Question

a drop of 10^ -6 kgs water carries 10^6 c charge what electric field should be applied to balance its weight (assume g is 10m/s^2 )​

Answer 1

Answer:

The electric field applied to balance the weight of the drop is 9.8 N/C.

Explanation:

Given that,

Mass of the drop, $m=10^{-6}\ kg$

Charge on the drop, $q=10^{-6}\ C$

The electric force acting on the drop is balanced by its weight.

Its expression is given by :

Electric force, F = qE

Force of gravity, F = mg

qE = mg

⇒ $E=\dfrac{mg}{q}$

⇒ $E=\dfrac{10^{-6}\times 9.8}{10^{-6}}$

⇒ E = 9.8 N/C

So,

The electric field applied on the drop is 9.8 N/C.

Answer 2

hi

hope it helps you and mark as brainliest