Find the electric field at an arbitrary position

BPBAIR
Messages
11
Reaction score
0
Problem:

Use Gauss’s Law to find the electric field at an arbitrary position r in the hollow region (ρ =0) between two concentric conducting cylinders. The inner cylinder has radius a and bears a constant charge per unit length, λ, while the outer cylinder has inner radius b and bears the same charge per unit length.

Equation:
fe0654d94b9cbaea831ad3fde7aef4e5.png



Solution:
Not really sure, where to go with this from a geometric point of view.
 
Physics news on Phys.org
Ok, first of all, what symmetry can you use in this problem? Is it spherical, cylindrical, etc?

Also, just think of the definition of Gauss' Law, and see where you can apply it here.
 
Well it would have cylindrical symettry. And the field would be completely radial b/c the top and bottom components would cancel leaving only the integral for the sides. rho=0 for in between the two cylinders, which resembles a coaxial cable with a space between the two cylinders. If anyone could help me furtehr on this problem, I would appreciate it. I am confused on how to setup the integrand since it is between two cylinders.
 
BPBAIR said:
Well it would have cylindrical symettry. And the field would be completely radial b/c the top and bottom components would cancel leaving only the integral for the sides. rho=0 for in between the two cylinders, which resembles a coaxial cable with a space between the two cylinders. If anyone could help me furtehr on this problem, I would appreciate it. I am confused on how to setup the integrand since it is between two cylinders.

Ok. First of all, you need to construct a hypothetical gaussian surface over which you need to evaluate the surface integral, which in this case will be a cylinder because of the symmetry. First, can you find out what is the elemental area you're integrating over? Can you proceed from here to find the field?
 
figured it out. only the inner cylinder will contribute to the field. from there it is two simple intergrals and the gaussian turns out yo be [ lambda / (2*pi*epsilon*r) ] = E
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top