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attashe74
23 hours ago
11

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

e second object?
Physics
1 answer:
Yuliya22 [2.4K]22 hours ago
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

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