JEE Mains · Physics · STD 12 - 1. Electric charges and fields
As shown in the figure. a configuration of two equal point charges \(\left( q _0=+2 \mu C \right)\) is placed on an inclined plane. Mass of each point charge is \(20\,g\). 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 \(h = x \times 10^{-3}\,m\) The value of \(x\) is \(..........\).(Take \(\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\,Nm ^2 C ^{-2}, g=10\,ms ^{-1}\right)\)

- A \(200\)
- B \(300\)
- C \(400\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(300\)
Step-by-step Solution
Detailed explanation
For equilibrium along the plane \(mg \sin \theta=\frac{1}{4 \pi \epsilon_0} \times \frac{ q _0^2}{\left( h \operatorname{cosec} 30^{\circ}\right)^2}\) \(\therefore h ^2=\frac{1}{4 \pi \epsilon_0} \times \frac{ q _0^2}{ mg \operatorname{cosec} 30^{\circ}}\)…
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