Question

two charges are placed on the x axis. one of the charges (q1=+mC) is at x1=+cm and the other (q2=-21mC) is at x1=+9.0cm . find the net electric field (magnitude and direction) at (a) x=0 (b) x=+6.0cm

Answer 1

Hi! I'll give you hints, and you can ask questions if you have any. Others might be able to help more!

First, drawing pictures usually help. They help you visualize it and help you test how well you understand the problem. So try that!

Second, an electric field vector from any one charge is given by
\(E=k\dfrac q{r^2}\)
where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance between the spot and the charge. But you need the direction, too.
First, find the direction from the charge (look at your drawing) to the place where you want to know about the electric field. Consider that unit vector. Then multiply it by By what you get from the equation above. Notice that a negative sign will flip the direction!

Third, these electric field vectors add up to create the total electric field vector -- which is the electric field at that point!

Answer 2

Notice that they're on the same axis! So you can make your \(x\)-axis from left to right, and so all the electric field vectors will be simply left or right!