Sign and direction on charges, coulumbs law

[FlopDaNut]

the material is over coulombs law, and i am having a rough time understanding the business of the sign's of the charges and the direction of the charge at the same time. here is a problem from the text.:

Two charges are located on the x-axis with q1 = 2.3 x 10^-8 C at the orgin, and with q2 = -5.6 x 10^-8 C at x = 1.30 meters. Find the force exerted by these 2 charges on a third charge of q3 = 3.3 x 10^-8 C which is located at x=.24 m on the x axis.

ok so i set off to calculate the force exerted by each charge on q3, and then add them together to get the net charge.

for Q1 and Q3 the calculation is ===>

((9.0 x 10^9)(2.3 x 10^-8)(3.3 x 10^-8))/.24^2 = 1.18 x 10^-4 N

i don't understand what direction this charge is? since it is positive does it go to the right? but to the right of q1 or q3? it is a positive charge but can it have negative direction?

ok, so for the second part (q3 and q2) i have,,,,

((9.0 x 10^9)(3.3 x 10^-8)(-5.6 x 10^-8))/(1.3-.24)^2 = -1.48 x 10^-5 N

well, in the answer for this problem the calculation is set up exactly the same way, only they use a POSITIVE 5.6 x 10^-8 in their calculation for the charge of q2, even though in the problem it states that this charge is negative. i don't understand. since the charge for q3 is positive, these 2 charges are attracting each other. would this tend to increase the magnitude of the resultant force of one charge on the other?

im used to negative and positive indicateing direction on vectors, but here it seems to say that you will have negative charges in the negative direction, and likewise for positive?

can you please help clear this up at all? specifically with regards to the problem i stated and why they changed the sign of q2 in their calcualtion of the answer.

thanks

-chris

 

[srikanth.isro]

First thing to be remembered in calculating the force is don't use the sign of the charge in the equation. Try to see the direction in which the charge will move. In the first case the charge q3 will be repelled by charge q1 and it will move towards right with a force 1.18x10^-4N. In the second case don't use the negative sign for the charge in calculating the force. Now the force on the q3 due to q2 is again to the right because it is attraced by the negative charge in the right with a force of 1.48x10^-5N. The resultant force is just the sum of both these forces. The sign will be positive only since in both the times the charge is experiencing a force towards right side only

 

[FlopDaNut]

thank you very much it makes much more sense now, i also noticed that there are absolute value signs around the coloumbs law equation.

one more question:

how do you know that q2 is pulling q3 along the positive x-axis (positive force)
and not the other way around (q3 pulling q2 towards the origin (negative force))?

 

[srikanth.isro]

By Newton's 3rd law forces come in pairs only. q2 is pulling q3 along the positive x-axis and at the same time q3 is also pulling q2 along the -ve x-axis with the same magnitude. Acctually they are equal and opposite forces.