# Find the force between the two charges as well as an angle?

## The force of - 45μC at (0,- 4.0m) on 32μC at (9.0m,0) is? I worked out the force $F = \frac{k q Q}{r} ^ 2 = \frac{\left(9 \cdot {10}^{9}\right) \left(45 \cdot {10}^{-} 6\right) \left(32 \cdot {10}^{-} 6\right)}{9.9} ^ 2 = 0.13 N$ I have tried everything to get the angle, but I do not know how that works. I drew a diagram of what the problem looks like. The answer is 204 degrees, but how do they get that answer? Thanks very much in advance.

Jul 7, 2018

I got ${246}^{\circ}$, Ultrilliam got ${204}^{\circ}$.

#### Explanation:

The origin of the coordinate system and the 2 charges allow you to draw a triangle. The $32 \mu C$ charge would be attracted toward the $- 45 \mu C$ -- along the hypotenuse of the triangle.

Look at the angle at (9.0m,0). Use the tangent trig function. The opposite side is 4 m and the adjacent side is 9 m.

${\tan}^{-} 1 \left(\frac{4}{9}\right) = {23.9}^{\circ}$

Think of your drawing as being drawn on a map with north in the direction of the +y axis. The charge at (9.0m,0) will go towards the other charge.

Go to this web site:
Scroll down to the section Aircraft heading. It says "North is 0°, east is 90°, south is 180°, and west is 270°." From the charge at (9.0m,0) that would be a direction a bit south of west. So, a bit less than the heading that is west or ${270}^{\circ}$. Hmmm, I get ${270}^{\circ} - {23.9}^{\circ} = {246}^{\circ}$.

That is not what you said it should be. I do not know why. I will set the request for a double check.

Steve