Question

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= (kqQ)/r^2=((9*10^9)(45*10^-6)(32*10^-6))/(9.9)^2=0.13N`
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.

Answer 1

I got `246^@`, Ultrilliam got `204^@`.
(Follow the thread of comments to understand the difference in answers.)

Explanation:

The origin of the coordinate system and the 2 charges allow you to draw a triangle. The `32 muC` charge would be attracted toward the `-45muC` -- 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(4/9) = 23.9^@`

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:
https://en.wikipedia.org/wiki/Course_(navigation)
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^@`. Hmmm, I get `270^@ - 23.9^@ = 246^@`.

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

Steve