What Is the Electric Field at the Surface of a Geiger Counter?

[manenbu]

Homework Statement

Taken from Resnick and Halliday:

A Geiger counter has a metal cylinder 2.10 cm in diameter along whose axis is stretched a wire 1.34E-4 cm in diameter. If 855 V is applied between them, find the electric field at the surface of (a) the wire and (b) the cylinder.

Homework Equations

Gauss' Law, potential-field

The Attempt at a Solution

Field inside the counter is by Gauss' law:

$$dq = \epsilon_0 E dA = \epsilon_0 E 2 \pi\ r dL$$
rearranging and using $\lambda = \frac{dq}{dL}$:
$$E = \frac{\lambda}{\epsilon_0 2 \pi r}$$

now:
$$V = \int E dr = \frac{\lambda}{\epsilon_0 2 \pi}\int\frac{dr}{r}$$
solving from r1 to r2:
$$V = \frac{\lambda}{\epsilon_0 2 \pi}\ln{\frac{r_2}{r_1}}$$
solving for $\lambda$ and putting into the expression of E from before:
$$E = \frac{V}{r \ln{\frac{r_2}{r_1}}}$$
plugging in the numbers from the beginning of the post, I get for r1 and r1:
4214 V/m
66.06 MV/m

The answers given in the appendix of the book are 2 times my (8kV/m and 132 MV/m).
Where did I miss a factor of 2?

 

[cathode-ray]

Hi!

I'm not 100% sure but I think you are just calculating the electric field generated by one of the elements. In the cylinder, there will be a charge Q (or -Q), and in the wire a charge of opposite sign, so each one will generate an electric field, which inside the cylinder have the same direction. So the electric field through your Gaussian surface will be the sum of these two vectors, that have the same magnitude and direction. It's from here that comes the factor of 2 (because you need to multiply by two the final electric field).

I hope this helps.

 

[sigmaro]

i am sorry cathode-ray but i strongly disagree, this is the last thing to be said someone who is dealing with gauss' law, outer cylinder is perfectly symmetric that it won't have any effective electric field inside and it is clear if you take a cylinderic gaussian area.
there is just one weird notation about you calculation, the rest is ok i think book is wrong or you calculate wrong (i can't check i lost my calculator)
you don't have to deal with
dq = 2(pi)€E dA = €E2(pi)rdL
if you say dL then your gaussian area is like a disk and you can't count its area, it is close to zero but it is not an exact mistake.
Just put numbers again, be sure you don't do any mistakes

 

[cathode-ray]

I'm sorry. You're right sigmaro. I'm also studying electromagnetism, but definitively I need to study more. Thanks for the help and Merry Christmas.

Note: http://www.cartage.org.lb/en/themes/sciences/Physics/Electromagnetism/Electrostatics/Capacitors/Capacitors/Parallelplate/Example/Example.htm"

 

[gneill]

radii = diameters/2

 

[gomunkul51]

like gneill stated,

2.10 cm and 1.34E-4 cm are diameters and you need radii

*it didn't matter in the expression ln(R/r) because the 1/2 factor cancels out.