A5.5-µc point charge is located at x = 0.9 m, y = 3.4 m, and a -3.6-µc point charge is located at x = 2.3 m, y = -1.9 m.
(a) find the magnitude and direction of the electric field at x = -2.9 m, y = 1.2 m.
magnitude kn/c:
direction °:
(b) find the magnitude and direction of the force on a proton at x = -2.9 m, y = 1.2 m.
magnitude n :
direction ° :

The electric field magnitude= 2.27kN/C direction=

The force is direction=

Explanation:

Here we have an array of charged points, we have to calculate the net force in the point x=-2.9m and y=1.2m; so we need have to find the distance of all the charged point from this position.

to obtain the distance we have to do this:

fot the charge of 5.5uC

θ

and for the charge of -3.6uC

θ

Now that we have the distance and the angle, we can calculate the electric field; ausuming a positive charge:

The net force in X direction:

the net force in Y direction:

So the magnitud of the electric field is:

with a direction of:

θ, because is going down and to the left, we have to add 180, so the direction is=

to obtain the force of a proton we only have to multiply the Electric field times the charge of the proton, that is:

As we calculated the electric field assuming that the charge was a positive, the direction of the force will be the same.

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement