Electric flux, derivation of Gauss' Law and using Gauss' Law to find electric field.

00:00 In this video, we introduce electric flux, derive Gauss' Law and work an example of using Gauss' Law to find electric field inside and outside a uniformly charged spherical shell. 🧠 Access full flipped physics courses with video lectures and examples at https://www.zakslabphysics.com/ 00:29 Electric Flux derivation for constant electric field and area orientation: we introduce the concept of electric flux (the number of field lines penetrating a surface). Next, we derive a formula for electric flux when the electric field is uniform and the area has fixed orientation (in other words the area is flat so the normal vector always points in the same direction). 03:10 Electric Flux through an extended surface: we write down a formula for electric flux through an infinitesimal area element dA, then we extend our definition of electric flux to extended surfaces over which the electric field and the orientation of the area elements dA may change. So we arrive at an electric flux integral: the integral of E dot dA. 04:18 Example: electric flux integral for a point charge located at the center of a spherical surface. We compute the electric flux through a spherical surface with a charge of +q at the center. This integral gives us a preview of the main idea of Gauss' Law: the dot product in the integral is trivial because the electric field is always perpendicular to the area elements (i.e., the electric field is parallel to the normal vector for every dA). Next, the electric field magnitude is constant, so we can factor the electric field out of the integral. We are able to compute the total electric flux through the sphere and it turns out to be q/epsilon_0. 05:58 Derivation of Gauss' Law: working from the results of our first flux calculation (the flux through a spherical shell with a point charge at the center), we make three generalizations to arrive at Gauss' Law: first, the size of the spherical shell doesn't matter. Second, the shape of the closed surface doesn't matter: the flux will be the same through a closed surface regardless of the shape or size of the surface, or the location of the charge within the surface. Third, we compute the flux through the arbitrary closed surface due to a distribution of many point charges, and we obtain Gauss' Law: the flux through any closed surface is equal to q_enc/epsilon_0 where q_enc is the net charge enclosed within the surface. 11:30 Example: using Gauss' Law to find the electric field inside and outside a uniformly charged spherical shell of radius R and charge Q. We exploit symmetry to apply Gauss' Law in our example: choose a gaussian surface over which the electric field is perpendicular to the area elements and constant in magnitude. Because the problem has spherical symmetry, we choose a spherical shell for our Gaussian surface. The dot product in the flux integral is trivial because the electric field is perpendicular to the area elements, and we can factor E out of the integral because the electric field magnitude is constant over the gaussian surface. This means the remaining surface integral is just equal to the area of the gaussian surface! Finally, we are able to isolate the electric field after invoking Gauss' Law, and we find that the field outside the charged shell is the same as if a point charge was located at the center of the shell, while the field inside the shell is zero because there is zero enclosed charge for a Gaussian surface inside the sphere.