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Chapter 23 Gauss’ Law

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23.1 What is Physics?: Gauss’ law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface. Gauss’ law considers a hypothetical (imaginary) closed surface enclosing the charge distribution. This Gaussian surface, as it is called, can have any shape, but the shape that minimizes our calculations of the electric field.

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23.2: Flux Fig (a) A uniform airstream of velocity is perpendicular to the plane of a square loop of area A.(b) The component of perpendicular to the plane of the loop is v cos q,where q is the angle between v and a normal to the plane. (c) The area vector A is perpendicular to the plane of the loop and makes an angle q with v. (d) The velocity field intercepted by the area of the loop. The rate of volume flow through the loop is F= (v cos q)A. This rate of volume flow through an area is an example of a flux—a volume flux in this situation. Volume flux

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**23.3: Flux of an Electric Field**

The electric flux through a Gaussian surface is proportional to the net number of electric field lines passing through that surface. Fig A Gaussian surface of arbitrary shape immersed in an electric field. The surface is divided into small squares of area DA. The electric field vectors E and the area vectors DA for three representative squares, marked 1, 2, and 3, are shown. The exact definition of the flux of the electric field through a closed surface is found by allowing the area of the squares shown in Fig to become smaller and smaller, approaching a differential limit dA. The area vectors then approach a differential limit dA.The sum of Eq then becomes an integral: 180° > θ >90°

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**Example, Flux through a closed cylinder, uniform field:**

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**Example, Flux through a closed cube, Non-uniform field:**

Right face: An area vector A is always perpendicular to its surface and always points away from the interior of a Gaussian surface. Thus, the vector for any area element dA (small section) on the right face of the cube must point in the positive direction of the x axis. The most convenient way to express the vector is in unit-vector notation, Although x is certainly a variable as we move left to right across the figure, because the right face is perpendicular to the x axis, every point on the face has the same x coordinate. (The y and z coordinates do not matter in our integral.) Thus, we have Top face Right face Left face A=4 m2

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**Example, Flux through a closed cube, Non-uniform field:**

Top face Left face A=4 m2

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**The net charge qenc is the algebraic sum of all the**

23.4 Gauss’ Law: Gauss’ law relates the net flux of an electric field through a closed surface (a Gaussian surface) to the net charge qenc that is enclosed by that surface. The net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero. If qenc is positive, the net flux is outward; if qenc is negative, the net flux is inward.

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**Example, Relating the net enclosed charge and the net flux:**

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**Example, Enclosed charge in a**

non-uniform field:

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**23.5 Gauss’ Law and Coulomb’s Law:**

Figure 23-8 shows a positive point charge q, around which a concentric spherical Gaussian surface of radius r is drawn. Divide this surface into differential areas dA. The area vector dA at any point is perpendicular to the surface and directed outward from the interior. From the symmetry of the situation, at any point the electric field, E, is also perpendicular to the surface and directed outward from the interior. Thus, since the angle q between E and dA is zero, we can rewrite Gauss’ law as This is exactly what Coulomb’s law yielded.

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Applying Gauss’ Law Gauss’ Law is valid for the electric field of any system of charges or continuous distribution of charge. Although Gauss’ Law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations Particularly spherical, cylindrical, or plane symmetry Remember, the Gaussian surface is a surface you choose, it does not have to coincide with a real surface

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**Conditions for a Gaussian Surface**

Try to choose a surface that satisfies one or more of these conditions: The value of the electric field can be argued from symmetry to be constant over the surface The dot product can be expressed as a simple algebraic produce E dA because and are parallel The dot product is 0 because and are perpendicular The field can be argued to be zero everywhere over the surface

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**Field Due to a Spherically Symmetric Charge Distribution**

Insulating solid sphere of radius a has a uniform charge density ρ Select a sphere as the Gaussian surface For r> a

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**Spherically Symmetric Distribution**

Select a sphere as the Gaussian surface, r < a qin < Q qin = r (4/3pr3)

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**Spherically Symmetric Distribution**

Inside the sphere, E varies linearly with r E ® 0 as r ® 0 The field outside the sphere is equivalent to that of a point charge located at the center of the sphere

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**Field at a Distance from a Line of Charge**

Applying Gauss’ Law Field at a Distance from a Line of Charge Select a cylindrical charge distribution The cylinder has a radius of r and a length of l E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface

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**Field Due to a Line of Charge, cont**

The end view confirms the field is perpendicular to the curved surface The field through the ends of the cylinder is 0 since the field is parallel to these surfaces

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**23.7 Applying Gauss’ Law: Cylindrical Symmetry:**

Figure shows a section of an infinitely long cylindrical plastic rod with a uniform positive linear charge density l. Let us find an expression for the magnitude of the electric field E at a distance r from the axis of the rod. At every point on the cylindrical part of the Gaussian surface, must have the same magnitude E and (for a positively charged rod) must be directed radially outward. The flux of E through this cylindrical surface is

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**Field Due to a Plane of Charge**

must be perpendicular to the plane and must have the same magnitude at all points equidistance from the plane Choose a small cylinder whose axis is perpendicular to the plane for the Gaussian surface

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**Properties of a Conductor in Electrostatic Equilibrium**

The electric field is zero everywhere inside the conductor If an isolated conductor carries a charge, the charge resides entirely on its surface The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of s/eo On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest

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Property 1: Einside = 0 Consider a conducting slab in an external field E If the field inside the conductor was 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

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Property 1: Einside = 0 Before the external field was applied, free electrons are distributed throughout the conductor When the external field is applied, the electron 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

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**Property 2: 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

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**Property 2: Charge Resides on the Surface**

Since no net charge can be inside the surface, any net charge must reside on the surface Gauss’ Law does not indicate the distribution of these charges, only that it must be on the surface of the conductor There is no net charge on the cavity walls

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**Property 3: Field’s Magnitude and Direction**

The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/ε0 Choose a cylinder as the Gaussian surface The field must be perpendicular to the surface If there was a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium

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**Example, Spherical Metal Shell,**

Electric Field, and Enclosed Charge:

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**ρ Electric field inside the cavity ?**

Position vector from the center of the sphere to the center to the cavity Electric field inside the cavity ?

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ρ

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**23.9 Applying Gauss’ Law: Spherical Symmetry:**

A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell. If a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell. Fig The dots represent a spherically symmetric distribution of charge of radius R, whose volume charge density r is a function only of distance from the center. The charged object is not a conductor, and therefore the charge is assumed to be fixed in position. A concentric spherical Gaussian surface with r <R is shown.

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Homework Chapter 23 ( page 624 ) 34, 49, 51, 52, 61, 73, 74, 76

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**23.8 Applying Gauss’ Law: Planar Symmetry Non-conducting Sheet:**

Figure shows a portion of a thin, infinite, nonconducting sheet with a uniform (positive) surface charge density s. A sheet of thin plastic wrap, uniformly charged on one side, can serve as a simple model. We need to find the electric field a distance r in front of the sheet. A useful Gaussian surface is a closed cylinder with end caps of area A, arranged to pierce the sheet perpendicularly as shown. From symmetry, E must be perpendicular to the sheet and hence to the end caps. Since the charge is positive, E is directed away from the sheet. There is no flux through this portion of the Gaussian surface. Thus E.dA is simply EdA, and Here sA is the charge enclosed by the Gaussian surface. Therefore,

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**23.8 Applying Gauss’ Law: Planar Symmetry Two Conducting Plates:**

Figure 23-16a shows a cross section of a thin, infinite conducting plate with excess positive charge. The plate is thin and very large, and essentially all the excess charge is on the two large faces of the plate. If there is no external electric field to force the positive charge into some particular distribution, it will spread out on the two faces with a uniform surface charge density of magnitude s1. Just outside the plate this charge sets up an electric field of magnitude E =s1/e0. Figure 23-16b shows an identical plate with excess negative charge having the same magnitude of surface charge density. Now the electric field is directed toward the plate. If we arrange for the plates of Figs a and b to be close to each other and parallel (Fig c), the excess charge on one plate attracts the excess charge on the other plate, and all the excess charge moves onto the inner faces of the plates as in Fig c. With twice as much charge now on each inner face, the new surface charge density, s, on each inner face is twice s1.Thus, the electric field at any point between the plates has the magnitude

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**Example, Electric Field:**

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**23.9 Applying Gauss’ Law: Spherical Symmetry:**

A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell. If a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell. Fig The dots represent a spherically symmetric distribution of charge of radius R, whose volume charge density r is a function only of distance from the center. The charged object is not a conductor, and therefore the charge is assumed to be fixed in position. A concentric spherical Gaussian surface with r <R is shown.

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