E&M Griffiths 3.34

1. Nov 9, 2005

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?

2. Nov 10, 2005

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.

3. Nov 10, 2005

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.

4. Nov 10, 2005

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!