Chapter 24 Gauss Law

Transcription

1 Chapter 24 Gauss Law

2 A charge inside a box can be probed with a test charge q o to measure E field outside the box.

3 The volume (V) flow rate (dv/dt) of fluid through the wire rectangle (a) is va when the area of the rectangle is perpendicular to the velocity vector v and (b) is va cos φ when the rectangle is tilted at an angle φ. We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric flux Φ E. Volume flow rate through the wire rectangle.

4 (a) The electric flux through the surface = EA. (b) When the area vector makes an angle φ with the vector E, the area projected onto a plane oriented perpendicular to the flow is A perp. = A cos φ. The flux is zero when φ = 90 o because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle A flat surface in a uniform electric field.

5 The Porcupine Theory of Gauss Law Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills.

6 The Porcupine Theory of Gauss Law Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills. Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. How many porcupines are in the bag?

7 The Porcupine Theory of Gauss Law Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills. Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. How many porcupines are in the bag? Answer: Count the number of quills coming out through the cloth. Then, by dividing the number of quills by 30, we can tell how many porcupines are in the bag.

8 The Porcupine Theory of Gauss Law + Imagine that we have some porcupines with very long quills. Careful examination shows each porcupine has exactly 30 quills Now suppose that we hide a number of these porcupines in a loose-weave burlap bag. How many porcupines are in the bag? Answer: Count the number of quills coming out through the cloth. Then, by dividing the number of quills by 30, we can tell how many porcupines are in the bag. This is a good analogy to Gauss Law. The quill-count is the electric flux, the porcupines are electric charges, and the bag is a Gaussian surface surrounding the charges. Each charge Q has a quill count of Q/ε 0.

9 The Concept of Electric Flux The basic idea here is that the qualitative idea of field lines can be quantified by defining electric flux as a conserved quantity that is delivered by electric charge. Charge creates flux, and observing flux allows one to deduce the presence of charge. If an electric field is observed to be coming out of all walls of an opaque box as in (a), we can conclude that the box contains a net positive charge. If an electric field is observed to be going into all walls of an opaque box as in (b), we can conclude that the box contains a net negative charge. If equal amounts of electric field are observed to be entering and exiting the box as in (c), we can conclude that the box contains no net charge.

10 Conceptual Question 1 The box contains: (a) A single positive charge; (b) A single negative charge; (c) No charge; (d) A net positive charge; (e) A net negative charge.

11 Gaussian Surfaces A Gaussian surface is a closed virtual surface that surrounds a region of space that may contain a quantity of charge. In practice, Gaussian surfaces are useful primarily when the symmetry of the Gaussian surface matches the symmetry of the electric field that passes through the surface and of the charge distribution that produces it, and when the field lines are perpendicular to the surfaces that they cross.

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13 Matching the Field with the Gaussian Surface We would like to choose a closed Gaussian surface to assess the flux so that: 1.The electric field & surface are. 2.The electric field is constant or zero over each surface element. To do this, we must match the symmetries of the field and the closed surface.

14 Electric Flux Remark: When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive.

15 The Electric Flux of a Nonuniform Electric Field r r δ Φ e = E δ A r r Φ = δ Φ = E δ A e i i i i i r r Φ = dφ = E da e surface e surface For a non-uniform field, the flux can be calculated by breaking up the area A of interest into small surface elements da and integrating the E da flux increments over the surface.

16 Flux through a Curved Surface For a non-uniform field plus a curved surface, the same approach works. Break up the area A into small surface elements da and integrate the E da flux increments over the surface. When E da the flux contributions are zero. When E da the flux contributions are maximum.

17 Flux Surface Integral The flux is of particular interest when it surrounds a net charge. Therefore, we will focus on closed Gaussian surfaces. For that case, we write the flux integral as: Φ net electric r = E r da Here, the loop on the integral sign indicates a surface integral over a closed surface. Since a simple non-intersecting closed surface has a definite inside and outside, we take the vector direction of da as always pointing toward the outside. As we will see, this integral is related to the amount of charge enclosed by the surface.

18 Flux Φ E from a point charge q. The projection of an element of area da of a sphere of radius R UP onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 da. The same number of lines of flux pass thru each area element.

19 The projection of the area element da onto the spherical surface is da cos φ. Flux through an irregular surface.

20 Spherical Gaussian surfaces around (a) positive and (b) negative point charge.

21 Gauss Law Gauss Law can be used to calculate the magnitude of the E field vector:

22 Use the following recipe for Gauss s Law problems: 1. Carefully draw a figure - location of all charges, direction of electric field vectors E 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field is constant on the surface and the surface contains the point at which you want to calculate the field. 3. Write Gauss Law and perform dot product E da 4. Since you drew the surface in such a way that the magnitude of the E is constant on the surface, you can factor the E out of the integral. 5. Determine the value of Q encl from your figure and insert it into Gauss Law equation. 6. You plug the surface area of your Gaussian surface in for the integral of da.

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24 Gaussian surface Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

25 The solution of this problem lies in the fact that the electric field inside a conductor is zero and if we place our Gaussian surface inside the conductor (where the field is zero), the charge enclosed must be zero (+ q q) = 0. Find electric charge q on surface of hole in the charged conductor.

26 A Gaussian surface drawn inside the conducting material of which the box is made must have zero electric field on it (field inside a con-ductor is zero). If the Gaussian surface has zero field on it, the charge enclosed must be zero per Gauss's Law. The E field inside a conducting box (a Faraday cage ) in an electric field.

27 Electric field = zero (electrostatic) inside a solid conducting sphere Under electrostatic conditions the electric field inside a solid conducting sphere is zero. Outside the sphere the electric field drops off as 1 / r 2, as though all the excess charge on the sphere were concentrated at its center.

28 Consider a conducting spherical shell, which has an inner radius of R 1 = 0.07 m, an outer radius of R 2 = 0.09 m and contains a charge of Q 2 = +18 C. This shell, in turn, encloses a point charge of Q 1 = -6 C located at its center, as shown in the diagram to the right. Using Gauss Law, develop an expression which will predict the electric field strength for values of r <R 1 and evaluate that expression for a value of r = 0.05 m.

29 Consider a conducting spherical shell, which has an inner radius of R 1 = 0.07 m, an outer radius of R 2 = 0.09 m and contains a charge of Q 2 = +18 C. This shell, in turn, encloses a point charge of Q 1 = -6 C located at its center, as shown in the diagram to the right. What is the electric field strength for R 1 < r < R 2? Find the charges on the inner surface and the outer surface of the conducting shell. Using Gauss Law develop an expression which will predict the electric field strength outside of the spherical shell and evaluate that expression for a value of r = 0.12 m. On the diagram above sketch the electric field lines everywhere.

30 Consider a conducting spherical shell, which has an inner radius of R 1 = 0.07 m, an outer radius of R 2 = 0.09 m and contains a charge of Q 2 = +18 C. This shell, in turn, encloses a point charge of Q 1 = -6 C located at its center, as shown in the diagram to the right. On the graph below sketch the electric field strength as a function of distance from the center of the spherical shell.

31 A coaxial cylindrical Gaussian surface is used to find the electric field outside an infinitely long charged wire.

32 An infinitely long charged wire has a uniform charge per unit length λ. Find E as a function of r (the distance from the wire) End view

33 An Infinite Conducting Plate Here is a plate (plane) geometry, where the charges are evenly distributed on a flat surface. The surface charge density is σ The E field is everywhere perpendicular to the plate (again, if not, the charges will move until the part parallel to the surface is nullified). Use a gaussian surface that is parallel to E on the sides (so no flux through side surfaces), and closes inside the conductor (no flux through that end). On the remaining side, the area vector A is parallel to the E field. Conducting Surface

34 Electric field between two oppositely charged parallel conducting plates.

35 Nonconducting Sheet A nonconducting sheet with a uniform surface charge density has the same geometry as for the conducting plate, so use the same gaussian surface. The only difference is that now one end cannot close in a conductor, so there is electric flux out both ends. As you may expect, the resulting electric field is half of what we got before. r r E da = 2EA = σa ε 0 E = σ 2ε 0 Nonconducting Sheet of Charge

36 Conceptual Question 2 How do the magnitudes of the E- fields compare at the points shown (consider the plate to be infinite in size)? (a) E b >E c >E d >E e >E a ; (b) E b =E c >E d =E e >E a ; (c) E b =E c =E d =E e >E a ; (d) E b =E c =E d =E e =E a ; (e) Cannot tell without more information.

37 Capacitor Edge Effects Since the electrodes of a real parallel plate capacitor are not infinite, there are edge effects at the ends of the electrodes.

38 Volume charge density : ρ = charge / unit volume is used to characterize the charge distribution of a uniformly charged insulator. The electric field of a uniformly charged INSULATING sphere.

39 Summary for Uniformly Charged Objects Gauss Law states that the electric flux is proportional to the net charge enclosed by the surface, and the constant of proportionality is ε 0. In symbols, it is Φ E = q r r q enc E da = ε 0 ε 0 There are three geometries we typically deal with for charge spread throughout object: enc Geometry Charge Density Gaussian surface Electric field Linear Sheet or Plane Spherical λ = q/l σ = q/a ρ = q/v Cylindrical, with axis along line of charge Cylindrical, with axis along E. Spherical (radius r), with center on center of sphere E E = E = = λ 2πε 0 r σ ε 0 Conducting q 4πε r r R 0 2 Line of Charge E E = σ 2ε 0 Nonconducting q 4πε 0 R = 3 r < R r