Solving Point Charge Impact Time on Grounded Plane

[Euclid]

A point charge (q, mass m) is released from rest at a distance d from a grouned infinite conducting plane. How long does it take to hit the plane?
Answer pi*(d/q)*sqrt(2pi*eps m d)
This problem seemed easy to me at very, but it leads to a second order nonlinear equation
$$m\frac{d^2 z}{dt^2} = \frac{q^2}{16 \pi \epsilon_0 z^2}$$.

I tried using energy considerations to write v as v(z), I then solved for z as z(v), and integrated the above equation for v(t), putting in the limits 0 and infinity. This did not give the correct answer, although it appeared to be close. Any suggestions?

 

[Jelfish]

I haven't actually worked out the problem, so I'm not sure if the right side of your equation is correct. Just as a note incase you didn't do this - since the infinite conducting plane is grounded, you need to use the method of images to get your potential function.

 

[qbert]

Maple choked on the d.e. but, you should be able to show
v² = 2c(1/z - 1/d) with c = q²/(16 Pi Eps₀ m)
with conservation of energy or integrating the force equation once.
now to solve for time put solve for v and use the positive root,
the negative one leads to t<0.
thus $$t = \frac{1}{\sqrt{2c}} \int_d^0 \sqrt{ \frac{ dz}{ d - z}} \. d z$$
The substitution z = d cos²(theta) makes this doable.

 

[Euclid]

Yeah, that's what I indicated I tried above. It appeared to fail the first time I did it, but the second time it worked out.
Thanks!