(a) A charged oil drop is suspended in a uniform field of 300 V cm^-1 so that it neither rises nor falls. Find the charge on it, in coulomb, assuming its mass to be 9.75 * 10^-15 kg. (b) Find the number of electrons on the oil drop. About the author Natalia
Explanation: Given : 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 Reply
Explanation:
Given : 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