Question

Two pellets, each with a charge of 1 microcoulomb (10-6 C), are located 3 cm (0.03 m) apart. What is the electric force between them? What would be the mass of an object that would experience this same force in the Earth's gravitational field?

Answer 1

Answer:

Therefore the electric force between them is 10 N

Mass of the object that would experience the same force is 1.02 kg.

Explanation:

Coulomb's Law:

The magnitude of electrostatic force of repulsion or attraction between two stationary electrical charged particle is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

$\therefore F=k \frac{q_1q_2}{r^2}$

$q_1$, $q_2$ are two charges.

r is the distance between the charges q₁ and q₂ .

k=$9\times 10^{9}$  N m²/C²

Here $q_1=q_2=10^{-6} \ C$ , r = 0.03 m.

$\therefore F=9\times 10^{9} .\frac{10^{-6}\times10^{-6}}{(0.03)^2}$ N

      =10 N

Therefore the electric force between them is 10 N

The formula for gravitation force:

F=mg

m= mass

g= 9.8 m/s²

F=10 N

$\therefore 10 = 9.8 \times m$

$\Rightarrow m=\frac{10}{9.8}$

       =1.02 kg

Mass of the object that would experience the same force is 1.02 kg.