# Gauss's Law Problem

1. Jul 7, 2009

### jghlee

1. The problem statement, all variables and given/known data
So here's the question...

A cylindrical shell of radius 7.00 cm and length 240 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm radially outward from its axis (mesaured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

2. Relevant equations
Flux = EA = q/Eo

3. The attempt at a solution
I was able to get the answer at (a) but I'm seriously questioning myself after taking a look at (b). So, basically (b) states that the E inside the cylindrical shell is 0 since all the charges reside on the surface of the spherical shell (so there's no charge inside right?). I solved (a) by ....
(3.6e4)(2r(pi)x) = q/Eo... where x=19cm since that's the distance from the center of the cylindrical shell to the point at where the E is 36.0 KN/c.

So that's the right way to do it but here's what confuses me. Why do we use 19cm instead of (19-7=12cm) when there are no charges inside the cylindrical shell? Isn't the charges ONLY coming from the surface of the shell? So basically shouldn't the distance from the charge to the point where we find the E be 12cm?

Any help will be appreciated :D

2. Jul 7, 2009

### LowlyPion

You are correct there is no charge inside the shell.
Consider the area of the cylindrical surface. It increases with r2.

So charge density decreases with area. (Q/A) And so does the field effect from the charge decrease with 1/r2.

So whether it is a line charge with 0 radius but charge measured by unit length, or surface charge at some radius, the field effect at some radial distance from the axis and outside the surface will be the same.

Line charge:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c1
conducting cylinder:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c2

3. Jul 7, 2009

### jghlee

So, just to make sure I'm on the right page here, we can basically consider the conducting cylinder as a really thick line charge?

4. Jul 7, 2009

### LowlyPion

Well the equations certainly bear a remarkable similarity don't they?

5. Jul 7, 2009

### jghlee

thanks for the help LowlyPion

6. Jul 7, 2009

### ideasrule

Gauss' law:

$$\oint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A} = \frac{Q_{\mathrm{enclosed}}}{\varepsilon_0}$$

applies to any closed surface whatsoever. That means you don't need to care how the charges are distributed inside the surface; as long as you have a closed surface, like the surface of a sphere, cylinder, or cube, the equation holds.

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