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Copyright © 2009 Pearson Education, Inc. Applications of Gauss’s Law
Copyright © 2009 Pearson Education, Inc. Example Spherical Conductor A thin spherical shell of radius r 0 possesses a total net charge Q that is uniformly distributed on it. Calculate the electric field at points. (a) Outside the shell (r > r 0 ) and (b) Inside the shell (r < r 0 ) (c) What if the conductor were a solid sphere? By symmetry, clearly a SPHERICAL Gaussian surface is needed!!
Copyright © 2009 Pearson Education, Inc. Example Solid Sphere of Charge An electric charge Q is distributed uniformly throughout a nonconducting sphere, radius r 0. Calculate the electric field (a) Outside the sphere (r > r 0 ) and (b) Inside the sphere (r < r 0 ). Results Outside (r > r 0 ): E = Q/(4πε 0 r 2 ) Inside (r < r 0 ): E = (Qr)/(4πε 0 r 0 3 ) E = Q/(4πε 0 r 2 ) E = (Qr)/(4πε 0 r 0 3 ) Note!! E inside has a very different r dependence than E outside! By symmetry, clearly a SPHERICAL Gaussian surface is needed!!
Copyright © 2009 Pearson Education, Inc. Results α = (5Q)/(4π 0 r 0 5 ) Outside (r > r 0 ): E = Q/(4πε 0 r 2 ) Inside (r < r 0 ): E = (Qr 3 )/(4πε 0 r 0 5 ) A solid sphere of radius r 0 contains total charge Q. It’s volume charge density is nonuniform & given by ρ E = αr 2 where α is a constant. Calculate: ( a) The constant α in terms of Q & r 0. (b) The electric field as a function of r outside the sphere. (c) The electric field as a function of r inside the sphere. Example: Nonuniformly Charged Solid Sphere Note !! E inside has a very different r dependence than E outside! By symmetry, clearly a SPHERICAL Gaussian surface is needed!!
Copyright © 2009 Pearson Education, Inc. Example Long Uniform Line of Charge Results: E = λ/(2πε 0 R) A very long straight (effectively, ℓ ) wire of radius R has a uniform positive charge per unit length, λ. Calculate the electric field at points near (& outside) the wire, far from the ends. By symmetry, clearly a Cylindrical Gaussian Surface is needed!! Note!! E for the wire has a very different R dependence than E for the sphere!
Copyright © 2009 Pearson Education, Inc. Example Infinite Plane of Charge Charge is distributed uniformly, with a surface charge density σ (= charge per unit area = dQ/dA) over a very large but very thin nonconducting flat plane surface. Calculate the electric field at points near the plane. Results: E = σ /(2ε 0 ) A Cylindrical Gaussian surface was chosen, but here, the shape of the Gaussian surface doesn’t matter!! The result is independent of that choice!!!
Copyright © 2009 Pearson Education, Inc. Example Electric Field Near any Conducting Surface Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by E = σ/ε 0 where σ is the surface charge density on the surface at that point. A Cylindrical Gaussian Surface was chosen, but here, the shape of the Gaussian surface doesn’t matter!! The result is independent of that choice!!!
Copyright © 2009 Pearson Education, Inc. The difference between the electric field outside a conducting plane of charge & outside a nonconducting plane of charge can be thought of in two ways: A thin, flat charged conductor with surface charge density σ on each surface. For the conductor as a whole, the charge density is σ´ = 2σ 1. The E field inside the conductor is zero, so the flux is all through one end of the Gaussian cylinder. 2. The nonconducting plane has a total surface charge density σ, but the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.
Copyright © 2009 Pearson Education, Inc. Consider a static conductor of any shape with total charge Q. See figure Choose a Gaussian surface (dashed in figure) just below & infinitesimally close to the conductor surface. We just said that the electric field inside a static conductor must be zero. By Gauss’s Law, since the electric field inside the surface is zero, there must be no charge enclosed. So, All charge on a static conductor must be on the surface.
Copyright © 2009 Pearson Education, Inc. Now, consider a static conductor of any shape with total charge Q & an empty cavity inside See figure Choose a Gaussian surface (dashed in the figure) just outside & below, infinitesimally close to the surface of the cavity. Since it is inside the conductor, there can be no electric field there. So, by Gauss’s Law, there can be no charge there, so there is no charge on the cavity surface & All charge on a static conductor must be on it’s OUTER surface
Copyright © 2009 Pearson Education, Inc. Outline of Procedure for Solving Gauss’s Law Problems: 1. Identify the symmetry, & choose a Gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). 2. Sketch the surface. 3. Use the symmetry to find the direction of E. 4. Evaluate the flux by integrating. 5. Calculate the enclosed charge. 6. Solve for the field.
Copyright © 2009 Pearson Education, Inc. Electric Flux: Gauss’s Law: lll Gauss’s Law: A method to calculate the electric field. It is most useful in situations with a high degree of symmetry. Gauss’s Law: Applies in all situations. So, it is more general than Coulomb’s Law. As we’ll see, it also applies when charges are moving & the electric field isn’t constant, but depends on the time. As we’ll see It is one of the 4 basic equations of Electromagnetism. It is one of the 4 Maxwell’s Equations of Electromagnetism. Summary of the Chapter Recall the Theme of the Course!
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