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 th

Question

3 months ago

11

e second object?

1 answer:

Yuliya22 [3.3K] · Answer 1 · 2026-03-25T13:06:16+03:00

5 0

Answer:

The charge on the second particle is defined as

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

Explanation:

It is known that the force existence between two charges is expressed as

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

It is given that

$F = 7 N$

$r = 0.31 \mu m$

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

At this juncture, we find that

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

Now we ascertain

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