Question

Urgent Question!! Really would appreciate quick help.

The net force on the charge in the figure is zero. what is q?

Answer 1

See attached figure.

Answer 2

I broke it down into three parts: Force of the 2nC charges on the 1nC, force of the 2nC charges on q, and force of the 1nC on q. Is this correct? I got 1.9e-8 N, but it is not the right answer, not sure what I'm doing wrong. I'd appreciate any explanations.

Answer 3

Which charge has a net force of zero? It can't be the \(q\) charge, right?

Answer 4

Ah, I thought you were looking at the question. It must be the \(1\ [nC]\) charge that experiences no net force. All other will experience a net force regardless of what \(q\) is.

Answer 5

From your reply I think you are really confused! There are only a few things that are relevant here.

What you need:
Force of the 2nC charges on the 1nC
force of the q on 1nC (not the other way because you are looking at 1nC)

What you can throw away:
force of the 2nC charges on q

You would also need to resolve the forces into the horizontal and vertical components for the force of 2nC on the 1nC and you will find that the horizontal force of the 2nC on the left cancels the horizontal 2nC force on the right! And the resultant vertical component of the force is upwards!

So what you need now is a force to push downwards on the 1nC, and if I had to take a guess, I will guess that charge q would be positively charged!

Answer 6

If you want, just one option, you can use the math where the forces from each charge onto the \(1\ [nC]\) charge add up to \(\overrightarrow 0\). The math will be similar no matter how you do it, in the end.

And Festinger is right! Those \(2\ [nC]\) charges provide forces whose horizontal parts cancel out. His response outlines everything well.

Answer 7

Festinger, could you work out your method? I tried something like that and ended up with 1.9e-8C, a wrong answer.

Answer 8

Meh. The things I do for evil.
Let left be the positive x direction, and up be the positive y direction.

Distance between 2nC and the 1nC charge is
\[R=\sqrt{0.03^{2}+0.02^{2}}=0.0361m\]
Magnitude of the force 2nC pushes on 1nC is
\[|F|=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\]
Magnitude and direction of the force of q on 1nC is:
\[F_{y}=-\frac{(q)(1nC)}{4\pi\epsilon_{0}0.02^{2}}\] (minus because downwards)

The 2nC on the left pushes the +1nC to the top right, thus
\[F_{x}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}cos{\theta}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.03}{R}\]\[F_{y}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}sin{\theta}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.02}{R}\]

The 2nC on the right pushes the +1nC to the top left, thus
\[F_{x}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}cos{\theta}=-\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.03}{R}\]\[F_{y}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}sin{\theta}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.02}{R}\]

Question says 1nC has no net force.
\[F_{x}=\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.03}{R}+(-\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.03}{R})=0\]
\[F_{y}=0=+\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.02}{R}+\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.02}{R}-\frac{(q)(1nC)}{4\pi\epsilon_{0}0.02^{2}}\]

Which reduces to:
\[\frac{(q)(1nC)}{4\pi\epsilon_{0}0.02^{2}}=2*\frac{(2nC)(1nC)}{4\pi\epsilon_{0}R^{2}}\frac{0.02}{R}\]\[\frac{(q)}{0.02^{2}}=2*\frac{(2nC)}{R^{2}}\frac{0.02}{R}\]\[q=2*\frac{(2nC*0.02^{3})}{R^{3}}=+3.4*10^{-10}C\]

Answer 9

sorry, q should be 6.8*10^-10 C