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Chapter 22. Gauss’s Law. Goals for Chapter 22. To use the electric field at a surface to determine the charge within the surface To learn the meaning of electric flux and how to calculate it

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## Chapter 22

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**Chapter 22**Gauss’s Law**Goals for Chapter 22**• To use the electric field at a surface to determine the charge within the surface • To learn the meaning of electric flux and how to calculate it • To learn the relationship between the electric flux through a surface and the charge within the surface**Goals for Chapter 22**• To use Gauss’s law to calculate electric fields • Recognizing symmetry • Setting up two-dimensional surface integrals • To learn where the charge on a conductor is located**Charge and electric flux**• Positive charge within the box produces outward electric flux through the surface of the box.**Charge and electric flux**• Positive charge within the box produces outward electric flux through the surface of the box. More charge = more flux!**Charge and electric flux**• Negative charge produces inward flux.**Charge and electric flux**• More negative charge – more inward flux!**Zero net charge inside a box**• Three cases of zero net charge inside a box • No net electric flux through surface of box!**Zero net charge inside a box**• Three cases of zero net charge inside a box • No net electric flux through surface of box!**Zero net charge inside a box**• Three cases of zero net charge inside a box • No net electric flux through surface of box!**Zero net charge inside a box**• Three cases of zero net charge inside a box • No net electric flux through surface of box!**What affects the flux through a box?**• Doubling charge within box doubles flux.**What affects the flux through a box?**• Doubling charge within box doubles flux. • Doubling size of box does NOT change flux.**Uniform E fields and Units of Electric Flux**• F= E·A = EA cos(q°) • [F] = N/C ·m2 = Nm2/C For a UNIFORM E field (in space)**Example 22.1 - Electric flux through a disk**Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux?**Example 22.1 - Electric flux through a disk**Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux? F = E·A = EA cos(30°) A = pr2 = 0.0314 m2 F =54 Nm2/C**Example 22.1 - Electric flux through a disk**Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux? What if n is perpendicular to E? Parallel to E?**Electric flux through a cube**• An imaginary cube of side L is in a region of uniform E. Find the flux through each side…**Electric flux through a sphere**F = ∫ E·dA**Electric flux through a sphere**• = ∫ E·dA E = kq/r2 = 1/(4pe0) q/r2 and is parallel to dAeverywhere on the surface • = ∫ E·dA = E ∫dA = EA**Electric flux through a sphere**• = ∫ E·dA E = kq/r2 and is parallel to dAeverywhere on the surface • = ∫ E·dA = E ∫dA = EA For q = +3.0nC, flux through sphere of radius r=.20 m?**Gauss’ Law**• = ∫ E·dA =qenc e0 S**Gauss’ Law**• = ∫ E·dA =qenc e0 S Electric Flux is produced by charge in space**Gauss’ Law**• = ∫ E·dA =qenc e0 S You integrate over a CLOSED surface (two dimensions!)**Gauss’ Law**• = ∫ E·dA =qenc e0 S E field is a VECTOR**Gauss’ Law**• = ∫ E·dA =qenc e0 S Infinitesimal area element dA is also a vector; this is what you sum**Gauss’ Law**• = ∫ E·dA =qenc e0 S Dot product tells you to find the part of E that is PARALLEL to dA at that point (perpendicular to the surface)**Gauss’ Law**• = ∫ E·dA =qenc e0 S Dot product is a scalar: E·dA = ExdAx +EydAy + EzdAz = |E||dA|cosq**Gauss’ Law**• = ∫ E·dA =qenc e0 S The TOTAL amount of charge…**Gauss’ Law**• = ∫ E·dA =qenc … but ONLY charge inside S counts! e0 S**Gauss’ Law**• = ∫ E·dA =qenc e0 S The electrical permittivity of free space, through which the field is acting.**Why is Gauss’ Law Useful?**• Given info about a distribution of electric charge, find the flux through a surface enclosing that charge. • Given info about the flux through a closed surface, find the total charge enclosed by that surface. • For highly symmetric distributions, find the E field itself rather than just the flux.**Gauss’ Law for Spherical Surface…**• Flux through sphere is independent of size of sphere • Flux depends only on charge inside. • F= ∫ E·dA = +q/e0**Point charge inside a nonspherical surface**As before, flux is independent of surface & depends only on charge inside.**Positive and negative flux**• Flux is positive if enclosed charge is positive, & negative if charge is negative.**Conceptual Example 22.4**• What is the flux through the surfaces A, B, C, and D?**Conceptual Example 22.4**• What is the flux through the surfaces A, B, C, and D? FA = +q/e0**Conceptual Example 22.4**• What is the flux through the surfaces A, B, C, and D? FB = -q/e0 FA = +q/e0**Conceptual Example 22.4**• What is the flux through the surfaces A, B, C, and D? FB = -q/e0 FA = +q/e0 FC = 0 !**Conceptual Example 22.4**• What is the flux through the surfaces A, B, C, and D? FB = -q/e0 FA = +q/e0 FC = 0 ! FD = 0 !!**Applications of Gauss’s law**• Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface. • Under electrostatic conditions, E field inside a conductor is 0!**Applications of Gauss’s law**• Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface. • Under electrostatic conditions, E field inside a conductor is 0! WHY ???**Applications of Gauss’s law**• Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface. • Under electrostatic conditions, E field inside a conductor is 0! • Assume the opposite! IF E field inside a conductoris not zero, then … • E field WILL act on free charges in conductor • Those charges WILL move in response to the force of the field • They STOP moving when net force = 0 • Which means IF static, THEN no field inside conductor!**Applications of Gauss’s law**• Under electrostatic conditions, field outside ANY spherical conductor looks just like a point charge!**Example 22.6: Field of a line charge**• E around an infinite positive wire of charge density l? ·E = ? ·E = ? Charge/meter = l ·E = ? ·E = ?**Example 22.6: Field of a line charge**• E around an infinite positive wire of charge density l? Charge/meter = l ·E = ? • You know charge, you WANT E field (Gauss’ Law!) • Choose Gaussian Surface with symmetry to match charge distribution to make calculating ∫ E·dA easy!**Example 22.6: Field of a line charge**• E around an infinite positive wire of charge density l? Imagine closed cylindricalGaussian Surface around the wire a distance r away…

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