Two point charges -4 mu C and 4 mu C,constitu | vedclass

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(A) The dipole moment $p = q 	imes (2a)$. Here,$q = 4 	imes 10^{-6} \ C$ and the distance $2a = 18 \ cm = 0.18 \ m$.So,$p = (4 	imes 10^{-6} \ C) \

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$14.4$

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(A) The dipole moment $p = q 	imes (2a)$. Here,$q = 4 	imes 10^{-6} \ C$ and the distance $2a = 18 \ cm = 0.18 \ m$.So,$p = (4 	imes 10^{-6} \ C) 	imes (0.18 \ m) = 7.2 	imes 10^{-7} \ Cm$.The potential energy of a dipole in an electric field is given by $U = -pE \cos 	heta$.The work done in rotating the dipole from $	heta_1 = 0^{\circ}$ (equilibrium) to $	heta_2 = 180^{\circ}$ is $W = U_f - U_i = (-pE \cos 180^{\circ}) - (-pE \cos 0^{\circ})$.$W = pE - (-pE) = 2pE$.Substituting the values: $W = 2 	imes (7.2 	imes 10^{-7} \ Cm) 	imes (10^4 \ NC^{-1})$.$W = 14.4 	imes 10^{-3} \ J = 14.4 \ mJ$.

Two point charges $-4 \mu C$ and $4 \mu C$,constituting an electric dipole,are placed at $(-9, 0, 0) \ cm$ and $(9, 0, 0) \ cm$ in a uniform electric field of strength $10^4 \ NC^{-1}$. The work done on the dipole in rotating it from the equilibrium position through $180^{\circ}$ is: (in $mJ$)

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