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anonymous
 3 years ago
what is the electric field at a point outside a hollow sphere made up of dielectric material which contain a charge at its centre (as shown below)?assume whatever variables one need to determine it.(eg. dielectric const,polarisability,etc)
anonymous
 3 years ago
what is the electric field at a point outside a hollow sphere made up of dielectric material which contain a charge at its centre (as shown below)?assume whatever variables one need to determine it.(eg. dielectric const,polarisability,etc)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362062634477:dw let the inner radius be r1 and outer be r2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't like the question. Assume values for everything? The dielectric and the point outside? so say dielectric of material is 1, and then use \[\epsilon_0\] for outside? q = \[1.6 \times 10^{19} C\] ? and the distances are not specified. I don't think this is a calculation, more a computation of formula to understand concepts. http://hyperphysics.phyastr.gsu.edu/%E2%80%8Chbase/electric/elefie.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Iamgmg90 : well all i meant by saying that is that i don't really understand what variables this force might depend so instead of saying that "take dielectric constant to be K and polarisibility to be P, etc..", i am asking others to consider the variables with any identifier and help me with this situation... If you'd like specifications then...take distance between P and charge to be r where r1<r2<r and r1 ,r2 are inner and outer radii respectively.the shell is made up of material of dielectric const K,electric susceptibility X and polarisibility P.find the electric field at P due to charge.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0\[\vec{\mathbf E}=\frac{Q}{4\pi \varepsilon_or^2}\hat{\mathbf r}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well if we check a special case of this problem then lets consider the situation when r1 be almost equal to 0 and r2 to be almost equal to r then the force surely should be Q/(4*pi*eps*r*r) where eps is the absolute permittivity of the dielectric..hence it shouldn't be the above..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I disagree. This should be a simple application of Gauss' law, assuming that the point in question is outside the sphere.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And that the sphere is a Gaussian surface :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But yes, definitely see your point.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Inside the material the field will be weaker, as you mentioned, but inside the hollow cavity and outside the sphere the field is the above. There are discontinuities at the inside and outside surfaces because the dielectric will become polarized and there will be a surface charge density there.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What do you mean by 'and the sphere is a Gaussian surface' ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0gauss's law only applies to gaussian surfaces. The mention of a dielectric goes against it slightly. Unless some of the assumptions we can make is that it's gaussian, and that the volume of the sphere is 4/3pir^3, in any case we'd have to find E, and then integrate over that area.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Gauss' Law says that if you draw any surface you want, the flux through the walls is equal to the charge enclosed divided by epsilon 0. It doesn't matter if there are dielectrics or not. In order to derive the expression for the electric field, all you need is that the problem is spherically symmetric, which it is.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see the integral form of gauss's law is only for non symmetrical objects.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know what you mean by that either. Gauss' law is very general and applies to all of electrostatics. It just happens that it's only useful when there's a nice symmetry to the problem.

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362197414836:dw @UnkleRhaukus and @Jemurray3 is right, for a hollow sphere we can say that \[\Phi=EA\] Area for a symmetrical sphere is given by \(4 \pi r^2\), \[\Phi=E4 \pi r^2 = \frac{Q}{\epsilon_0}\] The electric field outside the sphere, Gaussian surface at r > R is identical all around the sphere, \[E=\frac{Q}{4 \pi r^2 \epsilon_0}\]

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.1It is known that the electric field of the highlighted area is zero, then there should be negative charges accumulate inside the surface so that the net charge is zero inside, this also cause positive charges to build up on the surface of the sphere because of the negative charges inside and with no charges on the highlighted area. dw:1362204565241:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@.Sam. The electric field inside the body of the sphere is only zero if it's a conductor. If it's a dielectric material, as is the case here, then the field is nonzero.

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, I'm aware of that, when the dielectric is placed in an electric field, electric charges do not flow through the material as they do in a conductor, but only slightly shift from their average equilibrium positions causing dielectric polarization. Because of dielectric polarization, positive charges are displaced toward the field and negative charges shift in the opposite direction. This creates an internal electric field which reduces the overall field within the dielectric itself. That's why I'm not saying that there's zero charges inside, but rather the 'net' charges is zero.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The first sentence of your last comment was "It is known that the electric field of the highlighted area is zero"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And the amount of negative charge that builds up on the inner surface depends on the dielectric properties of the material.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well while applying gauss' law we do get the same field as a point charge without the dielectric..but as i mentioned above if we take a special case when the hollow shell tends to a solid sphere,then this formula doesn't holds..i think the epsilon zero we are using is a special case for absolute space and should be changed to "epsilon"  the permeability of the dielectric medium, when both the charge and the point are inside the dielectric..thats the whole trouble i m having with this situation..which epsilon to take when both space and dielectric are partially occupying the region between the point at distance r and the charge..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why does the formula no longer hold when the shell tends to a solid sphere?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll cut to the chase  it does. If you want a straightforward way to think about it, use the idea of the displacement current. Gauss' law becomes: \[ \int \vec{D} \cdot \vec{dS} = Q_{free} \] where Qfree is the charge that is NOT induced by the polarized medium. In your case, the charge in the center is Qfree, everything else is not (it's called Qbound). Because of spherical symmetry, \[ D(r) = \frac{Q_{free}}{4\pi r^2} \] \[D = \epsilon E \] So as long as we are outside the dielectric, \[ D = \epsilon_0 E\] and so \[E = \frac{Q_{free}}{4\pi \epsilon_0 r^2} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Gauss' law is always true for electrostatics, regardless of the medium in which you find yourself. However, if you yourself are inside a dielectric, it can get complicated because the charge inside your Gaussian surface is not only the free charge but also the induced charge inside the dielectric. In such cases, the use of the displacement field (I called it displacement current earlier, my typo, sorry) is a little bit more general.
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