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Ch 24 – Gauss’s Law. Karl Friedrich Gauss (1777-1855) – German mathematician. Ch 24 – Gauss’s Law. Already can calculate the E-field of an arbitrary charge distribution using Coulomb’s Law. Gauss’s Law allows the same thing, but much more easily…

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## Karl Friedrich Gauss (1777-1855) – German mathematician

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**Ch 24 – Gauss’s Law**Karl Friedrich Gauss (1777-1855) – German mathematician**Ch 24 – Gauss’s Law**Already can calculate the E-field of an arbitrary charge distribution using Coulomb’s Law. Gauss’s Law allows the same thing, but much more easily… … so long as the charge distribution is highly symmetrical. Karl Friedrich Gauss (1777-1855) – German mathematician**Ch 24 – Gauss’s Law**For example, in Ch 23: We found the E-fields in the vicinity of continuous charge distributions by integration… booooo: E-field of charged disc (R>>x): Now, we’ll learn an easier way.**Ch 24.1 – Electric Flux**• Sounds fancy, but it’s not hard • Electric Fluxmeasures how much an electric field wants to “push through” or “flow through” some arbitrary surface area • We care about flux because it makes certain calculations easier.**Ch 24.1 – Electric Flux – Case 1**Easiest case: • The E-field is uniform • The plane is perpendicular to the field Electric Flux**Ch 24.1 – Electric Flux – Case 1**Easiest case: • The E-field is uniform • The plane is perpendicular to the field Electric Flux Flux depends on how strong the E-field is and how big the area is.**Ch 24.1 – Electric Flux – Case 2**Junior Varsity case: • The field is uniform • The plane is not perpendicular to the field**Ch 24.1 – Electric Flux – Case 2**Junior Varsity case: • The field is uniform • The plane is not perpendicular to the field Flux depends on how strong the E-field is, how big the area is, and the orientation of the area with respect to the field’s direction.**Ch 24.1 – Electric Flux – Case 2**And, we can write this better using the definition of the “dot” product. where:**Ch 24.1 – Electric Flux – Case 2**Quick Quiz: What would happen to the E-flux if we change the orientation of the plane?**Ch 24.1 – Electric Flux – Case 3**Varsity (most general) case: • The E-field is not uniform • The surface is curvy and is not perpendicular to the field**Ch 24.1 – Electric Flux – Case 3**Imagine the surface A is a mosaic of little tiny surfaces ΔA. Pretend that each little ΔA is so small that it is essentially flat.**Ch 24.1 – Electric Flux – Case 3**Then, the flux through each little ΔA is just: is a special vector. It points in the normal direction and has a magnitude that tells us the area of ΔA .**Ch 24.1 – Electric Flux – Case 3**So… to get the flux through the entire surface A, we just have to add up the contributions from each of the little ΔA’s that compose A.**Ch 24.1 – Electric Flux – Case 3**Electric Flux through an arbitrary surface caused by a spatially varying E-field.**Ch 24.1 – Electric Flux – Flux through a Closed Surface**• The vectors dAi point in different directions • At each point, they are perpendicular to the surface • By convention, they point outward**Electric Flux: General Definition**• E-Flux through a surface depends on three things: • How strong the E-field is at each infinitesimal area. • How big the overall area A is after integration. • The orientation between the E-field and each infinitesimal area.**Electric Flux: General Definition**Flux can be negative, positive or zero! -The sign of the flux depends on the convention you assign. It’s up to you, but once you choose, stick with it.**Quick Quiz: what are the three things on which E-flux**depends?**Ch 24.1 – Electric Flux – Calculating E-Flux**• The surface integral means the integral must be evaluated over the surface in question… more in a moment. • The value of the flux will depend both on the field pattern and on the surface • The units of electric flux are N.m2/C**Ch 24.1 – Electric Flux**• The net electric flux through a surface is directly proportional to the number of electric field lines passing through the surface.**EG 24.1 – Flux through a Cube**Assume a uniform E-field pointing only in +x direction Find the net electric flux through the surface of a cube of edge-length l, as shown in the diagram.**Ch 24.2 – Gauss’s Law**Gauss’s Law is just a flux calculation We’re going to build imaginary surfaces – called Gaussian surfaces – and calculate the E-flux. Gauss’s Law only applies to closed surfaces. Gauss’s Law directly relates electric flux to the charge distribution that creates it.**Ch 24.2 – Gauss’s Law**Gauss’s Law**Ch 24.2 – Gauss’s Law**Gauss’s Law The net E-flux through a closed surface Charge inside the surface**Ch 24.2 – Gauss’s Law**• In other words… • 1. Draw a closed surface around a some charge. • Set up Gauss’s Law for the surface you’ve drawn. • Use Gauss’s Law to find the E-field. You get to choose the surface – it’s a purely imaginary thing.**Quick Quiz**Which surface – S1, S2 or S3 – experiences the most electric flux?**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**Let’s calculate the net flux through a Gaussian surface. Assume a single positive point charge of magnitude q sits at the center of our imaginary Gaussian surface, which we choose to be a sphere of radius r.**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**At every point on the sphere’s surface, the electric field from the charge points normal to the sphere… why? This helps make our calculation easy.**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**At every point on the sphere’s surface, the electric field from the charge points normal to the sphere… why? This helps make our calculation easy.**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**Now we have: But, because of our choice for the Gaussian surface, symmetry works in our favor. The electric field due to the point charge is constant all over the sphere’s surface. So…**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**This, we can work with. We know how to find the magnitude of the electric field at the sphere’s surface. Just use Coulomb’s law to calculate the E-field due to a point charge a distance r away from the charge.**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**Thus: And, this surface integral is easy.**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**Therefore: But, we can rewrite Coulomb’s constant.**Ch 24.2 – Gauss’s Law – confirming Gauss’s Law**Therefore: But, we can rewrite Coulomb’s constant. Thus, we have confirmed Gauss’s law:**A few more questions**• If the electric field is zero for all points on the surface, is the electric flux through the surface zero? • If the electric flux is zero for a closed surface, can there be charges inside the surface? • What is the flux through the surface shown? Why?**EG 24.2 – Flux due to a Point Charge**• A spherical surface surrounds a point charge. • What happens to the total flux through the surface if: • the charge is tripled, • the radius of the sphere is doubled, • the surface is changed to a cube, and • the charge moves to another location inside the surface?**Ch 24.3 – Applying Gauss’s Law**Gauss’s Law can be used to (1) find the E-field at some position relative to a known charge distribution, or (2) to find the charge distribution caused by a known E-field. In either case, you must choose a Gaussian surface to use.**Ch 24.3 – Applying Gauss’s Law**• Choose a surface such that… • Symmetry helps: the E-field is constant over the surface (or some part of the surface) • The E-field is zero over the surface (or some portion of the surface) • The dot product reduces to EdA (the E-field and the dA vectors are parallel) • 4. The dot product reduces to zero (the E-field and the dA vectors are perpendicular)**EG 24.3 – Spherical Charge Distribution**• An insulating solid sphere of radius a has a uniform volume charge density ρ and carries total charge Q. • Find the magnitude of the E-field at a point outside the sphere • Find the magnitude of the E-field at a point inside the sphere**EG 24.4 – Spherical Charge Distribution**Find the E-field a distance r from a line of positive charge of infinite length and constant charge per unit length λ.**EG 24.5 – Spherical Charge Distribution**Find the E-field due to an infinite plane of positive charge with uniform surface charge density σ**Ch 24.4 – Conductors in Electrostatic Equilibrium**• In an insulator, excess charge stays put. • Conductors have free electrons and, correspondingly, have different electrostatic characteristics. • You will learn four critical characteristics of a conductor in electrostatic equilibrium. • Electrostatic Equilibrium – no net motion of charge.**Ch 24.4 – Conductors in Electrostatic Equilibrium**• Most conductors, on their own, are in electrostatic equilibrium. • That is, in a piece of metal sitting by itself, there is no “current.”**Ch 24.4 – Conductors in Electrostatic Equilibrium**Four key characteristics • The E-field is zero at all points inside a conductor, whether hollow or solid. • If an isolated conductor carries excess charge, the excess charge resides on its surface. • The E-field just outside a charged conductor is perpendicular to the surface and has magnitudeσ/ε0, where σ is the surface charge density at that point. • Surface charge density is biggest where the conductor is most pointy.**Ch 24.4 – Conductors (cont.) – Justifications**Einside = 0 • Place a conducting slab in an external field, E. • If the field inside the conductor were not zero, free electrons in the conductor would experience an electrical force. • These electrons would accelerate. • These electrons would not be in equilibrium. • Therefore, there cannot be a field inside the conductor.

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