Question 19 of 311

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 $20g$. 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
$\left(\text{Take}\dfrac{1}{4\pi {\epsilon }_{0}}=9\times {10}^{9}N{m}^{2}{C}^{-2},g=10m{s}^{-2}\right)$

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Answer: 300

For the condition of equilibrium, the Coulomb force is equal to $mg\sin \theta$. Hence,
$\dfrac{k{{q}_{0}}^{2}}{4{h}^{2}}=(mg\sin \theta ) \Rightarrow \dfrac{k{{q}_{0}}^{2}}{4{h}^{2}}=\left(20\times {10}^{-3}\right)\times 10\times \dfrac{1}{2}$
$\Rightarrow \dfrac{9\times {10}^{9}\times 4\times {10}^{-12}}{4{h}^{2}}=20\times {10}^{-3}\times 10\times \dfrac{1}{2} \Rightarrow h=\left(\sqrt{\dfrac{9}{100}}\right)m=0.3m=300\times {10}^{-3}m$
${h}^{2}=\left(\dfrac{9}{100}\right)\Rightarrow h=\left(\dfrac{3}{10}\right)m=0.3m$
$=300\times {10}^{-3}m$