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

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Answer & Solution

Correct Answer

(B) \(300\)

Detailed explanation

For equilibrium along the plane

\(mg \sin \theta=\frac{1}{4 \pi \epsilon_0} \times \frac{ q _0^2}{\left( h \operatorname{cosec} 30^{\circ}\right)^2}\)

\(\therefore h ^2=\frac{1}{4 \pi \epsilon_0} \times \frac{ q _0^2}{ mg \operatorname{cosec} 30^{\circ}}\)

\(=9 \times 10^9 \times \frac{\left(2 \times 10^{-6}\right)^2}{0.02 \times 10 \times 2}\)

\(\therefore h =3 \times 10^4 \times \frac{2 \times 10^{-6}}{0.2}\)

\(=0.3\,m\)

\(=300\,mm\)

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