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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 \(\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\,Nm ^2 C ^{-2}, g=10\,ms ^{-1}\right)\)

  1. A \(200\)
  2. B \(300\)
  3. C \(400\)
  4. D \(100\)
Verified Solution

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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