Question

The electric field required to just balance a liquid droplet of mass 2 X 10-9 kg and 9.8x10^8 microcoloumb ​

Answer 1

Answer:

Here, m=10;q=1.6×10

−19

C

Let E be the strength of the electric field required to just support the water drop. Then,

Ee=mg

or E=

e

mg

=

1.6×10

−19

10

−7

×9.8

=6.125×10

12

NC

−1