Question

Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is

Answer 1

`r=200m`

Step-by-step explanation:

From the question we are told that:

Charges:

`Q_1=8.0mC`

`Q_2=2.0mC`

`Q_3=8.mC`

Distance
`d=300m`

Generally the equation for Force is mathematically given by

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

Therefore

`F_(32)=F_(31)`

`(q_2)/((300-r)^2)=(q_1)/(r^2)`

`(2*10^(-3))/((300-r)^2)=(8*10^(-3))/(r^2)`

`r=2(300-r)`

`r=200m`