Question

A charged water drop of radius r is in equilibrium in an electric field. If charge on it is equal to charge on an electron, then intensity of electric field will be? (Density of water = d)

Answer 1

d=m/v

m=dv

m=4πr^3d/3

Now for equilibrium

qE=mg

E=mg/q

E=4πr^3d* 9.8/3*1.6*10^- 19

Answer 2

Answer:E=4πr^3d* 9.8/3*1.6*10^- 19

Explanation:

Force due to electric field = weight of the water drop

qE=mg

eE=d× 3/4π$r^{3}$ ×g

​Hence,

E=4π$r^{3}$ ×pg /3e

d=m/v

m=dv

m=4πr^3d/3

Now for equilibrium

qE=mg

E=mg/q

E=4πr^3d* 9.8/3*1.6*10^- 19

#SPJ2

​