Question

a charged Oil Drop of mass 2 mg is floating in space in an electric field of intensity 1000 N/C. The electric field is directed downward.What is the charge on the oil drop?​

Answer 1

Answer:

Electric field, E=3×10

4

V/m

Mass of the drop, M=9.9×10

−15

kg

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

∴ qE=Mg ⟹q=

E

Mg

∴ q=

3×10

4

9.9×10

−15

×10

=3.3×10

−18

C

Explanation:

May be it is useful to u

Answer 2

Answer:

Explanation: Electric field, E=3×10

4

V/m

Mass of the drop, M=9.9×10

−15

kg

Let q be the amount of the charge that the drop carries.

The coulomb force balances the gravitational force acting on the drop at equilibrium.

∴ qE=Mg ⟹q=

E

Mg

∴ q= 3×10  4 9.9×10  −15  ×10 =3.3×10  −18  C