Guass's Law for a charge distribution

[Demon117]

First, this is not a homework question, just something I've been confused about for some time. I understand how to use Guass's law in many ways but one thing I have always stumbled with is whether the E-field of a charge distribution should involve little r or big R in such an example:

In finding the E-field of a long cylindrical charge distribution with radius R and a uniform charge density $\rho$. If I want to find the E-field inside (r<R) such a distribution shouldn't the final answer contain a big R?

Here is my work. Since we have cylindrical symmetry we can consider a guassian cylinder of radius r inside about the cylindrical axis. The total charge Q is given by $Q=\rho V$ where V is the volume of the entire cylinder correct? (This is where I get confused) Or just the Guassian cylinder?

If the entire cylinder then the volume is $V=\pi R^{2}L$, where L is the length of the cylindrical volume. Then working out the rest of the problem we have:

$\oint EdA=E(2\pi r)L = \frac{Q}{\epsilon_{0}} = \frac{\rho \pi R^{2}L}{\epsilon_{0}}$
$E = \frac{\rho R^2}{2 \epsilon_{0} r}$

If the Guassian cylinder then the volume is $V=\pi r^{2}L$, then the E-field is just:

$E = \frac{\rho r}{2 \epsilon_{0}}$

Do you see my confusion? I need to understand this and I think it is just geometrical issue. . .

 

[ZapperZ]

You need to be aware of a very important aspect of Gauss's Law - that the field at any point r depends ONLY on the charges enclosed by the surface of size r (under highly symmetric situation). The charges outside the surface do not contribute to the E-field.

So in your example, you could have R = infinity if you wish, and you still have the same expression for the E-field, because it only cares about the amount of charges enclosed by your gaussian surface.

Zz.

 

[BruceW]

It's the Gaussian cylinder. Otherwise, there is no point to making the Gaussian surface in the first place.

 

[physwizard]

If you're confident you've done it right, you have to believe the results which you get!

 

[Pushpam Singh]

According to the Gauss's Law, Q is the charge inside the gaussian surface. So, the 2nd answer involving 'r' is right.