Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

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 is _________ . (Note that for three coplanar forces keeping a point mass in equilibrium,

F/sinθ is the same for all forces, where F is any one of the forces and θ is the angle between the other two forces)

solution : A point dipole of dipole moment P is now brought towards q from infinity so that charge moves away.

so initial potential energy, U_i = 0

final potential energy, U_f = potential energy of dipole charge system + potential energy of charge due to its weight.

⇒U_f = kqP/(2lsinα/2)² + mgh ......(1)

from triangle,

sum of all angles = 180°

⇒α + 90° - θ + 90° - θ = 180°

⇒α = 2θ ⇒θ = α/2 so, sin(α/2) = sinθ .......(2)

from figure, it is clear that, h = (2lsinα/2)sinθ

⇒h = 2l sinα/2 sinα/2 [ from eq (2) ]

= 2lsin²α/2 ......(3)

Now charge is in equilibrium at point B.

from sine rule,

⇒mg/sin(90° + α/2) = qE/sin(180° - 2θ)

⇒mg/cosα/2 = qE/sin2θ

⇒mg/cosα/2 = qE/sinα

⇒mg/cosα/2 = qE/2sinα/2 cosα/2

⇒2mgsinα/2 = qE

now, E = 2kp/(2lsinα/2)³

⇒2kpq/(2lsinα/2)³ = 2mgsinα/2

⇒kpq/(2lsinα/2)² = mg(2lsin²α/2)

⇒kpq/(2lsinα/2)² = mgh [ from eq (3) ]

from equation (1),

U_f = mgh + kpq/(2lsinα/2)²

therefore U_f = mgh + mgh = 2mgh

so, W = N × mgh = 2 mgh

⇒N = 2