Electricity problem (Coulomb's Law)

- osanseviero

Electricity problem (Coulomb's Law)

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- osanseviero

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- osanseviero

If the net force is 4* 10^-6 N, which is the value of q3?

- anonymous

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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## More answers

- osanseviero

q1 is negative, q2 is positive, so it should go to the right.

- osanseviero

I did F12 + F23 = Fn
Then I wrote all the value of each one, and tried to get out q3

- anonymous

and what charge is q3?

- osanseviero

That is what the problem is asking

- anonymous

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

- anonymous

screenshot the problem.

- osanseviero

It is in Spanish

- osanseviero

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?

- osanseviero

Since the force is positive, I guess that it is going to the right

- anonymous

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 ?

- anonymous

You see where it says "There is a positive point charge at the origin q3?"

- osanseviero

Oh, sorry :/

- anonymous

I can still figure it out whether is postie or not by knowing the net force and distance btw.
But it takes longer.

- anonymous

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.)

- anonymous

\[\frac{ k*q1q3 }{ (0.2m)^2 }+\frac{ kq2q3 }{ (0.3)^2 }=4*10^{-6}N\]
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

- osanseviero

\[q3 = \frac{ 4 \times 10^{-6} }{ 9\times 10 ^{9} } \times \left( \frac{ r1^{2} }{ q1 } + \frac{ r2^{2} }{ q2 }\right)\]
That looks right?

- anonymous

I got \[q3(\frac{ k*q1 }{ r1^2 }+\frac{ k*q2 }{ r2^2 })=4x10^{-6}N\]

- anonymous

4*10^-6

- anonymous

\[q3=\frac{ 4*10^{-6} N}{ \frac{ k*q1 }{ r1^2 }+\frac{ k*q2 }{ r^2 } }\]
let r1 be distance between q1 and q3
and r2 be distance between q2 and q3

- osanseviero

Wooh got the result. 3.16nC

- anonymous

forgot to write r2^2 on the second denominator fraction.

- anonymous

Using your equation?

- osanseviero

Nope, I did it again

- osanseviero

You did it right

- anonymous

Let me calculate to make sure.

- anonymous

Yes 3.2nC

- anonymous

Close enough.

- osanseviero

Yep, calculated it again and got 3.2. Thanks a lot for all the help

- anonymous

Ye, no problem.

- anonymous

Just remember that net force is change in force in the x and y component.
It is useful for the one that goes diagonal.

- osanseviero

Yep, that is the next problem

- osanseviero

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

- anonymous

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.

- osanseviero

Ok, thanks :)

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