A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole moment p is now brought towards q from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is N×(mgh), where g is the acceleration due to gravity, then the value of N is _________ . (Note that for three coplanar forces keeping a point mass in equilibrium, Fsin⁡θ is the same for all forces, where F is any one of the forces and θ is the angle between the other two forces)

Question

A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole moment p is now brought towards q from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is N×(mgh), where g is the acceleration due to gravity, then the value of N is _________ . (Note that for three coplanar forces keeping a point mass in equilibrium, Fsin⁡θ is the same for all forces, where F is any one of the forces and θ is the angle between the other two forces)

Infinity Learn · Accepted Answer

Ui=0Uf=kqP2ℓsin⁡α22+mgh ......(i) Now, from ΔOABα+90−θ+90−θ=180⇒α=2θ From ΔABC:h=2ℓsin⁡α2sin⁡θ h=2ℓsin⁡α2sin⁡α2⇒h=2ℓsin2⁡α2Now charge is in equilibrium at point B.So, using sine rule⇒mgsin⁡90+α2=qEsin⁡[180−2θ]⇒mgcos⁡α2=qEsin⁡2θ⇒mgcos⁡α2=qEsin⁡α⇒mgcos⁡α2=qE2sin⁡α2cos⁡α2⇒qE=mg2sin⁡α2⇒q2kp2ℓsin⁡α23=mg2sin⁡α2⇒kpq2ℓsin⁡α22=mgsin⁡α2×2ℓsin⁡α2⇒kpq2ℓsin⁡α22=mgh⇒ substituting this in equation (i) Uf=mgh+kpq2ℓsin⁡α22⇒Uf=2mghW=ΔU=Nmgh=N=2

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