Gauss Theorem:Electric field of an uniformly charged non-conducting spherical shell

Question

I want to know the electric field of an uniformly charged non-conducting spherical shell.

I know that in case of conductors(metals),the sphere can be shell or it can be solid,but in both the cases ,the charge resides at the surface,so I can easily get the electric field.

But,in case of non-conductors,if the sphere is solid,then the charge is all throughout distributed,so I can easily get the electric field.

But,I want to know if any spherical shell(empty inside) is non-conductor,then what is the electric field?

Answer 1

The electrostatic field depends only on the total charge distribution. If the charge distribution is known, as it is in your case, then you don't need to worry about the shape or conductivity of the structure supporting the charge.

The charge on a conducting solid sphere will, as you say, distribute evenly at the surface. If you by some other method manage to distribute the same charge evenly on a non-conducting sperical shell then you will see the same electric field: Zero inside and that of a point charge outside.

Answer 2

Appending to the answer given above, since I cannot post comments. Integrating in the onion peel manner, charge enclosed in the gaussian surface of a 'thin' shell taking care of limits (inner radius and beyond to whatever distance so desired) should yield you the answer to your question. Note the exponent of the distance parameter in your 'distance - electric field magnitude' function. Is it -1, -2, -3? Of course, this is just leads, providing bare skeleton and direction. The math of it is well. . .just that: the math. This should give you some mathematics to springboard from.

As for your wondering why the distinction between the two cases of conducting and non-conducting, you rightly mentioned how distributions seek least potential and stick as far as they can on conducting shells (like our hair on static!), but as for non-conducting, as you 'peel' inwards out, your charge within that peel increases proportional to the. . (is it linear, square, triple, check! :) ). That'll be all. My first 'post'. Hope I helped. :)

Answer 3

If you know Gauss's Law, apply it. Within the boundary, charge enclosed is zero. So, closed integral of E.dS is zero(i.e. flux). Now, we know that the field and area vectors are parallel, and area vector is non-zero. What is indeed zero is the magnitude of field vector, and there you go. Field inside the spherical shell is zero at any point. On drawing a spherical Gaussian surface outside, charge enclosed is the charge present on the shell(call it q) and again suitably applying Gauss's Law, we get the formula, which turns out to be the same, had you been working with a point charge placed at the geometrical center of this shell. Hope this helps. P.S. I am new here, so I still need to get used to LATEX and I don't have enough reputation to start commenting.

Answer 4

The Electric field outside a point of the shell=KQ/r^2 assuming Q=Charge on the shell and r=distance of the point from the Centre of the Sphere.This can be easily derived if you draw another Gaussian sphere of radius r enclosing the given Sphere.By Gauss Law, Closed integral of E.dA=Charge enclosed/Epsilon. And You can easily find out the Electric Field. Similarly at the surface as well, you can apply the same method and find out the electric Field which would be equal to: E=KQ/R^2 where R=radius of the spherical shell. Now For a point inside the sphere, E will not be equal to 0 if Sphere is non-conducting and it is not symmetrical or if an External E.Field exists. Otherwise, E.Field inside the shell=0 as no charge is present inside the shell. I am sorry, I am new here, and haven't learnt using LATEX. Anyone can help by editing. Hope this helps.

Answer 5

The electric field will be zero inside a spherical shell no matter if it is conducting or nonconducting, because according to Gauss's law $\Phi = \frac{Q_{enc}}{\epsilon}$, where $\Phi$ is the electric flux thru the Gaussian surface, and $Q_{enc}$ is the charge inside the Gaussian surface. So, as the charge inside the Gaussian surface is zero therefore $\Phi=0$, and as $\vec{E}$ is perpendicular to the surface, so $\vec{E}$ must be 0.