Question

Calculate the electric field strength required to just support a water drop of mass 10^-7 kg and having charge 1.6x10^-9 C?

Answer 1

Force due to electric field = Weight of water drop
Eq = mg

E = mg/q
= (10^-7 kg × 9.8 m/s^2) / (1.6 × 10^-19 C)
= 6.125 × 10^12 N/C

∴ Electric field strength required is 6.125 × 10^12 N/C

Answer 2

Answer:

The electric field strength required to just support a water drop is $6.125\times10^{12}\ N/C$

Explanation:

Given that,

Mass $m= 10^{-7}\ kg$

Charge $q = 1.6\times10^{-9}\ C$

Balance equation of the drop

$Eq = mg$

Where, E = electric field strength

q = charge

m = mass

g = acceleration due to gravity

$E = \dfrac{mg}{q}$

$E= \dfrac{10^{-7}\times9.8}{1.6\times10^{-19}}$

$E = 6.125\times10^{12}\ N/C$

Hence, The electric field strength required to just support a water drop is $6.125\times10^{12}\ N/C$