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Ch 24 – Gauss’s Law Karl Friedrich Gauss (1777-1855) – German mathematician Ch 24 – Gauss’s Law Already can calculate the E-field of an arbitrary charge distribution using Coulomb’s Law. Gauss’s Law allows the same thing, but much more easily… … so long as the charge distribution is highly symmetrical. Karl Friedrich Gauss (1777-1855) – German mathematician Ch 24 – Gauss’s Law For example, in Ch 23: We found the E-fields in the vicinity of continuous charge distributions by integration… booooo: R r x dE E-field of charged disc (R>>x): E 2 o Now, we’ll learn an easier way. Ch 24.1 – Electric Flux • Sounds fancy, but it’s not hard • Electric Flux measures how much an electric field wants to “push through” or “flow through” some arbitrary surface area • We care about flux because it makes certain calculations easier. Ch 24.1 – Electric Flux – Case 1 Easiest case: • The E-field is uniform • The plane is perpendicular to the field E E A Electric Flux Ch 24.1 – Electric Flux – Case 1 Easiest case: • The E-field is uniform • The plane is perpendicular to the field E E A Electric Flux Flux depends on how strong the E-field is and how big the area is. Ch 24.1 – Electric Flux – Case 2 Junior Varsity case: • The field is uniform • The plane is not perpendicular to the field E E A E A cos Ch 24.1 – Electric Flux – Case 2 Junior Varsity case: • The field is uniform • The plane is not perpendicular to the field E E A E A cos Flux depends on how strong the E-field is, how big the area is, and the orientation of the area with respect to the field’s direction. Ch 24.1 – Electric Flux – Case 2 E E A E A cos n̂ And, we can write this better using the definition of the “dot” product. E E A where: A Anˆ A A nˆ Ch 24.1 – Electric Flux – Case 2 E E A Quick Quiz: What would happen to the E-flux if we change the orientation of the plane? Ch 24.1 – Electric Flux – Case 3 E E Varsity (most general) case: A • The E-field is not uniform • The surface is curvy and is not perpendicular to the field A Ch 24.1 – Electric Flux – Case 3 E E Imagine the surface A is a mosaic of little tiny surfaces ΔA. A Pretend that each little ΔA is so small that it is essentially flat. A Ch 24.1 – Electric Flux – Case 3 E E Then, the flux through each little ΔA is just: E E A A A A is a special vector. It points in the normal direction and has a magnitude that tells us the area of ΔA . Ch 24.1 – Electric Flux – Case 3 E So… to get the flux through the entire surface A, we just have to add up the contributions from each of the little ΔA’s that compose A. E A A E E An n Ch 24.1 – Electric Flux – Case 3 E E E An E n A E E dA surface A Electric Flux through an arbitrary surface caused by a spatially varying E-field. Ch 24.1 – Electric Flux – Flux through a Closed Surface • The vectors dAi point in different directions – At each point, they are perpendicular to the surface – By convention, they point outward Electric Flux: General Definition E E dA surface Electric Flux: General Definition E E dA surface E-Flux through a surface depends on three things: 1. How strong the E-field is at each infinitesimal area. 2. How big the overall area A is after integration. 3. The orientation between the E-field and each infinitesimal area. Electric Flux: General Definition E E dA surface Flux can be negative, positive or zero! -The sign of the flux depends on the convention you assign. It’s up to you, but once you choose, stick with it. Quick Quiz: which little area experiences the most flux? Quick Quiz: what are the three things on which E-flux depends? Ch 24.1 – Electric Flux – Calculating E-Flux • The surface integral means the integral must be evaluated over the surface in question… more in a moment. • The value of the flux will depend both on the field pattern and on the surface • The units of electric flux are N.m2/C Ch 24.1 – Electric Flux • The net electric flux through a surface is directly proportional to the number of electric field lines passing through the surface. EG 24.1 – Flux through a Cube Assume a uniform E-field pointing only in +x direction Find the net electric flux through the surface of a cube of edgelength l, as shown in the diagram. Ch 24.2 – Gauss’s Law Gauss’s Law is just a flux calculation We’re going to build imaginary surfaces – called Gaussian surfaces – and calculate the E-flux. Gauss’s Law only applies to closed surfaces. Gauss’s Law directly relates electric flux to the charge distribution that creates it. Ch 24.2 – Gauss’s Law Gauss’s Law NET qenclosed E dA surface 0 Ch 24.2 – Gauss’s Law Gauss’s Law NET qenclosed E dA surface 0 The net E-flux through a closed surface Charge inside the surface Ch 24.2 – Gauss’s Law In other words… 1. Draw a closed surface around a some charge. 2. Set up Gauss’s Law for the surface you’ve drawn. 3. Use Gauss’s Law to find the E-field. NET qenclosed E dA surface 0 You get to choose the surface – it’s a purely imaginary thing. Quick Quiz Which surface – S1, S2 or S3 – experiences the most electric flux? Ch 24.2 – Gauss’s Law – confirming Gauss’s Law Let’s calculate the net flux through a Gaussian surface. Assume a single positive point charge of magnitude q sits at the center of our imaginary Gaussian surface, which we choose to be a sphere of radius r. NET E dA surface Ch 24.2 – Gauss’s Law – confirming Gauss’s Law At every point on the sphere’s surface, the electric field from the charge points normal to the sphere… why? This helps make our calculation easy. NET E dA surface Ch 24.2 – Gauss’s Law – confirming Gauss’s Law At every point on the sphere’s surface, the electric field from the charge points normal to the sphere… why? This helps make our calculation easy. NET E dA surface EdA cos surface EdA surface Ch 24.2 – Gauss’s Law – confirming Gauss’s Law Now we have: NET EdA surface But, because of our choice for the Gaussian surface, symmetry works in our favor. The electric field due to the point charge is constant all over the sphere’s surface. So… NET E dA surface Ch 24.2 – Gauss’s Law – confirming Gauss’s Law This, we can work with. NET E dA surface We know how to find the magnitude of the electric field at the sphere’s surface. Just use Coulomb’s law to calculate the E-field due to a point charge a distance r away from the charge. E point ch arg e ke q 2 r Ch 24.2 – Gauss’s Law – confirming Gauss’s Law Thus: NET ke q 2 dA r surface Ch 24.2 – Gauss’s Law – confirming Gauss’s Law Thus: NET ke q 2 dA r surface And, this surface integral is easy. 2 dA 4 r sphere Ch 24.2 – Gauss’s Law – confirming Gauss’s Law Therefore: NET ke q 2 (4 r 2 ) r But, we can rewrite Coulomb’s constant. ke 1 40 Ch 24.2 – Gauss’s Law – confirming Gauss’s Law Therefore: NET ke q 2 (4 r 2 ) r But, we can rewrite Coulomb’s constant. ke 1 40 Thus, we have confirmed Gauss’s law: NET q 0 A few more questions +Q + – – 3Q • If the electric field is zero for all points on the surface, is the electric flux through the surface zero? • If the electric flux is zero for a closed surface, can there be charges inside the surface? • What is the flux through the surface shown? Why? EG 24.2 – Flux due to a Point Charge A spherical surface surrounds a point charge. What happens to the total flux through the surface if: (A) (B) (C) (D) the charge is tripled, the radius of the sphere is doubled, the surface is changed to a cube, and the charge moves to another location inside the surface? Ch 24.3 – Applying Gauss’s Law Gauss’s Law can be used to (1) find the E-field at some position relative to a known charge distribution, or (2) to find the charge distribution caused by a known E-field. In either case, you must choose a Gaussian surface to use. Ch 24.3 – Applying Gauss’s Law Choose a surface such that… 1. Symmetry helps: the E-field is constant over the surface (or some part of the surface) 2. The E-field is zero over the surface (or some portion of the surface) 3. The dot product reduces to EdA (the E-field and the dA vectors are parallel) 4. The dot product reduces to zero (the E-field and the dA vectors are perpendicular) EG 24.3 – Spherical Charge Distribution An insulating solid sphere of radius a has a uniform volume charge density ρ and carries total charge Q. (A) Find the magnitude of the E-field at a point outside the sphere (B) Find the magnitude of the E-field at a point inside the sphere EG 24.3 – Spherical Charge Distribution EG 24.4 – Spherical Charge Distribution Find the E-field a distance r from a line of positive charge of infinite length and constant charge per unit length λ. EG 24.5 – Spherical Charge Distribution Find the E-field due to an infinite plane of positive charge with uniform surface charge density σ Ch 24.4 – Conductors in Electrostatic Equilibrium • In an insulator, excess charge stays put. • Conductors have free electrons and, correspondingly, have different electrostatic characteristics. • You will learn four critical characteristics of a conductor in electrostatic equilibrium. • Electrostatic Equilibrium – no net motion of charge. Ch 24.4 – Conductors in Electrostatic Equilibrium • Most conductors, on their own, are in electrostatic equilibrium. • That is, in a piece of metal sitting by itself, there is no “current.” Ch 24.4 – Conductors in Electrostatic Equilibrium Four key characteristics 1. The E-field is zero at all points inside a conductor, whether hollow or solid. 2. If an isolated conductor carries excess charge, the excess charge resides on its surface. 3. The E-field just outside a charged conductor is perpendicular to the surface and has magnitude σ/ε0, where σ is the surface charge density at that point. 4. Surface charge density is biggest where the conductor is most pointy. Ch 24.4 – Conductors (cont.) – Justifications Einside = 0 • Place a conducting slab in an external field, E. • If the field inside the conductor were not zero, free electrons in the conductor would experience an electrical force. • These electrons would accelerate. • These electrons would not be in equilibrium. • Therefore, there cannot be a field inside the conductor. Ch 24.4 – Conductors (cont.) – Justifications Einside = 0 • Before the external field is applied, free electrons are distributed evenly throughout the conductor. • When the external field is applied, charges redistribute until the magnitude of the internal field equals the magnitude of the external field. • There is a net field of zero inside the conductor. • Redistribution takes about 10-15s. Ch 24.4 – Conductors (cont.) – Justifications Charge Resides on the Surface • Choose a Gaussian surface inside but close to the actual surface • The electric field inside is zero (prop. 1) • There is no net flux through the gaussian surface • Because the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface Ch 24.4 – Conductors (cont.) – Justifications Charge Resides on the Surface • Since no net charge can be inside the surface, any net charge must reside on the surface • Gauss’s law does not indicate the distribution of these charges, only that it must be on the surface of the conductor Ch 24.4 – Conductors (cont.) – Justifications E-Field’s Magnitude and Direction • Choose a cylinder as the Gaussian surface • The field must be perpendicular to the surface – If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium E-Field’s Magnitude and(cont.) Direction Ch 24.4 – Conductors – Justifications E-Field’s Magnitude and Direction • The net flux through the surface is through only the flat face outside the conductor – The field here is perpendicular to the surface • Applying Gauss’s law A E EA 0 E 0 E-Field’s Magnitude and(cont.) Direction Ch 24.4 – Conductors – Justifications E-Field’s Magnitude and Direction • The field lines are perpendicular to both conductors • There are no field lines inside the cylinder EG 24.7 – Sphere inside a Spherical Shell A solid insulating sphere of radius a carries a uniformly distributed charge, Q. A conducting shell of inner radius b and outer radius c is concentric and carries a net charge of -2Q. Find the E-field in regions 1-4 using Gauss’s Law. Example P24.29 Consider a thin spherical shell of radius 14.0 cm with a total charge of 32.0 μC distributed uniformly on its surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm from the center of the charge distribution. k Q 8.99 10 32.0 10 E 7.19 M N 6 9 e 2 r 0.200 2 C Example P24.31 A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 μC. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder. 9 2ke 2 8.99 10 N m E r 2 C 2 2.00 106 C 7.00 m 0.100 m E EA cos E 2 r cos0 E 51.4 kN C ,radially outw ard E 5.14 104 N C 2 0.100 m 0.020 0 m 1.00 646 N m 2 C Example P24.35 A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face. 8.00 104 8.85 1012 7.08 107 C m 2 Q A 7.08 107 0.500 C 2 Q 1.77 107 C 177 nC Example P24.39 A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of λ, and the cylinder has a net charge per unit length of 2λ. From this information, use Gauss’s law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis. 0 qin qin E 2ke 3 6ke 3 radially outw ard r r 2 0 r 3

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