The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron:

Question

The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron: (A)  textem/g  (B)  textmg/e  (C)  textemg  (D)  textmg

Tardigrade Admin · Accepted Answer

Here: mass of water drop =m, so, weight of water drop =mg so, mg= electrostatic force applied by the field mg=qE=eE ⇒ E=(mg/e)

Q. The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron:

$em/g$

$mg/e$

$emg$

$mg$

Here: mass of water drop $= m,$ so, weight of water drop $= m g$ so, $m g =$ electrostatic force applied by the field $m g = qE = e E \Rightarrow E = \frac{m g}{e}$

Q. The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron:

em/g

mg/e

emg

mg

Here: mass of water drop =m, so, weight of water drop =mg so, mg= electrostatic force applied by the field mg=qE=eE⇒E=emg​

$em/g$

$mg/e$

$emg$

$mg$

Here: mass of water drop $= m,$ so, weight of water drop $= m g$ so, $m g =$ electrostatic force applied by the field $m g = qE = e E \Rightarrow E = \frac{m g}{e}$