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  • As shown in the figure, a configuration of two equal point charges <img alt="\mathrm{q_{0}=+2\mu C}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bq_%7B0%7D%3D&plus;2%5Cmu%20

As shown in the figure, a configuration of two equal point charges \mathrm{q_{0}=+2\mu C} is placed on an inclined plane. Mass of each point charge is \mathrm{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 \mathrm{h=x\times 10^{-3}m}. The value of x is _________.

(Take \mathrm{\frac{1}{4\pi\varepsilon _{0}}=9\times 10^{9}Nm^{2}C^{-2},g=10ms^{-2}})

Option: 1

300


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Point charge on equilibrium is at rest.

\begin{aligned} & \text { So } \quad \mathrm{F}_{\mathrm{e}}=\mathrm{mg} \sin \theta \\ & \frac{\mathrm{kq}_0 \cdot \mathrm{q}_0}{\mathrm{r}^2}=\mathrm{mg} \sin 30^{\circ} \\ \end{aligned}

\begin{aligned} & \frac{\mathrm{kq}_0^2}{\left(\frac{\mathrm{h}}{\sin 30^{\circ}}\right)^2}=\frac{\mathrm{mg}}{2} \\ & \frac{9 \times 10^9 \times\left(2 \times 10^{-6}\right)^2}{4 \mathrm{~h}^2}=\frac{20 \times 10^{-3} \times 10}{2} \\ \end{aligned}

\begin{aligned} & \frac{9 \times 4 \times 10^9 \times 10^{-12}}{4 \mathrm{~h}^2}=10^{-1} \\ & \mathrm{~h}^2=9 \times 10^{-2} \\ \end{aligned}

\begin{aligned} & \mathrm{~h}=0.3 \mathrm{~m}=300 \times 10^{-3} \mathrm{~m} \\ \end{aligned}

\begin{aligned} & \mathrm{x}=300 \\ & \end{aligned}


 

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