Recent content by Babybluedino

Finally! Thanks

Alright. 0.222² + 0.250² = 0.334 which is the resultant force of the first 2 charges. Then 0.334 - F3 = 0 → 0.334 = 2.5 / d² → d = 2.73 0.334 cos Θ = 0.222 → Θ = 48.3° 2.73 cos 48.3 = 1.82m 2.73 sin 48.3 = 2.04m

##k \frac {\frac 5 2 Qp} {d²} = \frac {kQp} {3² + 2²} → k \frac {\frac 5 2 Qp} {d²} = \frac {kQp} {13} →## ##\frac { 1.58QpK} {d} = \frac {kQp} {3.6} → 5.7QpK = kQp *d → d = 5.7##

In this case d = 5.7m and this is the distance from q to the new charge, right? Which is what I'm looking for. However, the answer in my book is x = -1.82 y = -2.04. Maybe I'm wrong, but that doesn't match.

I think I'm almost there. 0.222² + 0.250² = c² c = 0.335 0.335 = 2.5/ d² d = 2.73m 0.335 cos theta = 0.222 theta= 48.5 Then 2.73 cos 48.5 = 1.81m 2.73 sin 48.5 = 2.04m Is this answer correct?

I guess it will be there : I'm thinking of using something like -0.222 +Fe cos theta = 0. I mean I tried a lot of thing, but I can't find the solution. I miss something, but I don't know what.

I have another question about the same problem. Basically, I have a 2.5Q charge and I need to find where to place it to have a null resultant force on q. I know the charge should be between Q and -2Q. So the charge should be -x -y. I know that -(k|q x 2Q| / -3²) - (k|q x Q| / -2²) + Fe = 0...

Fx = k|q x 2Q| / -3² Fy = k|q x Q| / -2² Ah... I found. It seems that the variables confused me. Thanks!

kqQ doesn't mean the vector kqQ? I can't figure out how to get (-0,222i -0,250j).

I forgot to mention. The answer should be kqQ(-0,222i -0,250j)

Alright, I'm not sure about that. There's nothing to tell if the value of q and Q is the same.

I did it, but I'm not sure if I have to find the value of Q first and how I can get it.

Summary:: I try to find the resultant force on "q". I think I have to find the value of Q, but I'm not sure. I Know F1 = k|q * 2Q| / 3² and F2 = k|q * Q| / 2² Hi, this is my first post on this forum I hope I posted in the right section. I try to find the resultant force on "q". I think I...