Question

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?

Answer 1

Answer:

Charge on the other particle is given as

$q_2 = 1.74 \times 10^{-18} C$

Explanation:

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

$F = \frac{kq_1q_2}{r^2}$

here we know that

$F = 7 N$

$r = 0.31 \mu m$

$q_1 = 4.3 \times 10^{-5} C$

now we have

$7 = \frac{9 \times 10^9 (4.3 \times 10^{-5}q_2}{(0.31\times 10^{-6})^2}$

now we have

$q_2 = 1.74 \times 10^{-18} C$