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MCQMediumJEE 2023Coulomb's Law & Superposition Principle

JEE Physics 2023 Question with Solution

As shown in the figure, a configuration of two equal point charges (q0=+2 μCq_0 = + 2 \, \mu \text{C}) is placed on an inclined plane. Mass of each point charge is 20 g20 \, \text{g}. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height h=x×10−3 mh = x \times 10^{-3} \, \text{m}. The value of xx is _____.

  • A

    300300

  • B

    250250

  • C

    150150

  • D

    400400

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Two equal point charges with q0=2×10−6 Cq_0 = 2 \times 10^{-6} \, \text{C} and mass of each charge m=20×10−3 kgm = 20 \times 10^{-3} \, \text{kg} are on an inclined plane. The system is in equilibrium and the required height is written as h=x×10−3 mh = x \times 10^{-3} \, \text{m}.

Find: The value of xx.

For equilibrium, the electrostatic repulsion balances the component of weight along the plane.

Fe=mgsin⁡θF_e = mg \sin \theta

The electrostatic force between the charges is

Fe=kq02r2F_e = \frac{k q_0^2}{r^2}

Using the geometry stated in the solution,

r=hsin⁡θr = \frac{h}{\sin \theta}

So,

kq02(hsin⁡θ)2=mgsin⁡30∘\frac{k q_0^2}{\left(\frac{h}{\sin \theta}\right)^2} = mg \sin 30^\circ

Substituting the values,

9×109×(2×10−6)2(hsin⁡30∘)2=20×10−3×10×sin⁡30∘\frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{\left(\frac{h}{\sin 30^\circ}\right)^2} = 20 \times 10^{-3} \times 10 \times \sin 30^\circ

The extracted solution concludes that solving gives

h=300×10−3 mh = 300 \times 10^{-3} \, \text{m}

Hence,

x=300x = 300

Therefore, the correct option is A.

Force Balance Idea

Given: The charge remains at rest on the incline, so net force along the plane must be zero.

Find: The numerical value of xx.

The key idea is to compare two forces along the incline:

  • electrostatic repulsion between the charges,
  • component of gravitational force down the plane.

Thus,

kq02r2=mgsin⁡θ\frac{k q_0^2}{r^2} = mg \sin \theta

From the extracted working, the geometry relation used is

r=hsin⁡θr = \frac{h}{\sin \theta}

and θ=30∘\theta = 30^\circ. After substitution of all numerical values, the final result reported in the solution is

h=300×10−3 mh = 300 \times 10^{-3} \, \text{m}

So the required value is x=300x = 300.

Common mistakes

  • Using full weight mgmg instead of the component mgsin⁡θmg \sin \theta along the incline is incorrect because only the force parallel to the plane is balanced by electrostatic repulsion. Resolve forces along the plane before applying equilibrium.

  • Taking the electrostatic force distance incorrectly as hh directly can be wrong if the geometry gives a different separation. Use the relation provided by the figure or geometry, here stated as r=hsin⁡θr = \frac{h}{\sin \theta}.

  • Using mass as 2020 instead of 20×10−3 kg20 \times 10^{-3} \, \text{kg} gives a large error. Convert grams to kilograms before substitution.

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