}$$9 = q 	imes 750$$q = \frac{9}{750} = \frac{3}{250} = 0.012 	ext{ C}$$q = 1.2 	imes 10^{-2} 	ext{ C}$

An electric dipole of length $3 \text{ cm}$ is placed with its axis making an angle of $60^\circ$ with a uniform electric field of $5 \times 10^4 \text{ N/C}$. It experiences a torque of $9\sqrt{3} \text{ Nm}$. The magnitude of charge on the dipole is $(\sin 60^\circ = \frac{\sqrt{3}}{2})$

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