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AP Physics C E & M Gauss's Law 20160109 www.njctl.org 2
Gauss's Law Click on the topic to go to that section • Electric Flux • Gauss's Law • Sphere • Infinite Rod of Charge • Infinite Plane of Charge • Electrostatic Equilibrium of Conductors 3
Electric Flux Return to Table of Contents 4
Electric Flux An Electric Field, , represented by the Electric Field lines below, passes through a rectangular area of space, with cross sectional area, A. The field is perpendicular to the area. Define Electric Flux as the strength of the Electric Field times the area through which it passes: Visually, we use the number of field lines to represent the Electric Field strength so, the more lines passing through an area of space, the greater Electric Flux. 5
Electric Flux But what if the Electric Field is not perpendicular to the area of space that we're interested in finding out the Electric Flux? Find the Electric Flux through the slanted area of the below shape. A is the area of the slanted side and A 1 is the area of the vertical side. The Electric Field makes an angle θ with the vector dA which is normal (perpendicular) to the light blue surface area. 6
Electric Flux The number of field lines through surfaces A 1 and A is the same. The widths of each surface are the same, and their lengths are two sides of a triangle where A has the length of the hypotenuse. Since the flux through both surfaces is the same: The flux is dependent on the number of field lines, and proportional to the angle that the normal to the surface makes with the field . 7
Electric Flux Not all surfaces are ramps. Let's see how calculus will help in calculating the flux through an arbitrary shape. The flux through any area element ΔA i is: To find the total flux through this surface, shrink ΔA i to almost zero and sum all the ΔΦ Ei 's over the entire surface: Note the different symbol for the integral that circle means you are integrating over an entire surface it is called a surface integral . 8
1 A rectangular loop is held perpendicular to a uniform Electric Field. At what angle to its normal does the loop have to be turned through so the electric flux is one half of its original value? o A 15 o B 30 Answer o C 45 o D 60 o E 90 9
2 A rectangular loop is held parallel to a uniform Electric Field and is turned through an angle of 35 0 clockwise. What is the Electric Flux through the loop after it is turned? A 1.1 x 10 4 Nm 2 /C Answer B 7.8 x 10 3 Nm 2 /C C 1.1 x 10 8 Nm 2 /C D 7.8 x 10 7 Nm 2 /C E Zero 10
Electric Flux through a Closed Surface A closed surface is a shape that has an inside and an outside. And you can't get from the inside to the outside without passing through the surface. Not the most profound of physics definitions, but it leads to a very powerful method for finding electric fields Gauss's Law. For Electric Field lines passing through a closed surface, we define a sign convention: • Field lines going from the outside to the inside of a closed surface are given a negative sign. • Field lines going from inside to the outside of a closed surface are given a positive sign. The net Electric Flux through a closed surface is then equal to the number of field lines leaving minus the number of lines entering. 11
Charge and Electric Flux Here are three identical boxes with different combinations of charges within their closed surfaces. Each charge has the same magnitude and generates an electric field represented by the red field lines. + + Box A Box B Box C What is the Electric Flux through each box? 12
Charge and Electric Flux + + Box A Box B Box C Box A: The electric field lines point out of the box, therefore it has an outward electric flux (positive). Box B: The electric field lines point into the surface of the box, therefore it has an inward electric flux (negative). Box C: Has a net charge of zero within the box, and since the flow is the same but in opposite directions for each charge, the electric flux is zero. 13
Charge and Electric Flux An empty box is placed in an external Electric field. The amount of charge enclosed by the box is zero, and the net electric flux is zero because the number of field lines entering is equal to the number of field lines leaving. 14
Charge and Electric Flux +q +2q If the electric field doubles, then the number of field lines representing the field through the surface will also double. From this we can conclude that the net electric flux is directly proportional to the enclosed charge. 15
Charge and Electric Flux A positive charge is surrounded by two concentric spheres what is the relationship of the electric flux through each sphere? 16
Charge and Electric Flux The Electric Flux through each sphere is the same. As the distance from the charge increases, the Electric Field decreases as 1/r 2 , while the surface area that encloses the field lines increases as a factor of r 2 . Since Electric Flux is a product of the field strength and the surface area, these factors cancel out. The Electric Flux is independent of the size of the enclosing geometry . 17
3 A positive charge is enclosed by a spherical surface. What direction is the electric flux relative to the surface? A Tangent. B Always perpendicular. Answer D C From the outside to the inside of the surface. D From the inside to the outside of the surface. E The lines do not pass through the surface. 18
4 A charged particle is enclosed within a rectangular box and produces an Electric Flux of Φ through the box. What is the Electric Flux if the dimensions of the box are tripled? A 1/9 Φ Answer B 1/3 Φ C Φ D 3 Φ E 9 Φ 19
Gauss's Law Return to Table of Contents 20
Gauss's Law Electric Flux will now be used to derive Gauss's Law. Start with a positive charge, q, and surround it with a hypothetical spherical surface called a "Gaussian surface." This surface has no physical meaning it's being used for its symmetry properties, and is typically shown as a dashed line or a shaded surface. Gaussian surface 21
Gauss's Law For any segment on the surface of the sphere, ΔA i , its normal vector, dA i and the Electric Field, E are parallel. Furthermore, E is constant everywhere on the surface. The flux through this segment is: The total flux through the Gaussian surface is then: 22
Gauss's Law The magnitude of E at a distance r from a point charge is: Substitute this value in the flux equation found on the previous slide: This is one of the reasons that ε 0 is used instead of κ we get rid of 4π in Gauss's Law. 23
Gauss's Law One more step to show that works for any closed surface surrounding a charge. All three surfaces have the same Electric Flux through them visually, the same amount of Electric Field lines go through each surface. Since they all enclose the same charge, the above equation works, regardless of the shape of the enclosing surface. 24
Gauss's Law If there is more than one charge enclosed by the surface, you just need to add the charges algebraically and Gauss's Law applies. This is a powerful equation, but several assumptions are required in order to use it to solve problems using simple integrals. Here's the list: • The Electric field is constant (which could also equal zero) at certain segments of the chosen Gaussian surface. • The Electric field is perpendicular to the surface so that . or, • The Electric field is parallel to the surface so that . 25
Charge Density When finding the Electric field due to a charge distribution, the following definitions (from last chapter) will be useful. Linear charge density: Surface charge density: Volume charge density: 26
5 A charge of magnitude q, is at the center of a sphere of radius r. What is the electric flux at the surface of the sphere? A 5.6x10 5 Nm 2 /C B 6.5x10 5 Nm 2 /C Answer C 7.3x10 5 Nm 2 /C D 1.1x10 6 Nm 2 /C E 1.3x10 6 Nm 2 /C 27
6 You are trying to calculate the Electric Field outside a positively charged metal sphere of radius R. The Gaussian surface should be what kind of geometrical object? + + A Circle + + R + + B Cylinder Answer + + + + C Cube + D Sphere E Rectangle 28
7 You are trying to calculate the Electric Field outside a positive line of charge. The Gaussian surface should be what kind of geometrical object? A Circle B Cylinder Answer C Cube D Sphere E Rectangle 29
Sphere Return to Table of Contents 30
Symmetric Charge Distributions There are three common symmetric charge distributions that are used to illustrate Gauss's Law. They are: • Sphere (both conducting and insulated) • Line of charge (conducting and insulated) • Infinite plane of charge We'll start with the Sphere. 31
Insulated Sphere An insulated sphere of radius, R, is uniformly positively charged with a volumetric charge density, ρ and total charge Q. Find the Electric Field inside and outside the sphere. Start with the field outside the sphere. Construct a spherical Gaussian surface of radius, r. The Q R Electric Field is perpendicular and constant everywhere on the surface, so Gauss's Law is appropriate. r 32
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