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Electrostatics question

2023 · 11 Apr · Shift 1 · Q68
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Electrostatics question

2023 · 11 Apr · Shift 1 · Q68

JEE MainPhysicsElectrostaticsNumerical+4 / −1
As shown in the figure, a configuration of two equal point charges (q0=+2μC)\left(q_{0}=+2 \mu \mathrm{C}\right)(q0​=+2μC) is placed on an inclined plane. Mass of each point charge is 20 g20 \mathrm{~g}20 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 m\mathrm{h}=x \times 10^{-3} \mathrm{~m}h=x×10−3 m. The value of xxx is ‾\underline{\hspace{2cm}}​. (Take 14πε0=9×109 N m2C−2,g=10 m s−2\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}, g=10 \mathrm{~m} \mathrm{~s}^{-2}4πε0​1​=9×109 N m2C−2,g=10 m s−2 ) JEE Main 2023 (Online) 11th April Morning Shift Physics - Electrostatics Question 78 English
Numerical answer
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