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In order to determine the net force other charges have on a particular charge, you must take into account each individual electric force between the charge and the charge of interest. However some of the individual electric forces may be negative (if the charges are negative). Therefore when solving the net force each of these electric forces have on a particular charge, you can use cosine law since we know the direction and magnitude of each charge. However if the net force is in fact negative, how do we account for this if the cosine law always gives you the root of your answer (so it could be either positive or negative-but how would you know??)?

To put this in context, take this diagram for example The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = + 9.70 μC; the other two charges have identical magnitudes, but opposite signs: q2 = - 5.02 μC and q3 = + 5.02 μC. The distance between q1 and the other two charges is 2.00 m.So to determine the net force exerted on q1 from the other two charges, we know the individual charges between q1 and q2, and q1 and q3. We also know the angle, therefore we must use cosine law to solve for net force. But cosine law only doesn't tell you whether the force is negative or positive :( Does anyone have any thoughts on this?

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from charge 2: F= kq1q2/r^2 = -0.1094

from charge 3 = F= kq1q3/r^2 = 0.1094

Fnet^2 = (-0.1094)^2 + (0.1094)^2 - 2(-0.1094)(0.1094)(cos 46)

Fnet = 0.201 N <---i dont know what direction this is in? (positive or negative?)

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    $\begingroup$ Can you edit your question to show exactly what calculation you are doing when you use the cosine law. $\endgroup$ – John Rennie Oct 12 '14 at 6:28
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    $\begingroup$ Okay i added it in! $\endgroup$ – user5027 Oct 12 '14 at 6:38
  • $\begingroup$ There is always a sign ambiguity when you sum forces in this way. A diagram showing the force vectors would make it obvious. $\endgroup$ – Rob Jeffries Oct 12 '14 at 7:02
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My thought would be: don't use cosine rule to solve this problem. Instead resolve the forces due to each charge along the $x$ and $y$ axes and write them as vectors.

Hence:
$${\bf F}_{21} = -k\frac{q_{1} q_{2}}{r_{21}^{2}} [\cos(23^{\circ}) \hat{\bf x} + \sin(23^{\circ}) \hat{\bf y}]$$ $${\bf F}_{31} = -k\frac{q_1 q_3}{r_{31}^{2}} [\cos(23^{\circ}) \hat{\bf x} - \sin(23^{\circ}) \hat{\bf y}]$$ Where the initial minus signs just ensure that if the charges have the same sign, the force is repulsive.

The net force is ${\bf F} = {\bf F}_{21} + {\bf F}_{31}$. In your problem $r_{21} = r_{31}$ and $q_2=-q_3$, so

$${\bf F} = -k\frac{q_1 q_2}{r_{21}^{2}} [ (-\cos(23^{\circ}) + \cos(23^{\circ}))\hat{\bf x} +(\sin(23^{\circ}) + \sin(23^{\circ}))\hat{\bf y}] $$

This formula explicity yields the direction of the resultant as well as its magnitude. You will find it doesn't give the same magnitude as you find. This is because you have made a mistake in your use of the cosine rule by assigning signs to the sides in your formula. You should not have done this, only the modulus is valid. i.e. It should have been $$ F = [ (0.1094)^2 + (0.1094)^2 - 2(0.1094)(0.1094)\cos(46^{\circ}]$$

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