Why is the electric field intensity zero inside a charged isolated spherical Gaussian surface?

Answer 1

A charged sphere has uniform density of charge on its surface. Choose any point inside the surface. Draw a narrow cone with its pointed end at your chosen point, and extending out to the surface of the sphere. It will intersect the surface of the sphere in a tiny, almost ellipse-like shape. Then, returning to your inside chosen point and extend your cone in the opposite direction until it hits the sphere again. Now you have two patches on the sphere; the surface area of each patch is proportional to the square of the distance of each one to your inner point. Also, the electric field strength generated by the patch of charge on your point is inversely proportional to the square of the distance from the point. So the total field from the two tiny patches is the sum of two equal and opposite fields, and is therefore zero. You can see that for whatever direction you choose for this, you are always going to get zero field. So there is no electric field inside a spherical charged conductor.

A second approach is to integrate the electric field generated by this point charge around the sphere relative to a random point inside, and see that this (line integral) is zero. This is effectively the same as the above visualization.

E = Q / (4*pi*e*d^2):

E = the electric field due to charge Q

Q = charge

e = permativity of the medium

d = the distance from the point of interest

It's easier to think of this in two dimensions, then expand to three, so start with two dimensions and choose a random point within the circle, (r, cos (phi)). For the sake of explaining here, I'm chosing a point on the X axis of a unit circle centered at 0. Notice the E field created by charge p*dl (charge density times the unit length) at an angle of 30 degrees can be broken into a vector pointed in the -X direction, and -Y direction. Notice also, similar charge at -30 degrees can be broken into the -X direction and Y direction. The X component magnitudes will be equal and in the same direction, and the Y component magnitudes will be equal and in the opposite direction, so only the E field along the X axis remains, and is equivalent to E*cos (phi) dl. Since the surface is a circle, you'll have an equivalent point on each right side of (r, cos(phi)) and left side that will cancel in this manner.

Notice if you expanded two three dimensions, the same would hold true. Also notice the location of the point can truly be random and the above will still hold true (but make integrating messy!).

Now, if you peform the line integral for the above charge density on the surface of the circle to the right of the random point chosen (r, cos (phi)), and compare to the line integral to the left, the magnitudes will be exactly equal and opposite each other. Depending on how you set the line integral up, it can be very difficult.

Answer 2

The electric field inside a charged isolated spherical Gaussian surface is zero because of Gauss's law. The net electric flux passing through the surface is zero, which implies that the net charge enclosed by the Gaussian surface is also zero. Therefore, there is no electric field inside the surface.