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JEE Main 2023 (11 Apr, Shift 1) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2023 (11 Apr, Shift 1)

Q42JEE Main 2023 · 11 APR · SHIFT 1Medium

Electric Charge & Coulomb's Law

As shown in the figure, a configuration of two equal point charges (q0=+2 μC)(q_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 ____________.

(Take 14πε0=9×109 N m2C−2,g=10 m s−2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^{2}\text{C}^{-2}, g = 10\text{ m s}^{-2})

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 300

  1. Distance between the charges along the incline: r=hsin⁡30∘=2hr = \dfrac{h}{\sin 30^\circ} = 2h.
  2. Upper charge in equilibrium: kq02(2h)2=mgsin⁡30∘\dfrac{kq_0^2}{(2h)^2} = mg\sin 30^\circ.
  3. kq02=9×109×4×10−12=0.036kq_0^2 = 9 \times 10^9 \times 4 \times 10^{-12} = 0.036; mgsin⁡30∘=0.02×10×0.5=0.1 Nmg\sin 30^\circ = 0.02 \times 10 \times 0.5 = 0.1\text{ N}.
  4. 4h2=0.36⇒h=0.3 m=300×10−3 m4h^2 = 0.36 \Rightarrow h = 0.3\text{ m} = 300 \times 10^{-3}\text{ m}
  5. So x=300x = 300.
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