Electrostatic Force and Superposition Principal

[shadowfox745]

Homework Statement

Three charges are arranged as shown in the figure.

Figure
(Attached Below Could not insert the image, wouldn't work for me)


Find the magnitude of the electrostatic force on the charge at the origin. Answer in units of nN. What is the angle theta between the electrostatic force on the charge at the origin and the positive axis. Answer in degrees between -180 and 180 degrees measured from the positive X-axis, with counterclockwise positive.

Homework Equations

Coulombs Constant= 8.98755X10^9 N m^2/C^2
F=k((q'q)/r^2)

The Attempt at a Solution

So what I did was I found the Electrostatic force of the center charge and that of the bottom charge using F=K((q'q)/r^2) and I found it to be -2.115x10^-9 C which comes to be -2.115nC and I used the same equation to solve for the electrostatic Force between the center charge and the charge to the right and that came out to be -7.800x10^-10 C which turns to be .78Nc. I thought the Superposition Principal said to find the electrostatic for exerted to the center is just adding both forces together which gave me -2.895nC however when I inputed it on my online hwk it says it to be incorrect. So am I missing something here? And I also have no clue how to find the angle, and my book shows no example of it so can I get some hints on it.

 

[shadowfox745]

Nevermind I forgot the most stupidest thing to use vector geometry. Though I am still trying to figure out how to find the angle. Actually I am confused as to what angle they are asking to find.

 

[LowlyPion]

Nevermind I forgot the most stupidest thing to use vector geometry. Though I am still trying to figure out how to find the angle. Actually I am confused as to what angle they are asking to find.

Each of the forces on the charge at the origin are developing force vectors as you noted. When you add them together - the x component and the y component - you will have a triangle with the x and y components being the perpendicular sides. The angle that it forms with the X axis is what they want. Hence you know that Y / X is equal to tangent theta.

 

[shadowfox745]

Oh, I understand now, got it. Thanks! I guess summer really does a number on my basic skills.