# Thread: Physics Electric Forces and Electric Fields

1. ## Physics Electric Forces and Electric Fields

There are 3 charges fixed in place. The charge at the coordinate origin has a value of q1=+8x10^-6C. The other two charges have identical magnitudes, but opposite signs: q2= -5x10^-6C and q3=+5x10^-6C.

a.) Determine the net force (magnitude and direction) exerted on q1 by the other two charges.

b.) If q1 had a mass of 1.5g and it were free to move, what would be its acceleration?

So in the drawing q1 is in the center. q2 is 1.3m away at 23degrees above the x-axis. q3 is 1.3m away at 23degrees below the x-axis.

I need help ASAP. I am so lost on this problem!!!

When I tried to work it, the farthest I got was the F1=2.13x10^-1N but I know you have to split that into x and y components but I don't remember how to do that... so help please!!!

2. Originally Posted by TutorGirl08
There are 3 charges fixed in place. The charge at the coordinate origin has a value of q1=+8x10^-6C. The other two charges have identical magnitudes, but opposite signs: q2= -5x10^-6C and q3=+5x10^-6C.

a.) Determine the net force (magnitude and direction) exerted on q1 by the other two charges.

b.) If q1 had a mass of 1.5g and it were free to move, what would be its acceleration?

So in the drawing q1 is in the center. q2 is 1.3m away at 23degrees above the x-axis. q3 is 1.3m away at 23degrees below the x-axis.

I need help ASAP. I am so lost on this problem!!!

When I tried to work it, the farthest I got was the F1=2.13x10^-1N but I know you have to split that into x and y components but I don't remember how to do that... so help please!!!
a) As you already know, the force is a vector, so you need to take vector components. I presume you know how to find the magnitude using Coulomb's Law because you have that correct, so I'll just remind you how to take components of a vector.

For q2, let the force between q1 and q2 be F. This force is at 23 degrees above the +x axis and is directed outward from the origin. Both force components will be positive since the angle is in the first quadrant.

$F_x = F \cdot cos(23^o)$
$F_y = F \cdot sin(23^o)$

For q3, again let the force be F. Here the angle is 23 degrees below the -x axis, putting the angle in quadrant III. Thus both components are negative:
$F_x = -F \cdot cos(23^o)$
$F_y = -F \cdot sin(23^o)$

-Dan