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**Applications of Gauss’s Law**

Figure Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2). b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero. c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

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**Example Spherical Conductor**

A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Calculate the electric field at points . (a) Outside the shell (r > r0) and (b) Inside the shell (r < r0) (c) What if the conductor were a solid sphere? By symmetry, clearly a SPHERICAL Gaussian surface is needed!! Figure Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3. Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2). b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero. c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

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**Example Solid Sphere of Charge Results**

An electric charge Q is distributed uniformly throughout a nonconducting sphere, radius r0. Calculate the electric field (a) Outside the sphere (r > r0) and (b) Inside the sphere (r < r0). By symmetry, clearly a SPHERICAL Gaussian surface is needed!! Results Outside (r > r0): E = Q/(4πε0r2) Inside (r < r0): E = (Qr)/(4πε0r03) E = (Qr)/(4πε0r03) Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2). b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03). E = Q/(4πε0r2) Note!! E inside has a very different r dependence than E outside!

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**Example: Nonuniformly Charged Solid Sphere**

By symmetry, clearly a SPHERICAL Gaussian surface is needed!! A solid sphere of radius r0 contains total charge Q. It’s volume charge density is nonuniform & given by ρE = αr2 where α is a constant. Calculate: (a) The constant α in terms of Q & r0. (b) The electric field as a function of r outside the sphere. (c) The electric field as a function of r inside the sphere. Results α = (5Q)/(4π0r05) Outside (r > r0): E = Q/(4πε0r2) Inside (r < r0): E = (Qr3)/(4πε0r05) Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2 dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05. b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05. Note!! E inside has a very different r dependence than E outside!

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**Long Uniform Line of Charge**

Example Long Uniform Line of Charge 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! Solution: If the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere. Therefore, a cylindrical gaussian surface will allow the easiest calculation of the field. The field is parallel to the ends and constant over the curved surface; integrating over the curved surface gives E = λ/2πε0R. Results: E = λ/(2πε0R)

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**Infinite Plane of Charge**

A Cylindrical Gaussian surface was chosen, but here, the shape of the Gaussian surface doesn’t matter!! The result is independent of that choice!!! 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. Solution: We expect E to be perpendicular to the plane, and choose a cylindrical gaussian surface with its flat sides parallel to the plane. The field is parallel to the curved side; integrating over the flat sides gives E = σ/2ε0. Results: E = σ /(2ε0)

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**Electric Field Near any Conducting Surface**

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/ε0 where σ is the surface charge density on the surface at that point. Solution: Again we choose a cylindrical gaussian surface. Now, however, the field inside the conductor is zero, so we only have a nonzero integral over one surface of the cylinder. Integrating gives E = σ/ε0. A Cylindrical Gaussian Surface was chosen, but here, the shape of the Gaussian surface doesn’t matter!! The result is independent of that choice!!!

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**σ´ = 2σ outside a conducting plane of charge**

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: 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. 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

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**All charge on a static conductor must be on the surface.**

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.

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**All charge on a static conductor must be on it’s OUTER surface**

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

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**Outline of Procedure for Solving Gauss’s Law Problems:**

Identify the symmetry, & choose a Gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). Sketch the surface. Use the symmetry to find the direction of E. Evaluate the flux by integrating. Calculate the enclosed charge. Solve for the field.

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**It is one of the 4 basic equations of Electromagnetism.**

Summary of the Chapter 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 Maxwell’s Equations of Electromagnetism. Recall the Theme of the Course!

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Charles Allison © 2000 Chapter 22 Gauss’s Law HW# 5 : Chap.22: Pb.1, Pb.6, Pb.24, Pb.27, Pb.35, Pb.46 Due Friday: Friday, Feb 27.

Charles Allison © 2000 Chapter 22 Gauss’s Law HW# 5 : Chap.22: Pb.1, Pb.6, Pb.24, Pb.27, Pb.35, Pb.46 Due Friday: Friday, Feb 27.

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