Discontinuity in electrical field.

[peripatein]

Hi,

Homework Statement

I'd like to find the electrical field at the center of a hole made in a hollow sphere of radius R₀ with uniform charge density σ. The radius of the disc-shaped hole is a << R₀.

Homework Equations

I know that the electrical field of the sphere is Q/r² for r>R and 0 for r<R. I also know that the electrical field of the disc-shaped hole is -2πσ(1-r/SQRT(a²+r²).
I am advised NOT to calculate the electrical field at the center of the hole directly, but to use the above data and examine the value as R -> R₀, once as R>R₀ and once as R<R₀.

The Attempt at a Solution

As R->R₀, the sphere itself would contribute 4πσ to the electrical field, whereas the contribution of the hole should be -2πσ, hence the electrical field at the center of the disc-shaped hole should be 2πσ.
I am quite positive this is not how it should be done, moreover am inclined to believe that is not the correct value of the requested field.
I'd appreciate some guidance, please.

 

[rude man]

A "disc-shaped hole"? The hole has to have a third dimension - how thick is it?

The hole is going to skew the E field everywhere including for r > R_0 so you can't use a Gaussian surface directly.

Need to know the details of this hole ...

 

[peripatein]

All that is given is that the hole is small, with radius a much smaller than that of the sphere, and that it is to be considered a small planar disc. I hope this settles the difficulty.
May you please assist me now?

 

[haruspex]

the electrical field of the disc-shaped hole is -2πσ(1-r/SQRT(a²+r²).

That had me puzzled, but I think you're using r for more than one purpose. In the equation above, r is the distance from the centre of the hole, right?
The 2πσ result looks right to me. I would expect it to be the average of the values for r just less than R and r just greater than R for elsewhere on the sphere.

 

[haruspex]

A "disc-shaped hole"? The hole has to have a third dimension - how thick is it?

It's just a circular patch cut out of a spherical shell.

 

[peripatein]

Right, haruspex. But is the final result I obtained indeed the value of the electrical field right at the center of the hole? Examining it from below and above, as I would any limit, would not suggest discontinuity, whereas I believe that is what the question implies. I could be wrong. What value did YOU get for the electrical field at that point?

 

[peripatein]

I have since received no reply. Should I infer that none of you is able/willing to help?

 

[SammyS]

I have since received no reply. Should I infer that none of you is able/willing to help?

No. I'm sure someone will help.

Let me read all the posts & get back, so I don't repeat anything.

 

[SammyS]

Right, haruspex. But is the final result I obtained indeed the value of the electrical field right at the center of the hole? Examining it from below and above, as I would any limit, would not suggest discontinuity, whereas I believe that is what the question implies. I could be wrong. What value did YOU get for the electrical field at that point?

For an infinitesimally thin spherical shell, the E field is discontinuous at the surface. The field is zero just inside, and finite just outside.

The field due to an infinitesimally thin disc, is also discontinuous at the surface of the disc.

(Maybe these discontinuities cancel each other.)


I suggest modeling the situation described in this as a superposition.

Consider a complete spherical shell of radius, R. This shell has a uniform charge density per unit area of σ . Look at the field due this shell both immediately inside of the shell and immediately outside of the shell.

Next consider a thin disc of radius a. This disc has a uniform charge density per unit area of -σ . (Yes, that's negative .) Look at the field due this shell directly on either side of the disc.​

Now, place the disc on the surface of the shell, and superimpose the fields.

 

[peripatein]

But you seem to have not read my attempt at solution at all, as this is precisely what I was trying to accomplish. I was simply asking for feedback on my attempt and whether I indeed succeeded in finding the correct value.

 

[haruspex]

Examining it from below and above, as I would any limit, would not suggest discontinuity, whereas I believe that is what the question implies. I could be wrong. What value did YOU get for the electrical field at that point?

Discontinuities can only arise where there is some charge density. Within the hole (except right at its edge) the field function must be smooth. Yes, I get the same answer.

 

[SammyS]

But you seem to have not read my attempt at solution at all, as this is precisely what I was trying to accomplish. I was simply asking for feedback on my attempt and whether I indeed succeeded in finding the correct value.

I did read your attempt. Are you sure you read my post ?

Your values for the E field are not correct.

 

[peripatein]

The values were confirmed by my textbook, so they are correct. Writing this I am referring strictly to the spherical shell and disc-shaped hole. I was asking for your evaluation of my attempt to find the electrical field at the center of the disc-shaped hole. I have read your reply and did make use of a minus sigma charge density in my expressions as well as the superposition principle. Haruspex claimed the expression I got, namely 2*pi*sigma, was correct. Do you disagree, Sammy?

 

[SammyS]

The values were confirmed by my textbook, so they are correct. Writing this I am referring strictly to the spherical shell and disc-shaped hole. I was asking for your evaluation of my attempt to find the electrical field at the center of the disc-shaped hole. I have read your reply and did make use of a minus sigma charge density in my expressions as well as the superposition principle. Haruspex claimed the expression I got, namely 2*pi*sigma, was correct. Do you disagree, Sammy?

The magnitude of the E field near a uniformly distributed surface charge with charge density, σ, where σ is in units of C/m² is ##\displaystyle \ E=\frac{\sigma}{2\,\varepsilon_0 }\ .\ ##

Your expression for the E field of [itex]\ 2\pi\sigma\ [/itex] doesn't have the correct units.

Furthermore, if a spherical shell of radius, R, has a uniform surface charge with charge density, σ, then the total charge on the shell is ##\displaystyle \ Q=4\pi R^2\sigma\ .\ ## The electric field a distance, r, from the center of the shell is ##\displaystyle \ E=\frac{4\pi R^2\sigma}{4\pi r^2\,\varepsilon_0 }=\frac{R^2\sigma}{r^2\,\varepsilon_0 }\,,\ ## for r > R.

 

[haruspex]

. Haruspex claimed the expression I got, namely 2*pi*sigma, was correct. Do you disagree, Sammy?

I should have clarified that I was accepting your 4πσ for outside the sphere elsewhere as correct. I should only have confirmed that it would half as much in the centre of the hole.