An electric dipole is placed at an angle of 3 | vedclass

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(D) Given: $	heta = 30^\circ$,$E = 2 	imes 10^5 \, 	ext{NC}^{-1}$,$	au = 4 \, 	ext{Nm}$,and $l = 2 \, 	ext{cm} = 0.02 \, 	ext{m}$.The torque ex

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$2 \, 	ext{mC}$

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(D) Given: $	heta = 30^\circ$,$E = 2 	imes 10^5 \, 	ext{NC}^{-1}$,$	au = 4 \, 	ext{Nm}$,and $l = 2 \, 	ext{cm} = 0.02 \, 	ext{m}$.The torque experienced by an electric dipole in an electric field is given by the formula: $	au = pE \sin 	heta$,where $p = ql$.Substituting $p = ql$ into the torque equation: $	au = qlE \sin 	heta$.Rearranging to solve for $q$: $q = \frac{	au}{El \sin 	heta}$.Substituting the given values: $q = \frac{4}{2 	imes 10^5 	imes 0.02 	imes \sin(30^\circ)}$.Since $\sin(30^\circ) = 0.5$: $q = \frac{4}{2 	imes 10^5 	imes 0.02 	imes 0.5} = \frac{4}{2 	imes 10^3} = 2 	imes 10^{-3} \, 	ext{C}$.Therefore,$q = 2 \, 	ext{mC}$.

An electric dipole is placed at an angle of $30^\circ$ with an electric field intensity of $2 \times 10^5 \, \text{NC}^{-1}$. It experiences a torque equal to $4 \, \text{Nm}$. The charge on the dipole,if the dipole length is $2 \, \text{cm}$,is:

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