Electric field of a hollow cylinder

In summary: That looks good.Let ##R \to 0##...In summary, the individual's method for finding the electric field due to one small ring involved treating it as many thin rings, using the equation for a thin ring and representing it infinitesimally. However, the individual realized they made a mistake in their calculation and corrected it by integrating from the point of observation, resulting in a more accurate and valid answer.
  • #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.
 

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  • #2
Zack K said:
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.
Yes, It's not correct.

In the diagram you provided (see immediately below) it looks like ##w## may be the point of observation on the ##x\text{-axis}##. That is to say, you intend to evaluate the electric field at ##x=w\,.##

241771


In that case your result should depend on ##w## .

In the Relevant equation you give,
##\displaystyle E_{ring}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}} \ ,##​
what do the variables represent? In particular, what does ##z## represent?
 
  • #3
SammyS said:
In particular, what does zzz represen
My bad, ##z## represents an arbitrary location, ##Q## is the charge of the ring and ##R## is the radius of the ring.

SammyS said:
In that case your result should depend on w
Yes that's what bothers me, my answer doesn't depend on a distance.

I think my mistake lies some where from the transition from
##E_{ring}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}## to ##dE=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q x}{(R^2+x^2)^{3/2}}##

Something that comes to mind, is that my distance vector is represented as ##\vec r= \langle w-x, 0, 0 \rangle##, but in my transition from ##E## to ##dE## I stated above, I labelled the distance as simply ##x##. Should it maybe be ##w-x## instead of just ##x##?
 
  • #4
Zack K said:
My bad, ##z## represents an arbitrary location, ##Q## is the charge of the ring and ##R## is the radius of the ring.Yes that's what bothers me, my answer doesn't depend on a distance.

I think my mistake lies some where from the transition from
##E_{ring}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}## to ##dE=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q x}{(R^2+x^2)^{3/2}}##

Something that comes to mind, is that my distance vector is represented as ##\vec r= \langle w-x, 0, 0 \rangle##, but in my transition from ##E## to ##dE## I stated above, I labelled the distance as simply ##x##. Should it maybe be ##w-x## instead of just ##x##?
Yes, ##z## is something like that. Actually, ##z## is the distance from the point of observation to the plane containing the ring. That's the x-coordinate of your ##\vec r ## vector. Right?

##x## represents the location a slice of your cylinder (a ring) and ##w## is the location of the point of observation. (Both locations are on the x-axis.)

How far is each slice from the point of observation?
 
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  • #5
SammyS said:
How far is each slice from the point of observation?
Isn't it how I represented it? As ##w-x##?

I'll try integrating this and see what I get.
 
  • #6
Zack K said:
Isn't it how I represented it? As ##w-x##?

I'll try integrating this and see what I get.
What ##d## ?

No. It's ##(w-x)##, is it not?
 
  • #7
SammyS said:
What ##d## ?

No. It's ##(w-x)##, is it not?
Yes sorry I changed it, ##d-w## is from another problem stuck in my head.

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

Giving me: ##E_x=\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\left[\frac{1}{\sqrt{R^2+(w-L)^2}}-\frac{1}{\sqrt{R^2+w^2}}\right]##

Looks like it makes more sense, but doesn't look pretty. How exactly do I test to see if this makes sense? I can't think of any limit tests that will clarify if the result is valid.
 
  • #8
Zack K said:
Yes sorry I changed it, ##d-w## is from another problem stuck in my head.

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

Giving me: ##E_x=\frac{1}{4\pi\epsilon_0}\frac{Q}{L}\left[\frac{1}{\sqrt{R^2+(w-L)^2}}-\frac{1}{\sqrt{R^2+w^2}}\right]##

Looks like it makes more sense, but doesn't look pretty. How exactly do I test to see if this makes sense? I can't think of any limit tests that will clarify if the result is valid.
That looks good.

Let ##R \to 0## .
 
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FAQ: Electric field of a hollow cylinder

What is the formula for calculating the electric field of a hollow cylinder?

The electric field of a hollow cylinder can be calculated using the formula E = λ/2πε₀r, where λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the center of the cylinder.

How does the electric field of a hollow cylinder vary with distance from the center?

The electric field of a hollow cylinder follows an inverse relationship with distance from the center. As the distance increases, the electric field strength decreases.

Is the electric field inside a hollow cylinder zero?

Yes, the electric field inside a hollow cylinder is zero. This is because the electric field is created by the charge on the surface of the cylinder, and there is no charge inside the cylinder.

How does the electric field of a hollow cylinder compare to that of a solid cylinder?

The electric field of a hollow cylinder is stronger than that of a solid cylinder of the same radius and charge. This is because the charge is spread out over a larger surface area in a hollow cylinder, resulting in a higher linear charge density.

Can the electric field of a hollow cylinder be affected by external charges?

Yes, the electric field of a hollow cylinder can be affected by external charges. External charges can either attract or repel the charges on the surface of the cylinder, altering the overall electric field in the surrounding area.

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