Question

I Just have one question and a little bit of help. I'm going to write out the work I have and what the answer is supposed to be. I need to know where I went wrong.

Answer 1

A charge is fixed in place. A charge -q is released form rest a distance d from Q. Determine the velocity of -q as a function of its distance x from Q. You will need to write out Newton's 2nd law, apply the chain rule and then integrate both sides of the equation appropriately.

Answer 2

\[F = m \dfrac{dv}{dt}\]
\[F=mv\frac{dv}{dx}\]
\[\int Fdx=\int mvdv\]
\[\int Fdx=\frac {mv^{2}}{2}\]
\[\frac{Qq}{4\pi \epsilon_{0}}\int \frac{1}{x^{2}}dx=\frac{mv^{2}}{2}\]
its obviously not x because the answer is this:
\[v=\sqrt{\frac{Qq}{2\pi \epsilon _{0}m}(\frac{1}{x}-\frac{1}{d})}\]

Answer 3

@ganeshie8 think you could help me?