As shown in figure. a configuration of two equal point charges left( q _0=+2 μ C right) is placed o

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1. Electric Charges and Fields

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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

$100$

(JEE MAIN-2023)

Solution

For equilibrium along the plane

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

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

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

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

$=0.3\,m$

$=300\,mm$

Standard 12

Physics

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