Electricity problem (Coulomb's Law)

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Electricity problem (Coulomb's Law)

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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If the net force is 4* 10^-6 N, which is the value of q3?
Is q3 attracting with q1? and is q3 repelling with q2? I need to know if the point charges are positive or negative.

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q1 is negative, q2 is positive, so it should go to the right.
I did F12 + F23 = Fn Then I wrote all the value of each one, and tried to get out q3
and what charge is q3?
That is what the problem is asking
We need to know charge of q3 to know whether it goes left of q1 or toward q1 and also whether it goes toward q2 or right of q2
screenshot the problem.
It is in Spanish
P1. Se ordenan tres cargas puntuales a lo largo del eje de las x. La carga q1=-4.5nC está en x=0.2m, y la carga q2 = 2.5 nC, en x=-0.3m. Hay una carga puntual positiva q3 en el origen. a) ¿Cuál debe ser el valor de q3 para que la fuerza neta sobre esta carga puntual tenga una magnitud de 4x10-6 N? Traduction: There are 3 charges in the x axis. Then they give all the values. What is the value to the force be 4x10^-6 N?
Since the force is positive, I guess that it is going to the right
Um. I google translated it. Three point load along the x axis are arranged . Q1 = -4.5nC the load is at x = 0.2m , and q2 = 2.5 nC charge at x = -0.3m . There is a positive point charge at the origin q3 . a) What should be the value of q3 to the net force on the point charge has a magnitude of 4x10-6 N ? Translation : There are 3 charges in the x axis . Then They give all the values ​​. What is the value to the force be 4x10 ^ -6 N ?
You see where it says "There is a positive point charge at the origin q3?"
Oh, sorry :/
I can still figure it out whether is postie or not by knowing the net force and distance btw. But it takes longer.
Let be k be the electric force constant. Since q2 and q3 are positive, they repel, so q3 goes to the right. Since q3 and q1 are opposite charged of each other, q3 attracts q1 so it gets near q1 (goes tot he right also.)
\[\frac{ k*q1q3 }{ (0.2m)^2 }+\frac{ kq2q3 }{ (0.3)^2 }=4*10^{-6}N\] where q1=4.5nC or 0.0000000045 C and q2=2.5nC=0.0000000025 C use the decimals it make it faster and plug in k
\[q3 = \frac{ 4 \times 10^{-6} }{ 9\times 10 ^{9} } \times \left( \frac{ r1^{2} }{ q1 } + \frac{ r2^{2} }{ q2 }\right)\] That looks right?
I got \[q3(\frac{ k*q1 }{ r1^2 }+\frac{ k*q2 }{ r2^2 })=4x10^{-6}N\]
4*10^-6
\[q3=\frac{ 4*10^{-6} N}{ \frac{ k*q1 }{ r1^2 }+\frac{ k*q2 }{ r^2 } }\] let r1 be distance between q1 and q3 and r2 be distance between q2 and q3
Wooh got the result. 3.16nC
forgot to write r2^2 on the second denominator fraction.
Using your equation?
Nope, I did it again
You did it right
Let me calculate to make sure.
Yes 3.2nC
Close enough.
Yep, calculated it again and got 3.2. Thanks a lot for all the help
Ye, no problem.
Just remember that net force is change in force in the x and y component. It is useful for the one that goes diagonal.
Yep, that is the next problem
If I want the net force to be 0, I would just do F13 = F23 and use r1 = r and r2 = (0.5 - r). I need to get r
It depends on the force to set them equal to each other If both are pushing left or right, you cannot set them equal to each other. If there are forces on the left and right and net force is zero, you can make the opposite forces equal to each other. Yea r1=r and r2=(0.5-r) is important to use if you do not know the distance of one of the charges and know the other one.
Ok, thanks :)

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