Question

Two particles, one with charge − 7.97 μC and one with charge 3.55 μC, are 6.59 cm apart. What is the magnitude of the force that one particle exerts on the other?

Answer 1

Answer:

58.6 N

Explanation:

We are given that

$q_1=-7.97\mu C=-7.97\times 10^{-6} C$

$q_2=3.55\mu C=3.55\times 10^{-6} C$

Using $1\mu C=10^{-6} C$

$r=6.59 cm=6.59\times 10^{-2} m$

$1 cm=10^{-2} m$

The magnitude of force that one particle exerts on the other

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

Where $k=9\times 10^9$

Substitute the values

$F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}$

F=58.6 N