Velocity of a charged particle

[Kosta1234]

Homework Statement

Velocity of charged particle

Relevant Equations

$$ \vec E \cdot \vec A = \rho (r) dV / \varepsilon _0 $$

Hi.
I will be glad if you can help me with a small problem.

I got a cylinder with a density $$ \rho (r) = b \cdot e^{\lambda r} $$, with radius $$ R $$.
If I let go a charge q < 0 from $$ R_0 $$ when $$ R_0 > R $$ , what will be the velocity that the charge will hit the axis of the cylinder.So I did the following:
I found the Electiric Field in space:

$$ \vec E = \frac {b}{\varepsilon _0 r} \cdot \frac {(\lambda r -1)\cdot e^{\lambda r} +1}{\lambda ^2} \hat r $$ when $$ r<R$$
$$ \vec E = \frac {b}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2} \hat r $$ when $$ r>R$$so that

$$ \vec F = \frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda r -1)\cdot e^{\lambda r} +1}{\lambda ^2} \hat r $$ when $$ r<R$$
$$ \vec F = \frac {qb}{\varepsilon _0 r} \cdot \frac {{{(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2} \hat r $$ when $$ r> R $$and now I want to use the second Newton's law:

$$ F = \frac {dv}{dt} $$
the problem is that I got an 'r' in the integral, can I solve the differential equation just like:

$$ \frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2})dt = dv $$

$$ (\frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2}) t dt = dx $$
$$ t^2 / 2 = \frac {dx}{(\frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2})} $$I'm a little bit confused because the force is changing along the distance

Thanks!. sorry that the latex didn't work in the end.. I've not idea what's wrong there

 

[BvU]

Hi,

sorry that the latex didn't work in the end..

Doesn't matter: at that point you were off the rails already

First of all, check your relevant equation: you can't have a differential on one side only !

Then: is the problem statement correct ? How does a charge reach the axis instead of bumping into the cylinder at $R$ ?

Then: as you say, the force changes. Don't we have something more useful to consider in such a case ?

Would you do all this match for a charge q > 0 when asked for the ultimate speed, or use a different approach ?

 

[BvU]

$$ \vec F = \frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2} \hat r $$ when $$ r> R $$and now I want to use the second Newton's law:

$$ F = \frac {dv}{dt} $$
the problem is that I got an 'r' in the integral, can I solve the differential equation just like:

$$ \frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2})dt = dv $$

$$ (\frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2}) t dt = dx $$
$$ t^2 / 2 = \frac {dx}{(\frac {qb}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2})} $$I'm a little bit confused because the force is changing along the distance

Thanks!. sorry that the latex didn't work in the end.. I've not idea what's wrong there Three {{{ instead of one

 

[Kosta1234]

I forgot to say that the cylinder is not solid.

and yes, my relevant equation is:

$$ \int \vec E \cdot d \vec A = \iiint \frac {\rho (r) dV}{\varepsilon _0 } $$

I tried a different approach now..
to calculate the potential in space, and from that the electric potential energy.

and then $$ \Delta U = \frac {1}{2}mv^2 $$
so that is:
$$ q \cdot ( \phi (0) - \phi(R_0) ) = \frac {1}{2}mv^2 $$

I think there is a problem with because there will be $$ ln(r) $$ in the phrase of the potential, and I can't put there r=0

 

[BvU]

Muuuuuch better ! And you know the potential at the axis, don't you ?

 

[Kosta1234]

Hmm

If I got those two:

$$ \vec E_1 = \frac {b}{\varepsilon _0 r} \cdot \frac {(\lambda r -1)\cdot e^{\lambda r} +1}{\lambda ^2} \hat r $$ when $$ r<R $$
$$ \vec E_2 = \frac {b}{\varepsilon _0 r} \cdot \frac {(\lambda R -1)\cdot e^{\lambda R} +1}{\lambda ^2} \hat r $$ when $$ r>R $$so I can calculate the potential

$$ \phi (0) = -\int_{R_0}^{R} {E_2 \cdot dr} - \int_R^0 E_1 \cdot dr $$
$$ \phi (R_0) = -\int_{\infty}^{R_0} E_2 \cdot dr $$
$$ U = q(\phi(0) - \phi(R_0) = 1/2mv^2 $$

What do you say?

 

[BvU]

And you know the potential at the axis, don't you ?

Perhaps that was a bit optimistic with the given charge density.

I agree with the relevant equation. Left is E A -- how do you work out the right for r = R ?

 

[BvU]

Left is E A -- how do you work out the right for r = R ?

never mind -- I'm a bit slow, but I'm with you. $\phi$ is the way to go.

 

[BvU]

But not towards infinity -- this is one of the cases where V can't be zero at infinity.

 

[Kosta1234]

Oh ok. so I can say that in general point $$ \phi(R_0) = 0 $$ so the answer will be:
$$ q \cdot \phi(0) = 1/2mv^2 $$ ?

 

[BvU]

I think so, yes. Or at 0, or at $R$ -- all you need is a potential difference

 

[Kosta1234]

Thank you!