# Coulombs Law. Calculate Charge Magnitude and Direction

## Homework Statement

Three charges are arranged as shown in the diagram below. Their magnitudes are:
q1 = +2.5 x 10^-17 C
q2 = +3.0 x 10^-17 C
q3 = +3.5 x 10^-17 C

(its a right angle triangle, with 50 cm as the hypotenuse, 40 cm as the horizontal, and 30 cm as the vertical)

Figure image: http://img80.imageshack.us/img80/8029/fig.png [Broken]

Calculate the direction and magnitude of the force acting on q1.

Fe = kq1q2/r2

## The Attempt at a Solution

F(q1q2)= 2.7x10^-23N [Equation above]
F(q1q3)=8.75x10^-23 [Equation above]

I used polar... trig. Do I have to show my work D:? Its hell on text.

The answer i obtained was 1.1x10^-22 N [Up 12 degrees Left] Could someone verify?

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verty
Homework Helper
At its heart, this is a pretty straightforward trig problem. I did get the same answer.

At its heart, this is a pretty straightforward trig problem. I did get the same answer.
Thank you for verifying.

I'm working on this same problem. I've calculated individual force that q2 and q3 are inflicting on q1, and I've calculated the angles of A, B, and C of the triangle. I know I now have to use trig:

2F = force of q2 on q1
3F = force of q3 on q1

Fnet^2 = (2F)^2 + (3F)^3 - 2(2F)(3F)cos(126.9 degrees)

To get the 126.9 degrees I added corners A and B in the triangle (90 and 36.9, respectively).

Now that gives me the sum of the forces q2 and q3 on q1, but I'm having trouble getting the exact angle that these forces (in summation) are acting on q1. I see the OP has a direction of [UP 12 degrees left].

Am I taking the correct approach? And how can I get the direction?

Thanks

gneill
Mentor
Hi saiyaex, welcome to Physics Forums.

The direction is easiest determined by noting that like charges repel and unlike charges attract. Since in this problem all the charges have the same sign, all forces will be repulsive ones. Also keep in mind that electrostatic forces operate along the direction of the line joining the charges involved. So, for example, the force that the charge q2 exerts on q1 will be along the same direction as the hypotenuse, pointing up and to the left.

You might also note that the triangle, with all its dimensions given, allows you to immediately write down the required distances and sine and cosine values! No need to do any trig. So, for example, if the magnitude of the force exerted by q2 on q1 is F (as provided by Coulomb's law using the hypotenuse length as the distance), then the x-component of the force is -F*(40/50), and the y-component is F*(30/50).

Thanks gneill,

I just seem to be lost and/or over thinking this. I know the direction caused by the repulsive force of q3 is up. For q2 I broke it up into it's x and y components as you suggested (converting the cm lengths to meters):

x-component: -(2.7 x 10^-23)(0.4/0.5) = -2.6 x 10^-23
y-component: (2.7 x 10^-23)(0.3/0.5) = 1.62 x 10^-23

I'm confused on getting the exact degree to the left that the overall force is causing.

gneill
Mentor
Okay. Since you've got the x and y components, I take it that you're looking for a way to calculate an appropriate angle associated with the net force vector?

First add all the like-components; sum up the x-components and sum up the y-components for all the contributing forces.

The usual point of reference is the positive x-axis, with angles measured counterclockwise from there being deemed positive. Taking the vector that you've calculated for the force of q2 on q1 as an example you can do the following:

You can look at your components and determine in which quadrant the vector lies. In this case you've got (scaled for ease of use) x = -2.6 and y = 1.62. So it lies in the 2nd quadrant. If you've got a calculator with an atan2 function you can calculate the angle right from those values. Otherwise you'll have to take a bit of care with the standard trig functions.

Have a look at the attached figure.

You can calculate the angle α using the standard atan function with the absolute values of the x and y components: α = atan(|y|/|x|) = atan(1.62/2.6), then subtract it from π to give θ. Or for the second quadrant you might use acos to calculate it directly from the component values: θ = acos(x/sqrt(x2 + y2)).

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Thank you again gneill. I have now arrived at the same answer as the OP. I was a little confused with the quadrants but I figured it out.