Recent content by sebah

Oh I see. In my case the horizontal components cancel and the vertical components add. I will then get: $$E=\frac{2*y*Q}{4 \pi \epsilon_0 (x^2+y^2)^{\frac{3}{2}}}$$

Like I mentioned in the other reply I got the formula from R. Shankar and his yale course. LINK

That's exactly what I did. My formula looks just like the one derived by R. Shankar in his yale physics course. LINK

Not really I think. If x=0 which means that the distance between the charges goes to zero (a=0) I get E=0. Not sure why I am getting zero here though. Should be something like the electric field of 2Q at the origin.

I am having trouble solving the following problem. I am given two positive charges on the x axis: I know that the electric field strength at point P is ##E=150 \frac{V}{m}##, ##d=1.8m## and ##a=2.5m##. I want to find the charge of ##Q##. As far as I know, the electric field on the y-axis...