Three charged particles are arranged in a line,as shown in figure.Calculate the net electrostatic force on particle3 ( the -4.0`muC` on the right ) due to the other two charges.

Strategy `:` The net force exerted on 3by particle 1 is the resultant of the force `bar(F_(31))` exerted on 3 by particle 1 and the force `bar(F_(32))` exerted on 3 by particle 2, `bar(F) =bar(F)_(31)+bar(F)_(32)`

Explanation

The magnitude of these two forces are obtained using Coulomb's law
`F_(31)= k ( Q_(3)Q_(1))/(r_(31))= ((9.0 xx 10^(9) N.m^(2)//C^(2)) ( 4.0 xx 10^(-6) C ) (8.0 xx 10^(-6) C))/(( 0.50 m)^(2))= 1.15 N `
where `r_(31) = 0.50 m` is the distance from `Q_(3)` to `Q_(1)`
Similarly , `F_(32)= k ( Q_(3)Q_(1))/(r_(32))= ((9.0 xx 10^(9) N.m^(2)//C^(2)) ( 4.0 xx 10^(-6) C ) (3.0 xx 10^(-6) C))/(( 0.20 m)^(2))= 2.7 N `
Since we were calculatingthe magnitude of the forces,we omitted the signs of the charges .But we must be aware of them to get the direction of each force. Let the line joining the particles be the x-axis and we take it positive to the right . Then, because , `vec(F) _(31)` is repulsiveand `vec(F)_(32)` is attractive the directions of the forces are as shown in figure. `F_(31)` points in the positive x direction and`F_(32)` points in the negative x-direction.

The net force on particle 3 is then
`F= - F_(32) + F_(31) = - 2.7 N + 1.15 N = - 1.55 N `
The magnite due of the net force is 1.55 N , and it points to the left.