a liquid drop having 6 exçess electrons is kept stationary under electric field of 25.5kV m^-1.The density of liquid is 1.26×10^3 kg m^-3. The radius of the drop is (neglect buoyancy).

Question

priyar · Answer

@samigupta8 @ganeshie8 @MuH4hA

priyar · Answer

@agent0smith

priyar · Answer

@rvc @Michele_Laino @Vincent-Lyon.Fr

michele_laino · Answer

here on the liquid drop are acting two forces, namely its weight and the electric force, so we can write:
$$\Large mg = qE$$
where $q$ is the charge of the liquid drop, and, of course, $g$ is the gravity.
After a substitution, we get:
$$\Large \frac{{4\pi }}{3}{R^3}\delta g = 6eE$$
where $\delta$ is the density of the liquid, and $R$ is the radius of the liquid drop

samigupta8 · Answer

Simply apply qE=mg
Q is the charge of 6 electrons

priyar · Answer

hey I did the exact same thing ..! but answer didn't match.. so i wanted to make sur my method was correct atleast ! thanks @Michele_Laino and  @samigupta8 !