# Week-10.pptx

28 de May de 2023
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### Week-10.pptx

• 1. Week # 10 Gauss’s Law: Electric charge & electric flux, calculating electric flux, Gauss’s law, applications of Gauss’s law Book: “University Physics” by Young & Freedman 13th edition, page: P-725” Physics-II (Ph-1002), 4 (3, 3) 1
• 2. 22.1 Charge and Electric Flux, p-725 The word “flux” comes from a Latin word meaning “flow.”) 22.3 Three cases in which there is zero net charge inside a box and no net electric flux through the surface of the box. (a) An empty box with E=0 (b) A box containing one positive and one equal magnitude negative point charge. (c) An empty box immersed in a uniform electric field. (a) No charge inside box, zero flux (b) Zero net charge inside box, inward flux cancels outward flux. (c) No charge inside box, inward flux cancels outward flux. 2
• 3. 22.4 (a) A box enclosing a positive point charge +q (b) Doubling the charge causes the magnitude of E to double, and it doubles the electric flux through the surface. (c) If the charge stays the same but the dimensions of the box are doubled, the flux stays the same. The magnitude of E on the surface decreases by a factor ¼ of but the area through which “flows” increases by a factor of 4. 3 (a) A box containing a charge (b) Doubling the enclosed charge doubles the flux. (c) Doubling the box dimensions does not change the flux.
• 4. 22.2: Calculating Electric Flux, p-728 Flux: Fluid-Flow Analogy 4 Volume flow rate Volume flow rate through A It is dot product from Fig 25.5 b
• 5. 5 Application: Flux Through a Basking (stretch out) Shark’s Mouth Unlike aggressive carnivorous sharks such as great whites, a basking shark feeds passively on plankton in the water that passes through the shark’s gills as it swims. To survive on these tiny organisms requires a huge flux of water through a basking shark’s immense mouth, which can be up to a meter across. The water flux , the product of the shark’s speed through the water and the area of its mouth—can be up to 0.5 m3/sec (500 liters per second, or almost 5×105 gallons per hour). In a similar way, the flux of electric field through a surface depends on the magnitude of the field and the area of the surface (as well as the relative orientation of the field and surface).
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• 7. 7 Fig. 22.26
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• 9. 9 Example 22.1, p-730
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• 11. 11 Example 22.2
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• 15. 15 Gauss’s law is an alternative to Coulomb’s law. While completely equivalent to Coulomb’s law, Gauss’s law provides a different way to express the relationship between electric charge and electric field. It was formulated by Carl Friedrich Gauss (1777–1855), one of the greatest mathematicians of all time.
• 16. 16 22.10: Carl Friedrich Gauss helped develop several branches of mathematics, including differential geometry, real analysis, and number theory. The “bell curve” of statistics is one of his inventions. Gauss also made state-of-the-art investigations of the earth’s magnetism and calculated the orbit of the first asteroid to be discovered. Point Charge Inside a Spherical Surface Gauss’s law states that the total electric flux through any closed surface (a surface enclosing a definite volume) is proportional to the total (net) electric charge inside the surface. In Section 22.1 we observed this relationship qualitatively for certain special cases; now we’ll develop it more rigorously. We’ll start with the field of a single positive point charge +q. The field lines radiate out equally in all directions. We place this charge at the center of an imaginary spherical surface with radius (R) The magnitude of the electric field (E) at every point on the surface is given by:
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• 25. 25 Here examples from 22.5 to 22.10 (pages 737 to 741) Uni Phy 13th ed., are the applications of Gauss’s Law. Solve such examples
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• 34. 34 Example 22.8 Field between oppositely charged parallel conducting plates Two large plane parallel conducting plates are given charges of equal magnitude and opposite sign; the surface charge densities are +σ and –σ Find the electric field in the region between the plates.
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• 37. 37 Example 22.9 Field of a uniformly charged sphere: Positive electric charge (Q) is distributed uniformly throughout the volume of an insulating sphere with radius (R) Find the magnitude of the electric field at a point a distance from the center of the sphere. 22.22 The magnitude of the electric field of a uniformly charged insulating sphere. Compare this with the field for a conducting sphere (see Fig. 22.18,as Fig of Example 22.5).
• 38. 38 SOLUTION IDENTIFY and SET UP: As in Example 22.5, the system is spherically symmetric. Hence we can use the conclusions of that example about the direction and magnitude of E To make use of the spherical symmetry, we choose as our Gaussian surface a sphere with radius r concentric with the charge distribution.
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• 40. 40 Please consult books for detail of each article
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