What is the sign and magnitude of the unknown charge

[shashaeee]

From the attached image, the net force on the 1.0mC charge is zero.
What is the sign and magnitude of the unknown charge q?

Here is what I did:

F = k QQ / r²
F = 9.0x10⁹Nm²/C²*(0.002C)*(0.001C) / (0.05m)²

F = 7.2 x 10⁶N

Then to look for the unknown charge:

F = k QQ / r²
Q = F* r² / k Q
Q = 7.2 x 10⁶N * (0.03m)² / 9.0x10⁹Nm²/C²*(0.001C)

Q = 7.2 x 10^-4C or 0.72mC

It's sign will be negative as well

Is this correct?

 

[CFede]

I think you got this wrong.

In order to look for the sign of the charge, you should think about in what direction it should exert the force. Since the two bottom charges are postive, the "push" the central charge upwards, which means that, in order to get null net force, you need de upper charge to push downwards on the central one. What does this tell you about the sign of the charge?

As for the amgnitude of the force, you are thinking the right way but you forgot something. The charges at the bottom are not aligned with the other ones, which means that the forces excerted by this charges won't be in a vertical line, but will at some angle different from zero with the vertical. However this two forces do have a vertical component. You should compute this, since this is the force the upper charge has to counter.

 

[shashaeee]

Thanks for your reply!

So If I have to find the vertical component, I would do:

From the F= 7.2 x 10⁶ I had earlier,

Fy = 7.2 x 10⁶N cos 60° = 4232054N

So that should be the same force by the unknown charge

Q = 4232054N * (0.03m)² / k * (0.001C)
Q = 4.23 X 10^-4mC

and this time, the sign will be positive?

 

[CFede]

As you said, this time the sign would be possitive. However tou have overlookes a couple of things.

The triangle formed by the charges is not an equilateral one, so the angle is not 60º. Actually it doesn't matter the actual value of the angle, since you only want to know its cosine, but you already now the values of the three sides, so knowing the cosine is pretty straight forward. You are looking at the upper angle in that triangle so, the cosine should be 3/5.

On the other hand, you should also remember that there are 2 charges at the bottom so, if looked at what you did correctly, the force should be twice as much.

 

[shashaeee]

The 5cm in the picture wasn't given. I calculated it using Pythagorean theorem

 

[shashaeee]

I'm sorry, I'm not that sure, but from what I understand of your last message


Cos θ= 0.03m / 0.05m = 59°

Fy = 7.2 X 10⁶N cos 59° = 4323026N

And you said, the force will be twice as much: F = 8646051N

Q= F*r²kQ

Therefore, Q = 0.86mC (positive)?

 

[CFede]

As you computed cos(θ)=0.03/0.05=0.6. And that's it. So then

Fy=7.2 X 10e6 N cosθ=7.2 X 10e6 X 0.6 N

As you see you never get (because you don't want to) to know the exact value of the angle θ, but there's no need to, because you already know its cosine.

 

[shashaeee]

Thanks for your help!