Charged particle near a grounded plane

[Reshma]

This one is from Griffiths. A toughie!

A point charge q of mass m is released at a distance d from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?

I chose a distance 'x' as a height above the plane. So the force on 'q' is given by:
$$F = {-1\over 4\pi \epsilon_0}{q^2\over 4x^2}$$

And equal that to force equation:
$$F = m\frac{d^2 x}{d t^2}$$

So,
$$\frac{d^2 x}{d t^2} = \frac{-q^2}{16\pi \epsilon_0 x^2 m}$$

I need to solve this equation for 't'. Any clues?

 

[Reshma]

Oh, by the way, for once Griffiths has been kind enough to provide the solution for this problem .

$$t = \left({\pi d \over q}\right) \sqrt{2\pi \epsilon_0 md}$$

This should be helpful.

 

[topsquark]

This one is from Griffiths. A toughie!

A point charge q of mass m is released at a distance d from an infinite grounded conducting plane. How long will it take for the charge to hit the plane?

I chose a distance 'x' as a height above the plane. So the force on 'q' is given by:
$$F = {-1\over 4\pi \epsilon_0}{q^2\over 4x^2}$$

And equal that to force equation:
$$F = m\frac{d^2 x}{d t^2}$$

So,
$$\frac{d^2 x}{d t^2} = \frac{-q^2}{16\pi \epsilon_0 x^2 m}$$

I need to solve this equation for 't'. Any clues?

You will find the substitution y=ln x useful.

-Dan

 

[StatusX]

You can multiply both sides by 2 dx/dt. This will leave you with d/dt((dx/dt)²) on the left side and d/dt(something else) on the right side, which you can easily integrate to get a first order ODE.

 

[Reshma]

You will find the substitution y=ln x useful.

-Dan

Did not work .

You can multiply both sides by 2 dx/dt. This will leave you with d/dt((dx/dt)²) on the left side and d/dt(something else) on the right side, which you can easily integrate to get a first order ODE.

OK, I multiplied it by v = dx/dt on both sides.

$$v {dv\over dt} = \left({q^2\over 16\pi \epsilon_0 m}\right)\left({-1\over x^2}\right){dx\over dt}$$

Taking $A = {q^2\over 16\pi \epsilon_0 m}$

$${d\over dt}\left({1\over 2} v^2\right) = {d\over dt}\left({A\over x}\right)$$

So,
$${1\over 2} v^2 = {A\over x} + \mbox{constant}$$

I need to bring this 'v' in terms of 'x' to integrate. How do I proceed?

 

[Reshma]

Oh dear, no replies... .

I proceeded with evalutions and I think I figured out the answer.

When v = 0; x = d. So, constant = (A/d)
Hence,
$$v^2 = 2A\left({1\over x} - {1\over d}\right)$$

$${dx\over dt} = \sqrt{{2A\over d}}\sqrt{{d-x\over x}}$$

$$\int_0^d {\sqrt{x}\over \sqrt{d-x}}dx = \sqrt{{2A\over d}}t$$

I just need to solve this for 't' but I'm not able to find a suitable substitution for the integrand terms to evaluate them. Any CLUES?

 

[gandharva_23]

put d-x = t square . u will get integeral of the form integeral(dt/(d-t^2)^1/2) which is a standard integeral . i don't know there might be a shorter way ... but u ll get the answer through this

 

[Physics Monkey]

Hi Reshma,

You are on the right track. Along the lines gandharva suggested, you might try a substitution of the form $$x = d \sin^2{\theta}$$.

 

[Reshma]

Thanks, Physics Monkey and gandharva. I got the answer.