- #1

Zack K

- 166

- 6

- Homework Statement
- A thin-walled hollow circular glass tube, open at both ends, has a radius R and length L. The axis of the tube lies along the x axis, with the left end at the origin. The outer sides are rubbed with silk and acquire a net positive charge +Q distributed uniformly. Determine the electric field at a location on the x axis, a distance w from the origin. s

- Relevant Equations
- ##E_{ring}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}##

I uploaded a diagram of the problem.

I treated this as many thin rings and integrated it over the length. I placed my origin as in the same place as the uploaded picture.

Finding the electric field due to one small ring:

##\vec r =\langle w-x, 0, 0 \rangle## where ##x## is the distance of the thin ring from the origin.

##|\vec r|= w-x## ##\therefore \hat r=1##

Now this is the spot where I'm sure I made a mistake. So using the equation of a thin ring, and representing that infinitesimally, we get:

##dE=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q x}{(R^2+x^2)^{3/2}}##

##\Delta Q=Q\left(\frac{2\pi R\Delta x}{2\pi R L}\right)=Q\frac{\Delta x}{L}##

##dE=\frac{1}{4\pi\epsilon_0}\frac{(Q\frac{\Delta x }{L})x}{(R^2+x^2)^{3/2}}##

Integrating from ##x=0## to ##x=L##:

##E=\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\int_0^L \frac{x}{(R^2+x^2)^{3/2}}dx##

Solving that integral, we get:

##E=\frac{KQ}{2L}\left( -\frac{1}{\sqrt{R^2+L^2}}+\frac{1}{R} \right)##

This doesn't seem right at all, and I have no way of checking since my book doesn't have the answer for this. My classes are over and my professor is at some conference, and I can't use Gauss's Law to check my answer since the electric field at that point is pointing in different directions.

I treated this as many thin rings and integrated it over the length. I placed my origin as in the same place as the uploaded picture.

Finding the electric field due to one small ring:

##\vec r =\langle w-x, 0, 0 \rangle## where ##x## is the distance of the thin ring from the origin.

##|\vec r|= w-x## ##\therefore \hat r=1##

Now this is the spot where I'm sure I made a mistake. So using the equation of a thin ring, and representing that infinitesimally, we get:

##dE=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q x}{(R^2+x^2)^{3/2}}##

##\Delta Q=Q\left(\frac{2\pi R\Delta x}{2\pi R L}\right)=Q\frac{\Delta x}{L}##

##dE=\frac{1}{4\pi\epsilon_0}\frac{(Q\frac{\Delta x }{L})x}{(R^2+x^2)^{3/2}}##

Integrating from ##x=0## to ##x=L##:

##E=\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\int_0^L \frac{x}{(R^2+x^2)^{3/2}}dx##

Solving that integral, we get:

##E=\frac{KQ}{2L}\left( -\frac{1}{\sqrt{R^2+L^2}}+\frac{1}{R} \right)##

This doesn't seem right at all, and I have no way of checking since my book doesn't have the answer for this. My classes are over and my professor is at some conference, and I can't use Gauss's Law to check my answer since the electric field at that point is pointing in different directions.