Question

A charged oil is suspended in uniform field of 3×10^⁴V/m, so that it neither falls nor rises. The charge on the drop will be : (given the mass of the charge =9.9×10^–¹⁵ kg and g =10 m/s²)
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Answer 1

GIVEN :-

• Electric Field , E = 3 × 10⁴ V/m.
• Mass of the charge , M = 9.9 × 10-¹⁵ Kg.
• g = 10 m/s².

TO FIND :-

• The amount of charge will be on the Drop , q.

SOLUTION :-

As we know that in the Coulomb's law unlike charges attract each other and like charges repel each other . Now as we know that the coulomb's forces balance the gravitational force acts on the top of the equilibrium.

Now on applying the formula,

→ qE = Mg

• [ Transpose E to R.H.S ]

→ q = Mg/E

• [ Now substitute all the given values ]

→ q = (9.9 × 10-¹⁵ × 10)/3 × 10⁴

→ q = (3.3 × 10-¹⁵ )/10⁴

→ q = (3.3 × 10-¹⁵ × 10¹)/10⁴

→ q = (3.3 × 10-¹⁴)/10⁴

→ q = 3.3 × 10-¹⁴ - ⁴

→ q = 3.3 × 10-¹⁸.

Hence Amount of charges will be on the Drop is 3.3 × 10-¹⁸ Coulomb's.