Question

A water particle of mass 10.0 mg and having a charge of 1.50 ×10^-8 C stays suspended in the room. What is the magnitude of the electric field in the room? What is it's direction.

Answer 1

Solution:

Given:

• Mass of water particle= 10mg = 10 × 10^-6 = 10^-5 kg.

• Charge (q)= 1.50 × 10^-8 c.

To find :

• Electric field and it's direction :

Answer :

Here ,

F = qE = mg

•Because , the drop is suspended in room so the force of E must be equal to the weight of the drop .

=> E = mg/q

=> E =( 10^-5×10)/(1.50× 10^-8)

=> E = 10^4/1.50

=> E = 0.66 × 10^4 N/C .

• Direction of E is vertically upward . So , that it balances the weight of the drop.

Answer 2

F = qE = mg

=> E = mg/ q

=> E = 10×10^-6×10/1.50×10^-8

=>E= 10^-4/1.50×10^-8

=>E = 0.66667× 10^4