1.14.3. Field due to a uniformly charged thin spherical shell

Consider a point P outside the shell with radius vector r. To calculate E at P, we take the Gaussian surface to be a sphere of radius r and with centre O, passing through P. All points on this sphere are equivalent relative to the given charged configuration. (That is what we mean by spherical symmetry.) The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point. Thus, E and DS at every point are parallel and the flux through each element is E DS. Summing over all DS, the flux through the Gaussian surface is E × 4πr². The charge enclosed is σ × 4πR². By Gauss’s law, E × 4πr² = (σ × 4πR²) / ε₀.

To find the electric field E outside a uniformly charged thin spherical shell, we use Gauss's law, which relates the electric flux through a closed surface to the charge enclosed within it. We consider a point outside the shell and draw an imaginary spherical surface (our Gaussian surface) around the shell to apply Gauss's law. Since the shell is spherical and uniformly charged, the electric field at any point on this spherical surface is constant and directed radially outward. Therefore, the electric flux through the surface is simply the electric field E times the surface area of the sphere (E × 4πr²). The charge enclosed by our Gaussian surface is the total charge on the sphere, which can be calculated from the surface charge density times the surface area of the shell. Setting these equal via Gauss's law gives us the formula for the electric field: E = (σR²)/(ε₀r²). Thus, the electric field outside a uniformly charged shell mimics that of a point charge located at the center.

Imagine standing in a large basketball arena filled with people carrying equal-sized balloons. If you are outside the arena and you hear a loud 'pop' from one of the balloons, the sound (representing the electric field) you hear from that one balloon is similar to the electric field you experience at position P outside the sphere. The impact of that one balloon's pop can be felt uniformly from a distance, just like how the field behaves outside the charged shell as if the charge were concentrated at the center.

In this case, the Gaussian surface encloses no charge. The flux through the Gaussian surface, calculated as before, is E × 4πr². Gauss’s law then gives E × 4πr² = 0, i.e., E = 0 (r < R) that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.

To find the electric field E inside a uniformly charged thin spherical shell, we again apply Gauss's law. In this scenario, we consider a Gaussian surface that lies entirely within the shell. The key here is that all the charge of the shell is located on its surface, and no charge exists within the volume of the shell itself. Consequently, when calculating the charge enclosed by our Gaussian surface, we find that it is zero. Gauss's law then tells us that the electric flux E × 4πr² must also be zero, which leads us to conclude that E = 0 for any point inside the shell. This implies that there is no electric field within the hollow space of a charged shell, a fact that is fundamental in electrostatics.

Think of being inside a large, empty room (the inside of the shell) that is perfectly insulated from outside sounds. Regardless of how much noise is made outside (the charge), you won't hear any sound (electric field) inside the room. The silence inside the room represents the electric field being zero at all locations within the shell.

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