Physics · Electrostatics

JEE Advanced 2020 — Paper 2 — Question 4

A point charge qq of mass mm is suspended vertically by a string of length l. A point dipole of dipole moment p⃗\vec{p} is now brought towards qq 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)\mathrm{N} \times(\mathrm{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\frac{\mathrm{F}}{\sin \theta} is the same for all forces, where F is any one of the forces and θ\theta is the angle between the other two forces)

Question figure

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

mgsin⁡(90+α2)=2kqpr3sin⁡(180∘−α)\frac{m g}{\sin \left(90+\frac{\alpha}{2}\right)}=\frac{2 k q p}{r^{3} \sin \left(180^{\circ}-\alpha\right)}

mgsin⁡αcos⁡(α2)=2kqpr3\frac{m g \sin \alpha}{\cos \left(\frac{\alpha}{2}\right)}=\frac{2 \mathrm{kqp}}{\mathrm{r}^{3}}

Uelec =PE=kqpr2=mgr⁡sin⁡(α2)=mgh\mathrm{U}_{\text {elec }}=\mathrm{PE}=\frac{\mathrm{kqp}}{\mathrm{r}^{2}}=\operatorname{mgr} \sin \left(\frac{\alpha}{2}\right)=\mathrm{mgh}

U=Ugrav +Uelec =2mgh\mathrm{U}=\mathrm{U}_{\text {grav }}+\mathrm{U}_{\text {elec }}=2 \mathrm{mgh}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Potential Energy and Electric Potential