An object with charge 4.3x10-5 C pushes another object 0.31 micrometers away with a force of 7 N. What is the total charge of the second object?

Respuesta :

Answer:

Charge on the other particle is given as

[tex]q_2 = 1.74 \times 10^{-18} C[/tex]

Explanation:

As we know that the force between two charges is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we know that

[tex]F = 7 N[/tex]

[tex]r = 0.31 \mu m[/tex]

[tex]q_1 = 4.3 \times 10^{-5} C[/tex]

now we have

[tex]7 = \frac{9 \times 10^9 (4.3 \times 10^{-5}q_2}{(0.31\times 10^{-6})^2}[/tex]

now we have

[tex]q_2 = 1.74 \times 10^{-18} C[/tex]