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I thought about using

E = 2*pi*k*sigma (although isn't that just true for a charged disc?)

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I thought about using

E = 2*pi*k*sigma (although isn't that just true for a charged disc?)

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Defennder

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and do they mean point as in infinitesimal point?

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The electric field for a spherical shell of charge is discontinuous by the amount [blank] at a point where there's a surface charge density sigma.

The answer is either

epsilon naught / sigma

sigma / epsilon naught

epsilon naught / sigma^2

epsilon naught 2 / sigma^2 (I think the book meant to say epsilon naught ^2 / sigma^2)

sigma^2 / epsilon naught

I'm really not sure what they're asking or how to figure this out.

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gabbagabbahey

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The easiest way to show this is to use the fact that the electric field due to an infinite plane is [tex]\frac{\sigma}{2{\epsilon}_0} \hat{n}[/tex]

where [tex]\hat{n}[/tex] is the unit vector perpendicular to the surface, pointing away from the surface. (this can be obtained using Gauss's Law) . If you are close enough to any surface charge, it will appear to be an infinite plane. The electric field will point in one direction on one side of the surface, and in the opposite direction on the other side of the surface and so the difference is:

[tex]\vec{E}_{above}-\vec{E}_{below}=\frac{\sigma}{2{\epsilon}_0} \hat{n}-\frac{\sigma}{2{\epsilon}_0} (-\hat{n})=\frac{\sigma}{\epsilon_0}\hat{n}[/tex]

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My book defines the discontinuity of the electric field of an infinite disc/plane (I'm not quite sure what this means conceptually) as E = 2*pi*k*sigma = (2*pi) / (4*pi*epsilon naught) = sigma / (2*epsilon naught).

(so not simply sigma / epsilon naught)

What does this mean conceptually and how would it apply to a spherical shell?

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Defennder

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Yeah I guess that's probably what they're referring to. The one described by gabba (his full name is too awkward and difficult to type out) works for a separate case of an infinite sheet.

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gabbagabbahey

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The discontinuity should always be by an amount [tex]\sigma / {\epsilon}_0[/tex] . If your text gives a value of half that, then they must be referring to the difference between the value on one side of the surface, very very close to the surface and zero which I guess mean that the author of your text thinks the electric field right at the location of a charge is zero; which is incorrect (it is infinite).sphericalshell?

My book defines the discontinuity of the electric field of an infinite disc/plane (I'm not quite sure what this means conceptually) as E = 2*pi*k*sigma = (2*pi) / (4*pi*epsilon naught) = sigma / (2*epsilon naught).

(so not simply sigma / epsilon naught)

What does this mean conceptually and how would it apply to a spherical shell?

For the difference between the field above and below a surface; you can check that you should get sigma/e_0: In the case of a constant charge density, the field inside a spherical shell of radius R, will be zero (you can check this using gauss's law) and the field immediately outside the surface will be

[tex]\frac{\sigma R^2}{\epsilon _0 r^2} |_{r=R}=\frac{\sigma}{\epsilon_0}[/tex].

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Defennder

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My book defines the discontinuity of the electric field of an infinite disc/plane as sigma / (2*epsilon naught).

(so not simply sigma / epsilon naught)

since the field of a surface isnt increasing as it approaches the surface I see no reason it would be infinite at the surface. even in the case of a point change why would it be infinite? if it is infinite then what direction does it point?The discontinuity should always be by an amount [tex]\sigma / {\epsilon}_0[/tex] . If your text gives a value of half that, then they must be referring to the difference between the value on one side of the surface, very very close to the surface and zero which I guess mean that the author of your text thinks the electric field right at the location of a charge is zero; which is incorrect (it is infinite).

.

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gabbagabbahey

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Well, for a point charge,since the field of a surface isnt increasing as it approaches the surface I see no reason it would be infinite at the surface. even in the case of a point change why would it be infinite? if it is infinite then what direction does it point?

[tex]E \propto \frac{1}{r^2}[/tex]

where r is the distance from the charge. What happens as r approaches zero?

The argument for the plane is a little different: if we let [tex]\hat{n}[/tex] be the unit normal upward froma horizontal plane charge, then the field(s) will be:

[tex]\vec{E}_{above}=\frac{\sigma}{{\epsilon}_0} \hat{n}, \quad \vec{E}_{below}=\frac{-\sigma}{{\epsilon}_0} \hat{n}[/tex]

But what is the field

The fact that, classically, the electric field is infinite at the location of a charge is due the way in which the electric field is defined: "The electric field [tex]\vec{E}(\vec{r})[/tex] due to a collection of source charges is the force per unit charge experienced by a test charge located at [tex]\vec{r}[/tex]".

How exactly is one to put a test charge at exactly the same location as the source charge?

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