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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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!
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 • = ∫ 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 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
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? 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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