As shown in the figure, a configuration of two equal point charges $(q_{0} = + 2 μ C)$ 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
$(Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}, g = 10 m s^{- 2})$

Question

Aicharya · Accepted Answer

For the condition of equilibrium, the Coulomb force is equal to $m g \sin θ$ . Hence,

$\frac{k {q_{0}}^{2}}{4 h^{2}} = (m g \sin θ) \Rightarrow \frac{k {q_{0}}^{2}}{4 h^{2}} = (20 \times 10^{- 3}) \times 10 \times \frac{1}{2}$

$\Rightarrow \frac{9 \times 10^{9} \times 4 \times 10^{- 12}}{4 h^{2}} = 20 \times 10^{- 3} \times 10 \times \frac{1}{2} \Rightarrow h = (\sqrt{\frac{9}{100}}) m = 0.3 m = 300 \times 10^{- 3} m$

$h^{2} = (\frac{9}{100}) \Rightarrow h = (\frac{3}{10}) m = 0 .3 m$

$= 300 \times 10^{- 3} m$

As shown in the figure, a configuration of two equal point charges q 0 = + 2 μ C is placed on an…

Solution

As shown in the figure, a configuration of two equal point charges q 0 = + 2 &#956; C is placed on an…

Solution

As shown in the figure, a configuration of two equal point charges q 0 = + 2 μ C is placed on an…