Question

apply Gauss law to Show that for a uniformly charged sperical shell, the electric field inside the shell vanishes​

Answer 1

Answer:

Of course, yes. Well, this is what is great about guass theorem as it can solve such problems which cant seemingly or very hard to solve with traditional vector resolution methods.

The definition of guass theorem is that the total electric flux(surface integral of electric field over a closed surface) is always equal to the charge inclosed by it, upon permitivity.

Say Q is the charge inclosed by the surface and e is the permitivity of the medium. I wrote the equation in this picture below.

In the picture you can see the guass law.

Now to the problem,

Sphere is a symmetrical shape and the charge is present at the surface of the sphere.

Now Imagine a sphere which covers the actual metallic sphere. The electric field lines are passing through this imaginary sphere. All the things that are coming up will be related to this imaginary sphere which is inclosing our charge Q. By the way, we can also call the surface as guassian surface as strength of electric field at this surface stays same.

As sphere is a symmetrical shape, and charge density is uniform on the metal surface, this means that the electric field must also be symmetric(Rotation of metal sphere doesn't change the electric field at any fixed point on our imaginary surface as charge is uniformly distributed).

Since electric fields are symmetric, the electric flux will become E × 4πr^2, where E is the electric field and r is the radius of imaginary sphere(guassian surface). Its just electric field multiplied by area of the guassian surface.

Now, According to the guass theorem, we have,

4πr^2E = Q/e

Or,

E = 1/4πe × Q/r^2

Which is electric field due to point charge.

We proved it.

One more thing to notice that the electric field inside the sphere is zero as all the charge is present on the suface of the metal sphere. If you choose our imaginary sphere(guassian surface) inside the metal sphere then no charge is present inside the guassian surface so total electric flux is zero which implies that the electric field is also zero.

Answer 2

Answer:

Using Gauss's law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.

 

 

Consider a spherical Gaussian surface of radius r (›R), concentric with given shell. If is electric field outside the shell, then by symmetry, electric field strength has

same magnitude on the Gaussian surface and is directed radially outward. Also the directions of normal at each point is radially outward, so angle between and is zero at each point. Hence, electric flux through Gaussian surface =

 

Now, Gaussian surface is outside the given charged shell, so charge enclosed by the Gaussian surface is Q.

Hence, by Gauss's theorem

Thus, electric field outside a charged thin spherical shell is same as if the whole charge Q is concentrated at the center.

Graphically,

For r ‹ R, there is no strength of electric field inside a charged spherical shell.

For r › R, electric field outside a charged thin spherical shell is same as if the whole charge Q is concentrated at the center.

Explanation: