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Chapter 24

Chapter 24. Gauss ’ s Law. Gauss law, concept. EF lines come OUT of the surface  there are some positive charges inside. EF lines come INTO of the surface  there are some negative charges inside. Net flux outward  positive charges inside. Net flux inward  negative charges inside.

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Chapter 24

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  1. Chapter 24 Gauss’s Law

  2. Gauss law, concept EF lines come OUT of the surface  there are some positive charges inside EF lines come INTO of the surface  there are some negative charges inside

  3. Net flux outward  positive charges inside Net flux inward  negative charges inside zero Net flux  zero charge inside

  4. less EF lines  less charges inside More EF lines  more charges inside

  5. examples: is there any charge inside?

  6. Gauss Law: in words: “The amount of flux coming out of a closed surface is proportional to the charge enclosed.” The surface used in Gauss Law is closed, virtual, and with arbitrary shape and size. It is called “gaussian surface” Equation: flux ~ charge enclosed

  7. Flux: in words • The net flux through the surface is proportional to the net number of lines leaving the surface. • This net number of lines is the number of lines leaving the surface minus the number entering the surface.

  8. Flux: Mathematical Definition Units: N · m2 / C The surface integral means the integral must be evaluated over the surface in question. the value of the flux will depend both on the field pattern and on the surface. En is the component of the field perpendicular to the surface En

  9. Case of spherical symmetry • Example: • A positive point charge, q • We know that the EF is radial • We choose as a gaussian surface a sphere of radius r centered at the charge. • The magnitude of the electric field everywhere on the surface of the sphere is constant and normal: E=En = keq / r2 Section 24.2

  10. Gauss’ Law • Gauss’ Law can be used as an alternative procedure for calculating electric fields. • It is convenient for calculating the electric field of highly symmetric charge distributions ONLY, where: • there is a gaussian surface such that • the EF is perpendicular to the surface • the EF is constant all over the surface • However, Gauss law is valid for ALL systems, symmetric or not. Introduction

  11. Gaussian Surface, Example 2 • The charge is outside the closed surface with an arbitrary shape. • Any field line entering the surface leaves at another point. • Verifies the electric flux through a closed surface that surrounds no charge is zero. Section 24.2

  12. Applying Gauss’s Law • To use Gauss’s law, you want to choose a gaussian surface over which the surface integral can be simplified and the electric field determined. • Take advantage of symmetry. • Remember, the gaussian surface is a surface you choose, it does not have to coincide with a real surface. Section 24.3

  13. 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 of can be expressed as a simple algebraic product EdA because and are parallel. The dot product is 0 because and are perpendicular. The field is zero over the portion of the surface. If the charge distribution does not have sufficient symmetry such that a gaussian surface that satisfies these conditions can be found, Gauss’ law is not useful for determining the electric field for that charge distribution. Conditions for a Gaussian Surface Section 24.3

  14. Spherical Symmetry Field Due to a Spherically Symmetric Charge Distribution • Select a sphere as the gaussian surface. • For r >a  same as the EF for a point charge! Section 24.3

  15. Spherically Symmetric, cont. • Select a sphere as the gaussian surface, r < a. • qin < Q • qin = Section 24.3

  16. Spherically Symmetric Distribution, final • 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. Section 24.3

  17. Case of cylindrical symmetry EF due to a line of charge with uniform linear density λ (uniform) • Select a cylindrical gaussian surface. • The cylinder has a radius of r and a length of ℓ. • is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface. • E is parallel to the top and bottom (En=0) Section 24.3

  18. Field due to an infinite Plane of Charge with uniform density σ Plane Symmetry • is perpendicular to the plane and must have the same magnitude at all points equidistant from the plane. • Choose a small cylinder whose axis is perpendicular to the plane for the gaussian surface. • is parallel to the curved surface (En=0) • E is perpendicular to the top and bottom surface • The flux through each end of the cylinder is EA and so the total flux is 2EA. Section 24.3

  19. Field due to an infinite Plane of Charge with uniform density σ Plane Symmetry The EF does not depend on the distance to the plane! Section 24.3

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