Conceptual question using Coulomb

[unrstudent]

1. Homework Statement
Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in Figure P15.10.

http://www.webassign.net/sf5/p15_10.gif

thats 3 cm if its to small to see.

6.00 µC charge Correct answer 46.76N
Magnitude to Left

1.50 µC charge Correct answer 157.325N
Magnitude to Right

-2.00 µC charge correct answer 110.565N
Magnitude to leftF = K|q1||q2|/r^2
I have it, but I don't know why it worksThis what I have
F13=43.14N
F23=67.425N
F12=89.9N

the 6.00 C made sense Fnet=|-89.9+43.14|
The following were completely different
1.50 C should be Fnet=|-89.9+67.425|
but it not instead, I add the two to get 157.325
this does not make sense because unless 6.00 c was negative, the charges should cancel out like they did in first oneBasically, I got the exact opposite magnitudes and direction, i don't understand

 

[Delphi51]

1.50 C should be Fnet=|-89.9+67.425|

No, both are in the same direction (right) so no minus sign. The positive 6 charge pushes to the right and the negative 2 charge pulls to the right.

 

[gneill]

You're getting your force directions confused.

Naming the charges from left to right q1, q2, q3, then q2 will be pushed to the right by q1 and pulled to the right by q3. So your sums should be +89.9N + 67.4N = 157.3N. Similarly for q3, where q1 and q2 both PULL on q3.

 

[unrstudent]

Thank you so much,

it was right in front of me.

 

[rwisz]

why.
I would like to first clarify the question and provide some background information on Coulomb's Law. Coulomb's Law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This means that as the distance between two charges increases, the force between them decreases and vice versa.

Now, looking at the given figure, we can see that there are three charges, one with a magnitude of 6.00 µC, one with a magnitude of 1.50 µC, and one with a magnitude of -2.00 µC. The question asks for the magnitude and direction of the Coulomb force on each of these charges. To calculate this, we can use the equation F = K|q1||q2|/r^2, where K is the Coulomb constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

For the 6.00 µC charge, the distance between it and the other two charges is 3 cm, as given in the figure. Plugging in the values, we get F = (9x10^9)(6x10^-6)(1.50x10^-6)/(0.03)^2 = 46.76 N. This is the correct answer given in the question.

For the 1.50 µC charge, the distance between it and the 6.00 µC charge is also 3 cm, but the distance between it and the -2.00 µC charge is 6 cm. Using the same equation, we get F = (9x10^9)(1.50x10^-6)(6.00x10^-6)/(0.03)^2 = 89.9 N for the force between the 1.50 µC and 6.00 µC charges, and F = (9x10^9)(1.50x10^-6)(2.00x10^-6)/(0.06)^2 = 43.14 N for the force between the 1.50 µC and -2.00 µC charges. These two forces are in opposite directions, so to find the net force on the 1.50 µC charge, we need to subtract the magnitudes of the two forces, giving us