Question

A point charge Q is situated at point B on the ground. A point charge q of mass m is vertically dropped along line AB from a multi-storey building of height h. Find the position of the point charge q when it is in equilibrium.

पर्याय

• ${\displaystyle \sqrt{\frac{\text{qQ}}{4\pi {\epsilon }_0\text{mg}}}}$

• ${\displaystyle \frac{\text{Q}}{{\text{h}}^2}}$

• ${\displaystyle \frac{\text{qh}}{\text{m}}}$

• None of these

Answer 1

${\displaystyle \sqrt{\frac{\text{qQ}}{4\pi {\epsilon }_0\text{mg}}}}$

Explanation:

For equilibrium

mg = F_e or mg = ${\displaystyle \frac{\text{qQ}}{4\pi {\epsilon }_0{\text{y}}^2}}$

∴ Y = ${\displaystyle \sqrt{\left(\frac{\text{qQ}}{4\pi {\epsilon }_0\text{mg}}\right)}}$