Phys103-Lec.6 2.pptx physics103 selfstudy
- 1. PHYS 103 Unit 6 Electric Flux and Gauss’s Law Physics 103 course
- 2. CONTENT’S REFERENCE You can find the lecture topics in: Course Reference: Physics for scientists and Engineers, Raymond Serway & et al 10th edition, "the entire presentation has been extracted from this reference". This lecture is a part of chapter 24 from the following book “Physics for Scientists and Engineers (with Physics NOW and InfoTrac), Raymond A. Serway - Emeritus, James Madison University , Thomson Brooks/Cole © 2004, 6th Edition, 1296 pages” Note that,You can refer to the Word file titled “lecture 6-Electric Flux and Gauss’s Law” as an additional reference for this lecture. Lecture contents ref: https://personal.tcu.edu/ystrzhemechn/Classes/2017/Spring/Lectures/Chapter%2021.ppt **PLEASE NOTE, THE CONTENT IN THIS PRESENTATION IS JUST TO HELP YOU, AND IT IS NOT A SUBSTITUTE TO THE COURSE REFERENCES
- 3. TOPICS TO BE COVERED 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions
- 5. Notice: Electric flux is proportional to the number of electric field lines penetrating some surface.
- 6. ELECTRIC FLUX The electric flux is proportional to the number of electric field lines penetrating some surface If the surface under consideration is not perpendicular to the field, as show in Figure , the electric flux E given by The field lines may make some angle θ with the perpendicular to the surface The flux is a maximum (zero) when the surface is perpendicular (parallel) to the field cos EA E A E NOTE THAT, the flux through a surface of fixed area A has a maximum value EA when the surface is perpendicular to the field (when the normal to the surface is parallel to the field, that is, = 0°); The flux is zero when the surface is parallel to the field (when the normal to the surface is perpendicular to the field, that is, = 90°).
- 8. Therefore, our definition of flux given by () has meaning only over a small element of area. Consider a general surface divided up into a large number of small elements, each of area (). The variation in the electric field over one element can be neglected if the element is sufficiently small. ELECTRIC FLUX It is convenient to define a vector () whose magnitude represents the area of the ith element of the surface and whose direction is defined to be perpendicular to the surface element, as shown in Figure.
- 9. The electric field () at the location of this element makes an angle with the vector (). The electric flux () through this element is cos E i i i i i E A θ E A The general definition of electric flux: ∅𝑬= 𝐥𝐢𝐦 ∆ 𝑨𝒊 →𝟎 ∑⃗ 𝑬𝒊 .⃗ ∆ 𝑨𝒊=¿ ∫ 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 ❑ ⃗ 𝑬.⃗ 𝒅𝑨¿ This equation is a surface integral, which means it must be evaluated over the surface in question. In general, the value of () depends both on the field pattern and on the surface
- 10. ELECTRIC FLUX THROUGH CLOSED SURFACE A closed surface in an electric field. The area vectors () are, by convention, normal to the surface and point outward. The flux through an area element can be: 1) Positive for Element (1). 2) Zero for Element (2). 3) Negative for Element (3).
- 11. Note: The net flux through the surface is proportional to the net number of lines leaving the surface, where the net number means the number leaving the surface minus the number entering the surface. If more lines are leaving than entering, the net flux is positive. If more lines are entering than leaving, the net flux is negative.
- 12. Now we describe a general relationship between the net electric flux through a closed surface (often called a gaussian surface) and the charge enclosed by the surface. This relationship, known as Gauss’s law, is of fundamental importance in the study of electric fields. GAUSS’S LAW consider a positive point charge q located at the center of a sphere of radius r, as shown in Figure.
- 13. we know that the magnitude of the electric field everywhere on the surface of the sphere 𝐸 =𝑘𝑒 𝑞 𝑟 2 The field lines are directed radially outward and hence are perpendicular to the surface at every point on the surface. That is, at each surface point, is parallel to the vector representing a local element of area surrounding the surface point.Therefore, ⃗ 𝐸𝑖 . ⃗ 𝐴𝑖= 𝐸 ∆ 𝐴𝑖
- 14. we find that the net flux through the gaussian surface is
- 15. Studding a point charge located outside a closed surface. The number of lines entering the surface equals the number leaving the surface We have moved E outside of the integral because, by symmetry, E is constant over the surface and given by: 𝐸 =𝑘𝑒 𝑞 𝑟 2 Because the surface is spherical The net flux through the gaussian surface
- 16. we can write: we note that : Now consider several closed surfaces surrounding a charge q, as shown in Figure. Surface S1 is spherical, but surfaces S2 and S3 are not. Therefore, we conclude that the net flux through any closed surface surrounding a point charge q is given by and is independent of the shape of that surface.
- 17. CONCLUSION: GAUSS’ LAW Gauss’ Law: electric flux through any closed surface is proportional to the net charge Q inside the surface εo = 8.85 x 10-12 C2 /Nm2 : permittivity of free space The area in is an Φ imaginary Gaussian surface (does not have to coincide with the surface of a physical object) e 0 k π 4 1 /ε
- 18. Now consider a point charge located outside a closed surface of arbitrary shape, as shown in Figure. As you can see from this construction, any electric field line that enters the surface leaves the surface at another point. The number of electric field lines entering the surface equals the number leaving the surface. Therefore, we conclude that the net electric flux through a closed surface that surrounds no charge is zero.
- 19. Let us extend these arguments to two generalized cases: (1) that of many point charges. (2) that of a continuous distribution of charge. We once again use the superposition principle, which states that the electric field due to many charges is the vector sum of the electric fields produced by the individual charges. Therefore, we can express the flux through any closed surface a: where is the total electric field at any point on the surface produced by the vector addition of the electric fields at that point due to the individual charges.
- 20. ELECTRIC FLUX Consider the system of charges shown in Figure. The surface S surrounds only one charge, q1; hence, the net flux through S is: The flux through S due to charges q 2 , q 3 , and q 4 outside it is zero because each electric field line that enters S at one point leaves it at another. The surface surrounds charges q 2 and q 3; hence, the net flux through it is:
- 21. Finally, the net flux through surface is zero because there is no charge inside this surface. That is, all the electric field lines that enter at one point leave at another. Notice that charge q 4 does not contribute to the net flux through any of the surfaces because it is outside all of the surfaces. Gauss’s law, which is a generalization of what we have just described, states that the net flux through any closed surface is where in represents the net charge inside the surface and E represents the electric field at any point on the surface.
- 22. EXAMPLE: Consider a closed triangular box resting within a horizontal electric field of magnitude E 5 7.80 3 10^4 N/C as shown in the figure. Calculate the electric flux through the vertical rectangular surface.
- 23. A positive point charge q is located at the center of a sphere of radius r The positive point charge is directed radially outward by symmetry and is therefore normal to the surface at every point.Thus, E is parallel to dA at each point.Therefore, E = keq / r2 , Asphere = 4πr2 A d E E dA E 2 2 4 r r q ke q ke 4 0 q
- 24. Problem:
- 25. HOMEWORK The electric field on the surface of a 10-cm-diameter sphere is perpendicular to the sphere and has magnitude 47 kN/C.What’s the electric flux through the sphere?