Question

A 6.6- μC charge is placed at the origin and a second charge is placed on the x-axis at x = 0.1 m. If the resulting force on the second charge is 5.7 N in the positive x-direction, what is the value of its charge?

Answer 1

`q_2=9.59* 10^(-7)\ C`

Step-by-step explanation:

It is given that,

Charge at origin,
`q_1=6.6\ \mu C=6.6* 10^(-6)\ C`

Let second charge is
`q_2` and it is placed at x = 0.1 m

Resulting force on the second charge,
`F_2=5.7\ N`

The electric force acting on the charged particles is given by the following formula as :

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

`q_2=(Fx^2)/(kq_1)`

`q_2=(5.7* (0.1)^2)/(9* 10^9* 6.6* 10^(-6))`

`q_2=9.59* 10^(-7)\ C`

So, the second charge is
`9.59* 10^(-7)\ C`. Hence, this is the required solution.