Question

Two particles, one with charge -5.45 uC and one with charge 3.55 uC, are 4.34 cm apart. What is the magnitude of the force that one particle exerts on the other?

Answer 1

`F = 92.45 N`

Step-by-step explanation:

As we know that the force between two charge particles is given by

`F = (kq_1q_2)/(r^2)`

here we know that

`q_1 = 3.55 \mu C`

`q_2 = 5.45 \mu C`

now the distance between the two charges is

r = 4.34 cm

now from the formula of electrostatic force we will have

`F = ((9* 10^9)(3.55 \mu C)(5.45 \mu C))/(0.0434^2)`

`F = 92.45 N`

Answer 2

Force, F = 92.02 N

Step-by-step explanation:

It is given that,

Charge 1,
`q_1=-5.45\ \mu C=-5.45* 10^(-6)\ C`

Charge 2,
`q_2=3.55\ \mu C=3.55* 10^(-6)\ C`

Distance between two charges, d = 4.35 cm = 0.0435 m

We need to find the magnitude of the force that one particle exerts on the other. It is given by :

`F=k(q_1q_2)/(d^2)`

`F=9* 10^9* (-5.45* 10^(-6)\ C* 3.55* 10^(-6)\ C)/((0.0435\ m)^2)`

F = -92.02 N

So, the force acting between two particles is 92.02 N. Hence, this is the required solution.