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GROUP MEMBER’s • • • • • • Imran Gujjar Kaleem Ullah Dilawar Gull Sufyan Maqsood Muhammad Umair Ghufran Gauss’s Law Karl Friedrich Gauss (1777-1855) – German mathematician 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 Gauss’s Law We found the E-fields in the vicinity of continuous charge distributions by integration… booooo: R r x dE E-field of charged disc (R>>x): E 2 o Now, we’ll learn an easier way. Electric Flux • Sounds fancy, but it’s not hard • Electric Flux measures 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. Electric Flux – Case 1 Easiest case: • The E-field is uniform • The plane is perpendicular to the field E E A Electric Flux – Case 1 Easiest case: • The E-field is uniform • The plane is perpendicular to the field E E A Electric Flux Flux depends on how strong the E-field is and how big the area is. Electric Flux – Case 2 Junior Varsity case: • The field is uniform • The plane is not perpendicular to the field E E A E A cos Electric Flux – Case 2 Junior Varsity case: • The field is uniform • The plane is not perpendicular to the field E E A E A cos 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. Electric Flux – Case 2 E E A E A cos And, we can write this better using the definition of the “dot” product. where: A A nˆ Electric Flux – Case 2 Quick Quiz: What would happen to the E-flux if we change the orientation of the plane? 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 E dA surface E-Flux through a surface depends on three things: 1.How strong the E-field is at each infinitesimal area. 2.How big the overall area A is after integration. 3.The orientation between the E-field and each infinitesimal area. Electric Flux: General Definition E E dA surface 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. 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 • 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. Gauss’s Law Gauss’s Law NET qenclosed E dA surface 0 The net E-flux through a closed surface Charge inside the surface Gauss’s Law In other words… 1. Draw a closed surface around a some charge. 2. Set up Gauss’s Law for the surface you’ve drawn. 3. Use Gauss’s Law to find the E-field. NET qenclosed E dA surface 0 You get to choose the surface – it’s a purely imaginary thing. 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. 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. 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. 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… Gauss’s Law – confirming Gauss’s Law This, we can work with. NET E dA surface 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. Gauss’s Law – confirming Gauss’s Law Thus: NET dA 4 r sphere ke q 2 dA r surface 2 And, this surface integral is easy. Gauss’s Law – confirming Gauss’s Law Therefore: NET ke q 2 (4 r 2 ) r But, we can rewrite Coulomb’s constant. NET Thus, we have confirmed Gauss’s law: q 0 Flux due to a Point Charge • A spherical surface surrounds a point charge. • What happens to the total flux through the surface if: (A) the charge is tripled, (B) the radius of the sphere is doubled, (C) the surface is changed to a cube, and (D) the charge moves to another location inside the surface? 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. Applying Gauss’s Law • Choose a surface such that… 1. Symmetry helps: the E-field is constant over the surface (or some part of the surface) 2. The E-field is zero over the surface (or some portion of the surface) 3. 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)