Question

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitude of the force that one particle exerts on the other? Two new particles, which have identical positive charge q3, are placed the same 0.0209 meters apart, and the force between them is measured to be the same as that between the original particles. What is q3?

Answer 1

Step-by-step explanation:

Given that,

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

Charge 2,
`q_2=4.39* 10^(-6)\ C`

Distance between charges, r = 0.0209 m

1. The electric force is given by :

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

`F=9* 10^9* (-5.45* 10^(-6)* 4.39* 10^(-6))/((0.0209)^2)`

F = -492.95 N

2. Distance between two identical charges,
`r=0.0209\ m`

Electric force is given by :

`F=(kq_3^2)/(r^2)`

`q_3=\sqrt{(Fr^2)/(k)}`

`q_3=\sqrt{(492.95* (0.0209)^2)/(9* 10^9)}`

`q_3=4.89* 10^(-6)\ C`

Hence, this is the required solution.